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A theory for generalized morphisms and beyond

Gang Hu
Abstract.

Some sorts of generalized morphisms are defined from very basic mathematical objects such as sets, functions, and partial functions. A wide range of mathematical notions such as continuous functions between topological spaces, ring homomorphisms, module homomorphisms, group homomorphisms, and covariant functors between categories can be characterized in terms of the generalized morphisms. We show that the inverse of any bijective generalized morphism is also a generalized morphism (of the same kind), and hence a generalized isomorphism can be defined as a bijective generalized morphism.

Galois correspondences are established and studied, not only for the Galois groups of the generalized automorphisms, but also for the “Galois monoids” of the generalized endomorphisms.

Ways to construct the generalized morphisms and the generalized isomorphisms are studied.

New interpretations on solvability of polynomials and solvability of homogeneous linear differential equations are introduced, and these ideas are roughly generalized for “general” equation solving in terms of our theory for the generalized morphisms.

Some more results are presented. For example, we generalize the algebraic notions of transcendental elements over a field and purely transcendental field extensions, we obtain an isomorphism theorem that generalizes the first isomorphism theorems (for groups, rings, and modules), and we show that a part of our theory is closely related to dynamical systems.

Key words and phrases:
Generalized morphisms; Galois theory; Galois correspondence; Constructions of morphisms; Solvability of equations; Isomorphism theorem; Transcendental elements
2020 Mathematics Subject Classification:
Primary 08A99, 08A35; Secondary 12H05, 18A99, 12F10, 13B05, 12F20, 20B25, 20B27

1. Introduction

The notions of generalized morphisms introduced in this paper develop from very basic mathematical objects such as sets, functions, and partial functions. As a consequence, many familiar mathematical notions are covered by the generalized morphisms. Specifically, continuous functions between topological spaces, ring homomorphisms, module homomorphisms, group homomorphisms, and covariant functors between categories can be characterized in terms of the generalized morphisms. Indeed, in these examples, the generalized morphisms preserve the structures of the mathematical objects involved. Basically, this is because we define the generalized morphisms to be commutative with the intrinsic operations on the mathematical objects involved. Moreover, we show that the inverse of any bijective generalized morphism is also a generalized morphism (of the same kind), and hence a generalized isomorphism can be defined as a bijective generalized morphism.

The notion of generalized morphisms in our theory is not the same as the notion of morphisms in category theory. In category theory, a morphism between two objects does not necessarily preserve the structure of objects, while a generalized morphism in our theory does. Nevertheless, since covariant functors preserve the structures of categories, all covariant functors can be characterized in terms of the generalized morphisms.

One may have encountered concepts similar to the notions of our generalized morphisms, e.g. the notion of homomorphisms between structures in the field of mathematical logic. However, we have not found any research with any content similar to ours.

Besides introducing the new notions, we establish Galois correspondences for the Galois groups of the generalized automorphisms as well as for the “Galois monoids” of the generalized endomorphisms. As is well known, in the Galois theory for infinite algebraic field extensions, Krull topology is put on Galois groups, and in differential Galois theory, differential Galois groups are endowed with Zariski topology. Then the fundamental theorem in each of the two theories tells us that there exists a Galois correspondence between the set of all closed subgroups of the Galois group of a Galois field extension (or a Picard-Vessiot extension) and the set of all intermediate (differential) fields. Now for our theory, we first study the lattice structures related to Galois correspondences. Then, we can determine under what conditions we can characterize Galois correspondences with topology (as is done in the above two Galois theories).

Moreover, we study ways to construct the generalized morphisms and the generalized isomorphisms, which we think to be important because the morphisms and isomorphisms preserve the structures of the mathematical objects involved.

In Part I of this paper, which consists of Sections 2 to 7, we address the above issues for the case where only single-variable functions are involved, and in Part II, we address the above issues for the case where multivariable functions or partial functions are involved.

In Part III, new interpretations on solvability of polynomials and solvability of homogeneous linear differential equations are introduced. The key idea is to “decompose” an equation QQ into n(+)n(\in{{\mathbb{Z}}^{+}}) equations Q1,,Qn{{Q}_{1}},\cdots,{{Q}_{n}} such that each solution of QQ can somehow be expressed in terms of the solutions of Q1,,Qn{{Q}_{1}},\cdots,{{Q}_{n}}, where for example in the case of a separable polynomial p(x)p(x) in Q:=(p(x)=0)Q:=(p(x)=0), the Galois group of each pi(x)p_{i}(x) in Qi:=(pi(x)=0)Q_{i}:=(p_{i}(x)=0) is simple. Moreover, this idea is roughly generalized for “general” equation solving in terms of the theory developed in Parts I and II.

Some more results are presented in Part IV. For example, in Section 19, we obtain analogues of the well-known fact that the Galois group of an irreducible polynomial acts transitively on its roots, and in Section 21, we generalize the algebraic notions of transcendental elements over a field and purely transcendental field extensions. Moreover, in Section 22, we obtain an isomorphism theorem which generalizes the first isomorphism theorems (for groups, rings, and modules). And in Section LABEL:App_to_dyn, we show a close relation between a part of our theory and dynamical systems.

To help the reader better understand the structure of the paper, we simplify the table of contents as follows.

Part I. Theory for the case of unary functions

Section 2. Basic notions and properties

Section 3. Galois correspondences

Section 4. Lattice structures of objects arising in Galois correspondences on a TT-space SS

Section 5. Topologies employed to construct Galois correspondences

Section 6. Generalized morphisms and isomorphisms from a T1T_{1}-space to a T2T_{2}-space

Section 7. Constructions of the generalized morphisms and isomorphisms

Part II. Theory for the case of multivariable total or partial functions

Section 8. Basic notions for the case of multivariable (total) functions

Section 9. Basic notions for the case of (multivariable) partial functions

Section 10. Basic properties and more notions

Section 11. Galois correspondences

Section 12. Lattice structures of objects arising in Galois correspondences on a TT-space SS

Section 13. Topologies employed to construct Galois correspondences

Section 14. Constructions of the generalized morphisms and isomorphisms

Part III. Solvability of equations

Section 15. A solvability of polynomial equations

Section 16. A solvability of homogeneous linear differential equations

Section 17. A possible strategy for equation solving

Part IV. Other topics and future research

Section 18. Dualities of operator semigroups

Section 19. Fixed sets and transitive actions of EndT(S)\operatorname{End}_{T}(S) and AutT(S)\operatorname{Aut}_{T}(S)

Section 20. Two questions on Galois TT-extensions and normal subgroups of Galois TT-groups

Section 21. \mathcal{F}-transcendental elements and \mathcal{F}-transcendental subsets

Section 22. A generalized first isomorphism theorem

Section 23. On topological spaces

Section LABEL:App_to_dyn. On dynamical systems

Section LABEL:Other_topics. Other topics for future research

The contents of Part I are roughly described section by section as follows.

Let TT be a set of functions from a set DD to DD. If the composite of any two elements in TT still lies in TT, then we call TT an operator semigroup on DD (cf. Definition 2.1.1). Let UDU\subseteq D. Then we call fTIm(f|U)\bigcup\nolimits_{f\in T}{\operatorname{Im}(f{{|}_{U}})} the TT-space generated by UU, where Im(f|U)\operatorname{Im}(f{{|}_{U}}) denotes the image of the restriction of ff to UU (cf. Definition 2.2.1). TT-spaces represent mathematical objects whose structures are preserved under generalized morphisms. A critical notion in this theory is TT-morphism, which is defined as a map σ\sigma from a TT-space SS to a TT-space such that σ\sigma commutes with every element of TT; that is, aS\forall a\in S and fTf\in T, σ(f(a))=f(σ(a))\sigma(f(a))=f(\sigma(a)) (cf. Definition 2.3.1). From these simple definitions much follows. For example, any ring homomorphism between B[u]B[u] and B[v]B[v] with field BB fixed pointwisely can be characterized in terms of a TT-morphism (cf. Proposition 2.3.2), and a map from a topological space to itself is continuous if and only if the map induces a TT-morphism (cf. Proposition 2.3.5). In Subsection 2.4, we show that the inverse of any bijective TT-morphism from a TT-space S1S_{1} to a TT-space S2S_{2} is a TT-morphism from S2S_{2} to S1S_{1}, and hence a TT-isomorphism can be defined as a bijective TT-morphism.

In Section 3, we establish the Galois correspondence between the Galois groups of a TT-space SS and the fixed subsets of SS under the actions of the TT-automorphisms of SS (cf. Corollary 3.2.6). Moreover, we shall find the Galois correspondence between the “Galois monoids” of a TT-space SS and the fixed subsets of SS under the actions of the TT-endomorphisms of SS (cf. Corollary 3.2.5).

To better understand these Galois correspondences, in Section 4, we study the lattice structures of those objects which arise in Corollaries 3.2.5 and 3.2.6. Results in Section 4 will be employed in Section 5.

In the fundamental theorem for infinite algebraic field extensions (resp. for differential field extensions), the Galois correspondences are characterized with Krull topology (resp. Zariski topology) (see e.g. [1, 3, 4, 5, 9]). In Section 5, we shall see when and how we can characterize Galois correspondences with topology.

Section 6 introduces the notions of θ\theta-morphism and θ\theta-isomorphism. As a generalization of the notion of TT-morphism, a θ\theta-morphism (cf. Definitions 6.1.1 and 6.1.8) is from a T1{{T}_{1}}-space to a T2{{T}_{2}}-space, where both T1{{T}_{1}} and T2{{T}_{2}} are operator semigroups and θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}}. We shall find that some more familiar mathematical notions can be characterized by θ\theta-morphisms (cf. Propositions 6.1.2, 6.1.3 and 6.1.10). In Subsection 6.2, we show that the inverse of any bijective θ\theta-morphism from a T1T_{1}-space S1S_{1} to a T2T_{2}-space S2S_{2} is a θ1\theta^{-1}-morphism from S2S_{2} to S1S_{1}, and thus a θ\theta-isomorphism can be defined as a bijective θ\theta-morphism.

Main results in Section 7 are about constructions of TT-morphisms and θ\theta-morphisms. Roughly speaking, for a TT-morphism or θ\theta-morphism from a TT-space SS, Subsections 7.1, 7.2 and 7.3 discuss how to construct the morphism by a map from a set UU which generates SS (i.e. S=fTIm(f|U)S=\bigcup\nolimits_{f\in T}{\operatorname{Im}(f{{|}_{U}})}). Moreover, in Subsection 7.4, we shall introduce a construction of TT-morphisms in terms of topology.

The contents of Part II are described as follows.

To make our theory more general, in Section 8, we generalize the notion of operator semigroup to incorporate functions of more than one variable (cf. Definition 8.1.2). And correspondingly, the concepts of TT-spaces, TT-morphisms and θ\theta-morphisms are generalized (cf. Definitions 8.2.1, 8.3.1, and 8.5.4). Then we shall find that ring homomorphisms, module homomorphisms, and group homomorphisms can be characterized in terms of TT-morphisms (cf. Propositions 8.3.4 to 8.3.7) or θ\theta-morphisms (cf. Propositions 8.5.7 to 8.5.11).

To further generalize our theory, in Section 9, we allow operators in TT to be partial functions (cf. Definition 9.1.4). And correspondingly, the notions of TT-spaces, TT-morphisms and θ\theta-morphisms are generalized (cf. Definitions 9.2.1, 9.3.1, and 9.5.5). Because operators in TT are now allowed to be partial functions, we can generate fields, differential fields, and categories as TT-spaces in a natural way (cf. Examples 9.2.2 to 9.2.4). Then we shall find that ring homomorphisms between fields, differential ring homomorphisms between differential fields, and covariant functors between categories can be characterized in terms of TT-morphisms (cf. Propositions 9.3.4 and 9.3.6) or θ\theta-morphisms (cf. Propositions 9.5.8 and 9.5.9).

For the case of (partial) functions of more than one variable, results obtained in Sections 2, 3, 4, 5 and 7 are generalized in Sections 10, 11, 12, 13 and 14, respectively. (Note that results obtained in Section 6 are generalized in Sections 8 and 9.)

Part III addresses the following issues.

A main goal of the classical Galois theory is to study the solvability by radicals of polynomial equations. In Section 15, however, we shall introduce another understanding of solvability of polynomial equations, which somehow corresponds to a composition series of the Galois group of the polynomial. And in Subsection 15.2 we shall explain why we introduce this notion of solvability. Roughly speaking, the central idea is that a separable polynomial p(x)p(x) can be “decomposed” into nn polynomials p1(x),,pn(x){{p}_{1}}(x),\cdots,{{p}_{n}}(x) such that the Galois group of each pi(x){{p}_{i}}(x) is simple and any root of p(x)p(x) can be expressed in terms of the roots of p1(x),,pn(x){{p}_{1}}(x),\cdots,{{p}_{n}}(x) (cf. Corollary 15.2.5).

Analogously, in Section 16, we shall introduce a new interpretation of solvability of homogeneous linear differential equations, which somehow corresponds to a “composition Zariski-closed series” of the differential Galois group of the equation. In Subsection 16.3 we shall explain why we introduce this notion of solvability. As what we did in Section 15, our main approach is to “decompose” a homogeneous linear differential equation L(Y)=0L(Y)=0 into nn homogeneous linear differential equations L1(Y)=0,,Ln(Y)=0{{L}_{1}}(Y)=0,\cdots,{{L}_{n}}(Y)=0 so that any solution of L(Y)=0L(Y)=0 can be expressed in terms of the solutions of L1(Y)=0,,Ln(Y)=0{{L}_{1}}(Y)=0,\cdots,{{L}_{n}}(Y)=0 (cf. Corollary 16.3.6).

Section 17 generalizes some results in Sections 15 and 16, and it roughly describes in terms of our theory a possible strategy for equation solving.

In Part IV we give some more results which we think to be important or deserve deeper research.

Section 18 is about dualities of operator semigroups. Let TT be an operator semigroup and let SS be a TT-space. We denote by EndT(S)\operatorname{End}_{T}(S) (resp. AutT(S)\operatorname{Aut}_{T}(S)) the set of all TT-endomorphisms of SS (resp. the set of all TT-automorphisms of SS). Then, roughly speaking, there is a duality between EndT(S)\operatorname{End}_{T}(S) and the maximum operator semigroup which “accommodates” EndT(S)\operatorname{End}_{T}(S) (i.e. σEndT(S)\forall\sigma\in\operatorname{End}_{T}(S), σ\sigma would still be a TT-endomorphism of SS if TT were extended to the maximum). Analogously, there is a duality between AutT(S)\operatorname{Aut}_{T}(S) and the maximum operator semigroup which “accommodates” AutT(S)\operatorname{Aut}_{T}(S).

In Section 19, we shall talk about fixed subsets of a TT-space SS under the actions of EndT(S)\operatorname{End}_{T}(S) and AutT(S)\operatorname{Aut}_{T}(S). And we shall obtain some transitive properties of the actions of EndT(S)\operatorname{End}_{T}(S) and AutT(S)\operatorname{Aut}_{T}(S) on SS, which are analogues of the well-known fact that the Galois group of an irreducible polynomial acts transitively on its roots.

Section 20 addresses an analogue of the issue which is normally the last part of the fundamental theorem of a Galois theory: the correspondence between the normal subgroups of the Galois group and the Galois extensions of the base field.

In Section 21, the algebraic notions of transcendental elements over a field and purely transcendental field extensions are generalized.

In Section 22, we shall obtain an isomorphism theorem which generalizes the first isomorphism theorems (for groups, rings, and modules).

In Section 23, we shall introduce for topological spaces some notions and properties related to our theory. And we shall employ Corollaries 3.2.5 and 3.2.6 to obtain Galois correspondences on topological spaces.

Let (M,S,Φ)(M,S,\Phi) be a dynamical system, where MM is a monoid, SS is the phase space and Φ\Phi is the evolution function. In Section LABEL:App_to_dyn, it is shown that (M,S,Φ)(M,S,\Phi) induces in a natural way an operator semigroup TT on SS such that TT contains the identity function and SS is a TT-space. Conversely, for any operator semigroup TT which contains the identity function, any TT-space SS induces a dynamical system (T,S,Φ)(T,S,\Phi) in a natural way. Hence we can apply our theory for operator semigroups to dynamical systems. In particular, we shall obtain Galois correspondences on dynamical systems.

Section LABEL:Other_topics gives some more suggestions on future research.

Part I. THEORY FOR THE CASE OF UNARY FUNCTIONS

In Part I of the article, we introduce the generalized morphisms where all functions involved are unary. The generalized morphisms preserve structures of mathematical objects which we call TT-spaces, where TT stands for a semigroup of functions with composition of functions as the binary operation.

2. Basic notions and properties

2.1. Operator semigroup TT

Sets which are closed under some operations are ubiquitous in mathematics. The following notion is related to this observation.

Definition 2.1.1.

Let DD be a set and let TT be a set of functions from DD to DD. If f,gT\forall f,g\in T, the composite fgTf\circ g\in T, then we call TT an operator semigroup on DD, and we call DD the domain of TT.

Remark.

It is clear that TT, possibly empty, is a semigroup with composition of functions as the binary operation.

Example 2.1.2.

Let BB be a subfield of a field FF and let B[x]B[x] be the ring of all polynomials (in one variable) over BB. Let

T={f:FF given by af(a)|f(x)B[x]}.T=\{{f}^{*}:F\to F\text{ given by }a\mapsto f(a)\,|\,f(x)\in B[x]\}.

(In particular, ff^{*} is a constant (polynomial) function if ff is a constant polynomial.) Then it is not hard to see that TT is an operator semigroup on FF. Note that the map which induces TT, i.e. τ:B[x]T\tau:B[x]\to T given by fff\mapsto{{f}^{*}}, is surjective, but it is not necessarily injective.

Remark.

Of course, there are other ways to define an operator semigroup on a field FF. For example, let T1T_{1} consist of only the identity function on FF and let T2T_{2} be the set of all ring homomorphisms from FF to FF, then both T1T_{1} and T2T_{2} are operator semigroups on FF. But unlike TT in Example 2.1.2, neither T1T_{1} nor T2T_{2} could lead us to any desirable result (in the remaining part of the article).

In fact and roughly speaking, to obtain desirable results, an operator semigroup should represent the intrinsic operations on the mathematical object involved.

Notation 2.1.3.

Throughout the paper, we denote by Id the identity function (on some set which is clear from the context).

As Example 2.1.2, the following two will lead us to desirable results.

Example 2.1.4.

Let XX be a topological space and let 𝒫(X)\mathcal{P}(X) be the power set of XX. Let

T={Id:𝒫(X)𝒫(X),Cl:𝒫(X)𝒫(X) given by AA¯},T=\{\operatorname{Id}:\mathcal{P}(X)\to\mathcal{P}(X),\operatorname{Cl}:\mathcal{P}(X)\to\mathcal{P}(X)\text{ given by }A\mapsto\overline{A}\},

where A¯\overline{A} is the closure of AA. Then TT has only two elements and it is an operator semigroup on 𝒫(X)\mathcal{P}(X).

Example 2.1.5.

Recall that a differential ring RR is a commutative ring with identity endowed with a derivation \partial. Then {n|n0}\{{{\partial}^{n}}\,|\,n\in{{\mathbb{N}}_{0}}\}, where 0:=Id{{\partial}^{0}}:=\operatorname{Id}, is an operator semigroup on RR.

Definition 2.1.6.

Let DD be a set and let GG be a set of functions from DD to DD. We call the intersection of all operator semigroups (on DD) containing GG the operator semigroup on DD generated by GG, and we denote it by G\left\langle G\right\rangle. If G={g}G=\{g\}, then we may denote G\left\langle G\right\rangle by g\left\langle g\right\rangle for brevity.

Remark.

G\left\langle G\right\rangle is the smallest operator semigroup (on DD) which contains GG.

2.2. TT-spaces and quasi-TT-spaces

Operator semigroups are used to generate TT-spaces defined as follows.

Definition 2.2.1.

Let TT be an operator semigroup on DD and let UDU\subseteq D. We call S:={f(u)|fT,uUS:=\{f(u)\,|\,f\in T,u\in U} the TT-space generated by UU and denote SS by UT{{\left\langle U\right\rangle}_{T}}. If U={u}U=\{u\}, then we also denote {u}T{{\left\langle\{u\}\right\rangle}_{T}} by uT{{\left\langle u\right\rangle}_{T}}.

Moreover, if BSB\subseteq S is also a TT-space generated by some subset of DD, then we say that BB is a TT-subspace of SS and write BSB\leq S or SBS\geq B.

Remark.

SS may have generators other than UU, and we may just call SS a TT-space.

Trivially, T={{\left\langle\emptyset\right\rangle}_{T}}=\emptyset is a TT-space and D={{\left\langle D\right\rangle}_{\emptyset}}=\emptyset is a \emptyset-space.

TT-spaces are mathematical objects whose structures are preserved under the generalized morphisms (to be defined later), and this claim will be shown later on the following examples.

Example 2.2.2.

Let BB be a subfield of a field FF. Let TT be the operator semigroup defined in Example 2.1.2. Then aF\forall a\in F, aT{{\left\langle a\right\rangle}_{T}} is the ring B[a]B[a]. In particular, if F/BF/B is an algebraic field extension, then aF\forall a\in F, aT{{\left\langle a\right\rangle}_{T}} is the field B(a)B(a).

Example 2.2.3.

Let XX be a topological space and let 𝒫(X)\mathcal{P}(X) be the power set of XX. If TT is defined as in Example 2.1.4, then 𝒫(X)T=𝒫(X){{\left\langle\mathcal{P}(X)\right\rangle}_{T}}=\mathcal{P}(X) is a TT-space.

Example 2.2.4.

Let RR be a differential ring with a derivation \partial. Let TT be the operator semigroup defined in Example 2.1.5. Recall that an ideal II of RR is a differential ideal if aIa\in I implies that (a)I\partial(a)\in I. Thus an ideal II of RR is a differential ideal of RR if and only if the TT-space IT=I{{\left\langle I\right\rangle}_{T}}=I. We will talk more about differential rings in Sections 6, 7 and 9.

We are going to develop some properties of the new notions. In Sections 2 to 7, unless otherwise specified, DD denotes a set and TT denotes an operator semigroup on DD.

First, the following example is used to prove the proposition right after it.

Example 2.2.5.

Let T={f:++T=\{f:{{\mathbb{Z}}^{+}}\to{{\mathbb{Z}}^{+}} given by xx+k|k+}x\mapsto x+k\,|\,k\in{{\mathbb{Z}}^{+}}\}. Then TT is an operator semigroup on +{{\mathbb{Z}}^{+}}, and 1T={2,3,4,}.{{\left\langle 1\right\rangle}_{T}}=\{2,3,4,\cdots\}.

Proposition 2.2.6.
  1. (a)

    A generator set of a TT-space may have no intersection with the TT-space.

  2. (b)

    It is possible that no subset of a TT-space may generate the TT-space.

  3. (c)

    DD is not necessarily a TT-space.

  4. (d)

    A subset SS of DD is not necessarily a TT-space even if it satisfies: f(a)Sf(a)\in S, aS\forall a\in S and fTf\in T.

Proof.

For (a): In Example 2.2.5,

{1}1T={1}{2,3,4,}=.\{1\}\bigcap{{\left\langle 1\right\rangle}_{T}}=\{1\}\bigcap\{2,3,4,\cdots\}=\emptyset.

For (b): In Example 2.2.5, no subset of 1T{{\left\langle 1\right\rangle}_{T}} may generate 1T{{\left\langle 1\right\rangle}_{T}}.

For (c): In Example 2.2.5, 1UT,U+1\notin{{\left\langle U\right\rangle}_{T}},\forall U\subseteq{\mathbb{Z}}^{+}, and hence +{{\mathbb{Z}}^{+}} is not a TT-space.

For (d): In Example 2.2.5, +{{\mathbb{Z}}^{+}} satisfies: f(a)+f(a)\in{{\mathbb{Z}}^{+}}, a+\forall a\in{{\mathbb{Z}}^{+}} and fTf\in T, but +{{\mathbb{Z}}^{+}} is not a TT-space. ∎

However, any TT-space SS must satisfy: f(a)Sf(a)\in S, aS\forall a\in S and fTf\in T, as shown below.

Proposition 2.2.7.

Let SS be a TT-space. Then aS\forall a\in S and fTf\in T, f(a)Sf(a)\in S. Hence AS\forall A\subseteq S, ATS{{\left\langle A\right\rangle}_{T}}\leq S, i.e. AT{{\left\langle A\right\rangle}_{T}} is TT-subspace of SS (Definition 2.2.1). In particular, STS{{\left\langle S\right\rangle}_{T}}\leq S.

Proof.

By Definition 2.2.1, UD\exists U\subseteq D such that UT=S{{\left\langle U\right\rangle}_{T}}=S. Let aSa\in S. Then gT\exists g\in T and uUu\in U such that g(u)=ag(u)=a. Let fTf\in T. Then by Definition 2.1.1, fgTf\circ g\in T, and thus f(a)=f(g(u))UT=Sf(a)=f(g(u))\in{{\left\langle U\right\rangle}_{T}}=S. Hence aS\forall a\in S and fTf\in T, f(a)Sf(a)\in S. Then AS\forall A\subseteq S, ATS{{\left\langle A\right\rangle}_{T}}\subseteq S, and hence by Definition 2.2.1, ATS{{\left\langle A\right\rangle}_{T}}\leq S. ∎

In light of Proposition 2.2.7 and (d) in Proposition 2.2.6, we introduce the following concept, whose use will be clear later.

Definition 2.2.8.

Let SDS\subseteq D. If STS{{\left\langle S\right\rangle}_{T}}\subseteq S, then we call SS a quasi-TT-space. Moreover, if BSB\subseteq S and both BB and SS are quasi-TT-spaces, then we call BB a quasi-TT-subspace of SS and write BqSB\leq_{q}S or SqBS\geq_{q}B.

By Proposition 2.2.7, the following is immediate.

Proposition 2.2.9.

Any TT-space is a quasi-TT-space.

However, the converse of Proposition 2.2.9 is not always true, as is implied by (d) in Proposition 2.2.6, unless we add a condition as follows.

Proposition 2.2.10.

Let SS be a quasi-TT-space. If SSTS\subseteq{{\left\langle S\right\rangle}_{T}}, then SS is a TT-space.

Remark.

If IdT\operatorname{Id}\in T, then SSTS\subseteq{{\left\langle S\right\rangle}_{T}}.

Proof.

If SSTS\subseteq{{\left\langle S\right\rangle}_{T}}, then ST=S{{\left\langle S\right\rangle}_{T}}=S because STS{{\left\langle S\right\rangle}_{T}}\subseteq S (by Definition 2.2.8), and hence SS is a TT-space (by Definition 2.2.1). ∎

Proposition 2.2.10 implies that, if IdT\operatorname{Id}\in T and SS is a quasi-TT-space, then SS is a TT-space, and hence in this case TT-space and quasi-TT-space are actually the same notion. In fact, we shall see that IdT\operatorname{Id}\in T in most of our examples.

Notation 2.2.11.

In this article, II denotes an index set unless otherwise specified.

Proposition 2.2.12.

The intersection of any family of quasi-TT-spaces is a quasi-TT-space.

Proof.

Let {Si|iI}\{{{S}_{i}}\,|\,i\in I\} be a family of quasi-TT-spaces and let S=iISiS=\bigcap\nolimits_{i\in I}{{{S}_{i}}}. Then iI\forall i\in I, STSiT{{\left\langle S\right\rangle}_{T}}\subseteq{{\left\langle{{S}_{i}}\right\rangle}_{T}} (since SSiS\subseteq{{S}_{i}}). And by Definition 2.2.8, iI\forall i\in I, SiTSi{{\left\langle{{S}_{i}}\right\rangle}_{T}}\subseteq{{S}_{i}}. Hence iI\forall i\in I, STSi{{\left\langle S\right\rangle}_{T}}\subseteq{{S}_{i}}. Therefore, STiISi=S{{\left\langle S\right\rangle}_{T}}\subseteq\bigcap\nolimits_{i\in I}{{{S}_{i}}}=S. ∎

Proposition 2.2.13.

The union of any family of quasi-TT-spaces is a quasi-TT-space.

Proof.

Let {Si|iI}\{{{S}_{i}}\,|\,i\in I\} be a collection of quasi-TT-spaces and let S=iISiS=\bigcup\nolimits_{i\in I}{{{S}_{i}}}. Then by Definitions 2.2.1 and 2.2.8,

ST=iISiT=iISiTiISi=S.{{\left\langle S\right\rangle}_{T}}={{\left\langle\bigcup\nolimits_{i\in I}{{{S}_{i}}}\right\rangle}_{T}}=\bigcup\nolimits_{i\in I}{{{\left\langle{{S}_{i}}\right\rangle}_{T}}}\subseteq\bigcup\nolimits_{i\in I}{{{S}_{i}}}=S.

For TT-spaces, we have

Proposition 2.2.14.

The union of any family of TT-spaces is a TT-space.

Proof.

Let {Si|iI}\{{{S}_{i}}\,|\,i\in I\} be a collection of TT-spaces and let S=iISiS=\bigcup\nolimits_{i\in I}{{{S}_{i}}}. By Definition 2.2.1, iI\forall i\in I, we can suppose Si=UiT{{S}_{i}}={{\left\langle{{U}_{i}}\right\rangle}_{T}}, where UiDU_{i}\subseteq D. Then S=iIUiT=iIUiTS=\bigcup\nolimits_{i\in I}{{{\left\langle{{U}_{i}}\right\rangle}_{T}}}={{\left\langle\bigcup\nolimits_{i\in I}{{{U}_{i}}}\right\rangle}_{T}} is a TT-space. ∎

However, the intersection of a set of TT-spaces is not necessarily a TT-space, as shown below.

Example 2.2.15.

Let

D={a0+a1π+a2π2++anπn|n0,ai0,i=0,,n}D=\{{{a}_{0}}+{{a}_{1}}\pi+{{a}_{2}}{{\pi}^{2}}+\cdots+{{a}_{n}}{{\pi}^{n}}\,|\,n\in{{\mathbb{N}}_{0}},{{a}_{i}}\in{{\mathbb{N}}_{0}},\forall i=0,\cdots,n\}.

Let f:DDf:D\to D be given by xπ+xx\mapsto\pi+x, let g:DDg:D\to D be given by xπxx\mapsto\pi x, let T={f,g}T=\left\langle\{f,g\}\right\rangle (Definition 2.1.6), and let A=0T1TA={{\left\langle 0\right\rangle}_{T}}\bigcap{{\left\langle 1\right\rangle}_{T}}. Then each element of AA is nonzero (because 01T0\notin{{\left\langle 1\right\rangle}_{T}}) and has the form a1π+a2π2++anπn{{a}_{1}}\pi+{{a}_{2}}{{\pi}^{2}}+\cdots+{{a}_{n}}{{\pi}^{n}} where n+n\in{{\mathbb{Z}}^{+}} and each ai0{{a}_{i}}\in{{\mathbb{N}}_{0}} (because each element of 0T{{\left\langle 0\right\rangle}_{T}} has the form). Assume that AA is a TT-space UT{{\left\langle U\right\rangle}_{T}}, where UDU\subseteq D. Because πA\pi\in A, we can tell that either 0U0\in U or 1U1\in U. However, it follows that either 0A0\in A or π+1A\pi+1\in A, a contradiction.

Nevertheless, we have

Proposition 2.2.16.

Let SS be the intersection of any family of TT-spaces. Then SS is a quasi-TT-space. Moreover, if IdT\operatorname{Id}\in T or more generally, SSTS\subseteq{{\left\langle S\right\rangle}_{T}}, then SS is a TT-space.

Proof.

SS is a quasi-TT-space by Propositions 2.2.9 and 2.2.12. If IdT\operatorname{Id}\in T or SSTS\subseteq{{\left\langle S\right\rangle}_{T}}, then by Proposition 2.2.10, SS is a TT-space. ∎

2.3. TT-morphisms

A generalized morphism which we call a TT-morphism between TT-spaces is only required to satisfy a simple but important condition.

Definition 2.3.1.

Let TT be an operator semigroup and let σ\sigma be a map from a TT-space SS to a TT-space. If fT\forall f\in T, σ\sigma commutes with ff on SS, i.e. σ(f(a))=f(σ(a))\sigma(f(a))=f(\sigma(a)), aS\forall a\in S and fTf\in T, then σ\sigma is called a TT-morphism.

Remark.
  1. (1)

    By Proposition 2.2.7, aS\forall a\in S and fTf\in T, f(a)Sf(a)\in S, and hence σ(f(a))\sigma(f(a)) is well-defined.

  2. (2)

    Note that if S=UTS={{\left\langle U\right\rangle}_{T}} and uUu\in U, then σ(u)\sigma(u) is not defined unless uUTu\in{{\left\langle U\right\rangle}_{T}}.

  3. (3)

    We regard the empty map \emptyset\to\emptyset as a TT-morphism from the TT-space \emptyset to \emptyset.

The following four results show that the notion of TT-morphism can characterize some familiar notions in mathematics.

Proposition 2.3.2.

Let BB be a subfield of a field FF. Let TT be the operator semigroup defined in Example 2.1.2. Then u,vF\forall u,v\in F, a map σ:uTvT\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}} is a ring homomorphism with BB fixed pointwisely if and only if σ\sigma is a TT-morphism.

Proof.

Clearly, uT{{\left\langle u\right\rangle}_{T}} and vT{{\left\langle v\right\rangle}_{T}} are the rings B[u]B[u] and B[v]B[v], respectively.

1. Sufficiency

Suppose that σ\sigma is a TT-morphism. Then by Definition 2.3.1, σ(f(a))=f(σ(a))\sigma(f(a))=f(\sigma(a)), fT\forall f\in T and auT=B[u]a\in{{\left\langle u\right\rangle}_{T}}=B[u].

Considering the case where fTf\in T is a constant polynomial function given by ac(B),aFa\mapsto c(\in B),\forall a\in F, we can tell that cB\forall c\in B, σ(c)(=σ(f(a))=f(σ(a)))=c\sigma(c)(=\sigma(f(a))=f(\sigma(a)))=c, i.e. BB is fixed pointwisely under σ\sigma.

Let a,buT=B[u]a,b\in{{\left\langle u\right\rangle}_{T}}=B[u]. Then p(x),q(x)B[x]\exists p(x),q(x)\in B[x] with p(u)=ap(u)=a and q(u)=bq(u)=b. Hence

σ(a+b)\sigma(a+b)

=σ(p(u)+q(u))=\sigma(p(u)+q(u))

=σ((p+q)(u))=\sigma((p+q)(u))

=σ(τ(p+q)(u))=\sigma(\tau(p+q)(u)) (τ:B[x]T\tau:B[x]\to T is defined in Example 2.1.2)

=τ(p+q)(σ(u))=\tau(p+q)(\sigma(u)) (since σ\sigma is a TT-morphism)

=(p+q)(σ(u))=(p+q)(\sigma(u))

=p(σ(u))+q(σ(u))=p(\sigma(u))+q(\sigma(u))

=τ(p)(σ(u))+τ(q)(σ(u))=\tau(p)(\sigma(u))+\tau(q)(\sigma(u))

=σ(τ(p)(u))+σ(τ(q)(u))=\sigma(\tau(p)(u))+\sigma(\tau(q)(u)) (since σ\sigma is a TT-morphism)

=σ(p(u))+σ(q(u))=\sigma(p(u))+\sigma(q(u))

=σ(a)+σ(b)=\sigma(a)+\sigma(b).

And

σ(ab)\sigma(ab)

=σ(p(u)q(u))=\sigma(p(u)q(u))

=σ((pq)(u))=\sigma((pq)(u))

=σ(τ(pq)(u))=\sigma(\tau(pq)(u))

=τ(pq)(σ(u))=\tau(pq)(\sigma(u)) (since σ\sigma is a TT-morphism)

=(pq)(σ(u))=(pq)(\sigma(u))

=p(σ(u))q(σ(u))=p(\sigma(u))q(\sigma(u))

=τ(p)(σ(u))τ(q)(σ(u))=\tau(p)(\sigma(u))\tau(q)(\sigma(u))

=σ(τ(p)(u))σ(τ(q)(u))=\sigma(\tau(p)(u))\sigma(\tau(q)(u)) (since σ\sigma is a TT-morphism)

=σ(p(u))σ(q(u))=\sigma(p(u))\sigma(q(u))

=σ(a)σ(b)=\sigma(a)\sigma(b).

Therefore, σ\sigma is a ring homomorphism with BB fixed pointwisely.

2. Necessity

Suppose that σ\sigma is a ring homomorphism with BB fixed pointwisely. Then

cB\forall c\in B, σ(c)=c\sigma(c)=c, and

a,buT=B[u]\forall a,b\in{{\left\langle u\right\rangle}_{T}}=B[u], σ(a+b)=σ(a)+σ(b)\sigma(a+b)=\sigma(a)+\sigma(b) and σ(ab)=σ(a)σ(b)\sigma(ab)=\sigma(a)\sigma(b).

Let zuT=B[u]z\in{{\left\langle u\right\rangle}_{T}}=B[u] and let fTf\in T. Because ff is a polynomial function, it is not hard to see that σ(f(z))=f(σ(z))\sigma(f(z))=f(\sigma(z)), as desired. ∎

Remark.

Proposition 2.3.2 justifies the definition of TT in Example 2.1.2: we could not obtain such a desirable result if TT were defined otherwise.

Then for ring isomorphisms, we have

Corollary 2.3.3.

Let BB be a subfield of a field FF, let TT be the operator semigroup defined in Example 2.1.2, and let u,vFu,v\in F be algebraic over BB with deg(min(B,u))=deg(min(B,v))\deg(\min(B,u))=\deg(\min(B,v)), where deg(min(B,))\deg(\min(B,*)) denotes the degree of the minimal polynomial of * over BB. Then a map σ:uTvT\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}} is a ring isomorphism with BB fixed pointwisely if and only if it is a TT-morphism.

Proof.

By Proposition 2.3.2, σ\sigma is a ring homomorphism with BB fixed pointwisely if and only if it is a TT-morphism. So it suffices to show that the ring homomorphism σ\sigma is bijective.

Since uu is algebraic over BB, uT(=B[u]=B(u)){{\left\langle u\right\rangle}_{T}}\,(=B[u]=B(u)) is a field. It follows that the ring homomorphism σ\sigma must be injective. Moreover, since deg(min(B,u))=deg(min(B,v))\deg(\min(B,u))=\deg(\min(B,v)), σ\sigma must be surjective. ∎

Recall that every finite field extension is algebraic and every finite separable field extension is simple. So by Corollary 2.3.3, we immediately have the following, which is for later use.

Corollary 2.3.4.

Let F/BF/B be a finite separable field extension. Let TT be the operator semigroup defined in Example 2.1.2. Then FF is a TT-space generated by a single element of FF. Moreover, a map σ:FF\sigma:F\to F is a field automorphism with BB fixed pointwisely if and only if it is a TT-morphism.

Note that every ring homomorphism discussed above involves only one polynomial ring (B[x]B[x] in Example 2.1.2). We will treat ring homomorphisms which involve two (distinct) polynomial rings in Subsection 6.1. Moreover, ring homomorphisms involving polynomial rings in more than one variable will be treated in Sections 8 and 9.

The following may be a little surprising.

Proposition 2.3.5.

Let XX be a topological space, let 𝒫(X)\mathcal{P}(X) be the power set of XX, and let T={IdT=\{\operatorname{Id}, Cl:𝒫(X)𝒫(X)\operatorname{Cl}:\mathcal{P}(X)\to\mathcal{P}(X) given by AA¯}A\mapsto\overline{A}\}. Let a map p:XXp:X\to X induce a map p:𝒫(X)𝒫(X){p}^{*}:\mathcal{P}(X)\to\mathcal{P}(X) as follows.

A𝒫(X)\forall A\in\mathcal{P}(X) that is closed in XX, let p(A)=p(A)¯{p}^{*}(A)=\overline{p(A)}, where p(A)p(A) ((with a slight abuse of notation)) denotes the set {p(x)|xA}\{p(x)\,|\,x\in A\}. And A𝒫(X)\forall A\in\mathcal{P}(X) which is not closed in XX, let p(A)=p(A){p}^{*}(A)=p(A).

Then pp is continuous if and only if p{p}^{*} is a TT-morphism.

Remark.

If A=A=\emptyset, then p(A)={p(x)|xA}=p(A)=\{p(x)\,|\,x\in A\}=\emptyset, and hence p()=p()¯={p}^{*}(\emptyset)=\overline{p(\emptyset)}=\emptyset.

Proof.

As in Examples 2.1.4 and 2.2.3, TT is an operator semigroup on 𝒫(X)\mathcal{P}(X) and 𝒫(X)T=𝒫(X){{\left\langle\mathcal{P}(X)\right\rangle}_{T}}=\mathcal{P}(X) is a TT-space.

1. Necessity

Suppose that pp is continuous. To prove that p{p}^{*} is a TT-morphism, it suffices to show that p(f(A))=f(p(A)){p}^{*}(f(A))=f({p}^{*}(A)), A𝒫(X)\forall A\in\mathcal{P}(X) and fTf\in T. The equation holds when f=Idf=\operatorname{Id}. So we only need to show that p(Cl(A))=Cl(p(A)){p}^{*}(\operatorname{Cl}(A))=\operatorname{Cl}({p}^{*}(A)), A𝒫(X)\forall A\in\mathcal{P}(X).

If A=A¯A=\overline{A}, by the definitions of p{p}^{*} and Cl\operatorname{Cl},

Cl(p(A))=Cl(p(A)¯)=p(A)¯=p(A)=p(Cl(A)),\operatorname{Cl}({p}^{*}(A))=\operatorname{Cl}(\overline{p(A)})=\overline{p(A)}={p}^{*}(A)={p}^{*}(\operatorname{Cl}(A)),

as desired.

Since pp is continuous, p(A¯)p(A)¯p(\overline{A})\subseteq\overline{p(A)} (see e.g. [6]). Hence p(A¯)¯p(A)¯\overline{p(\overline{A})}\subseteq\overline{p(A)}. On the other hand, p(A¯)¯p(A)¯\overline{p(\overline{A})}\supseteq\overline{p(A)} because p(A¯)p(A)p(\overline{A})\supseteq p(A). So p(A¯)¯=p(A)¯\overline{p(\overline{A})}=\overline{p(A)}. Thus if AA¯A\neq\overline{A}, then

Cl(p(A))=Cl(p(A))=p(A)¯=p(A¯)¯=p(A¯)=p(Cl(A)),\operatorname{Cl}({p}^{*}(A))=\operatorname{Cl}(p(A))=\overline{p(A)}=\overline{p(\overline{A})}={p}^{*}(\overline{A})={p}^{*}(\operatorname{Cl}(A)),

as desired.

2. Sufficiency

Suppose that p{p}^{*} is a TT-morphism. To prove that pp is continuous, we only need to show that p1(A)=p1(A)¯{{p}^{-1}}(A)=\overline{{{p}^{-1}}(A)}, A𝒫(X)\forall A\in\mathcal{P}(X) with A=A¯A=\overline{A}.

Assume A𝒫(X)\exists A\in\mathcal{P}(X) such that A=A¯A=\overline{A} and p1(A)p1(A)¯{{p}^{-1}}(A)\neq\overline{{{p}^{-1}}(A)}. Let B=p1(A)B={{p}^{-1}}(A). Then BB¯B\neq\overline{B} and p(B¯)Ap(\overline{B})\nsubseteq A (because otherwise B=B¯B=\overline{B}). Hence

p(Cl(B))=p(B¯)=p(B¯)¯p(B¯)A,{{p}^{*}}(\operatorname{Cl}(B))={{p}^{*}}(\overline{B})=\overline{p(\overline{B})}\supseteq p(\overline{B})\nsubseteq A,

but

Cl(p(B))=Cl(p(B))=p(B)¯A¯=A\operatorname{Cl}({p}^{*}(B))=\operatorname{Cl}(p(B))=\overline{p(B)}\subseteq\overline{A}=A

because p(B)A.p(B)\subseteq A. Therefore, Cl(p(B))p(Cl(B))\operatorname{Cl}({p}^{*}(B))\neq{p}^{*}(\operatorname{Cl}(B)), which is contrary to the assumption that p{p}^{*} is a TT-morphism. ∎

The map pp in Proposition 2.3.5 is from a topological space to itself. We will treat continuous functions between different topological spaces in Subsection 6.1.

Several properties of TT-morphisms are as follows. We shall employ them later.

Proposition 2.3.6.

Let σ\sigma be a TT-morphism from S1{{S}_{1}} to S2{{S}_{2}}. Then ImσqS2\operatorname{Im}\sigma\leq_{q}{S_{2}}, i.e. Imσ\operatorname{Im}\sigma is a quasi-TT-subspace of S2S_{2} (Definition 2.2.8). Moreover, if IdT\operatorname{Id}\in T or more generally, ImσImσT\operatorname{Im}\sigma\subseteq{{\left\langle\operatorname{Im}\sigma\right\rangle}_{T}}, then ImσS2\operatorname{Im}\sigma\leq{{S}_{2}}, i.e. Imσ\operatorname{Im}\sigma is a TT-subspace of S2S_{2} (Definition 2.2.1).

Proof.

By Definition 2.3.1, aS1\forall a\in{{S}_{1}} and fTf\in T, f(σ(a))=σ(f(a))Imσf(\sigma(a))=\sigma(f(a))\in\operatorname{Im}\sigma, and thus σ(a)TImσ{{\left\langle\sigma(a)\right\rangle}_{T}}\subseteq\operatorname{Im}\sigma. Hence ImσTImσ{{\left\langle\operatorname{Im}\sigma\right\rangle}_{T}}\subseteq\operatorname{Im}\sigma, and so by Definition 2.2.8, Imσ\operatorname{Im}\sigma is a quasi-TT-space.

Moreover, if IdT\operatorname{Id}\in T or ImσImσT\operatorname{Im}\sigma\subseteq{{\left\langle\operatorname{Im}\sigma\right\rangle}_{T}}, then by Proposition 2.2.10, Imσ\operatorname{Im}\sigma is a TT-space, and thus ImσS2\operatorname{Im}\sigma\leq{{S}_{2}}. ∎

Definition 2.3.7.

Let SS be a TT-space. We call a TT-morphism from SS to SS a TT-endomorphism of SS, and we denote by EndT(S){}_{T}(S) the collection of all TT-endomorphisms of SS.

Proposition 2.3.8.

Let SS be a TT-space. Then EndT(S)\operatorname{End}_{T}(S) constitutes a monoid, which we still denote by EndT(S)\operatorname{End}_{T}(S), with composition of functions as the binary operation.

Remark.

We regard the empty function \emptyset\to\emptyset as the identity function Id{}_{\emptyset}. So EndT()={Id}\operatorname{End}_{T}(\emptyset)=\{\text{Id}_{\emptyset}\}.

Proof.

The identity map on SS lies in EndT(S)\operatorname{End}_{T}(S), and so we only need to show that EndT(S)\operatorname{End}_{T}(S) is a semigroup with composition of functions as the binary operation.

Let σ1,σ2EndT(S){{\sigma}_{1}},{{\sigma}_{2}}\in\operatorname{End}_{T}(S), let fTf\in T and let aSa\in S. Then

σ1σ2(f(a))=σ1(f(σ2(a)))=f(σ1(σ2(a))).{{\sigma}_{1}}\circ{{\sigma}_{2}}(f(a))={{\sigma}_{1}}(f({{\sigma}_{2}}(a)))=f({{\sigma}_{1}}({{\sigma}_{2}}(a))).

So σ1σ2{{\sigma}_{1}}\circ{{\sigma}_{2}} commutes with ff on SS, and hence σ1σ2EndT(S){{\sigma}_{1}}\circ{{\sigma}_{2}}\in\operatorname{End}_{T}(S).

Composition of functions is associative. Therefore, EndT(S)\operatorname{End}_{T}(S) is a monoid with composition of functions as the binary operation. ∎

Then the following is obvious.

Proposition 2.3.9.

Let SS be a TT-space. Then the intersection of any family of submonoids of EndT(S)\operatorname{End}_{T}(S) is again a submonoid of EndT(S)\operatorname{End}_{T}(S).

2.4. TT-isomorphisms

Definition 2.4.1.

Let SS, S1{{S}_{1}} and S2{{S}_{2}} be TT-spaces.

Let σ\sigma be a TT-morphism from S1{{S}_{1}} to S2{{S}_{2}}. If σ\sigma is bijective, then we call σ\sigma a TT-isomorphism from S1{{S}_{1}} to S2{{S}_{2}}. We denote by IsoT(S1,S2){}_{T}({{S}_{1}},{{S}_{2}}) the family of all TT-isomorphisms from S1{{S}_{1}} to S2{{S}_{2}}.

Moreover, if σ\sigma is a TT-isomorphism from SS to itself, then we call σ\sigma a TT-automorphism of SS. We write AutT(S){}_{T}(S) for the collection of all TT-automorphisms of SS.

To justify Definition 2.4.1, we have

Proposition 2.4.2.

Let S1{{S}_{1}} and S2{{S}_{2}} be TT-spaces, let σIsoT(S1,S2)\sigma\in\operatorname{Iso}_{T}({{S}_{1}},{{S}_{2}}) and let σ1{{\sigma}^{-1}} be the inverse map of σ\sigma. Then σ1IsoT(S2,S1){{\sigma}^{-1}}\in\operatorname{Iso}_{T}({{S}_{2}},{{S}_{1}}).

Proof.

By Definition 2.3.1, fT\forall f\in T and aS1a\in{{S}_{1}}, σ(f(a))=f(σ(a))\sigma(f(a))=f(\sigma(a)), and so f(a)=σ1(f(σ(a)))f(a)={{\sigma}^{-1}}(f(\sigma(a))). Then fT\forall f\in T and bS2b\in{{S}_{2}}, f(σ1(b))=σ1(f(b))f({{\sigma}^{-1}}(b))={{\sigma}^{-1}}(f(b)). Hence σ1{\sigma}^{-1} is a TT-morphism from S2S_{2} to S1S_{1}, and so σ1IsoT(S2,S1){{\sigma}^{-1}}\in\operatorname{Iso}_{T}({{S}_{2}},{{S}_{1}}). ∎

Proposition 2.4.3.

Let SS be a TT-space. Then AutT(S)\operatorname{Aut}_{T}(S) constitutes a group, which we still denote by AutT(S)\operatorname{Aut}_{T}(S), with composition of functions as the binary operation.

Proof.

Let σAutT(S)\sigma\in\operatorname{Aut}_{T}(S) and let σ1{{\sigma}^{-1}} be the inverse map of σ\sigma. By Proposition 2.4.2, σ1AutT(S){{\sigma}^{-1}}\in\operatorname{Aut}_{T}(S).

Let σ1,σ2AutT(S){{\sigma}_{1}},{{\sigma}_{2}}\in\operatorname{Aut}_{T}(S). Then by Proposition 2.3.8, σ1σ2EndT(S){{\sigma}_{1}}\circ{{\sigma}_{2}}\in\operatorname{End}_{T}(S). Hence σ1σ2AutT(S){{\sigma}_{1}}\circ{{\sigma}_{2}}\in\operatorname{Aut}_{T}(S) because both σ1{{\sigma}_{1}} and σ2{{\sigma}_{2}} are bijective.

Moreover, composition of functions is associative and the identity map on SS lies in AutT(S)\operatorname{Aut}_{T}(S). Therefore, AutT(S)\operatorname{Aut}_{T}(S) is a group with composition of functions as the binary operation. ∎

Then the following, which is for later use, is obvious.

Proposition 2.4.4.

Let SS be a TT-space. Then the intersection of any family of subgroups of AutT(S)\operatorname{Aut}_{T}(S) is again a subgroup of AutT(S)\operatorname{Aut}_{T}(S).

2.5. Galois TT-extensions

The remaining part of Section 2 will be used later to study e.g. Galois correspondences. The concept of Galois extensions in the classical Galois theory is generalized as follows.

Definition 2.5.1.

Let SS be a TT-space and let HEndT(S)H\subseteq\operatorname{End}_{T}(S). Let

SH={aS|σH,σ(a)=a}.{{S}^{H}}=\{a\in S\,|\,\forall\sigma\in H,\sigma(a)=a\}.

We call SS a Galois TT-extension of SH{{S}^{H}}, and we say that SH{{S}^{H}} is fixed ((pointwisely)) under the action of HH on SS.

More generally, let KSK\subseteq S. Then we denote by KH{{K}^{H}} the set {aK|σH,σ(a)=a}\{a\in K\,|\,\forall\sigma\in H,\sigma(a)=a\}.

In particular, if H={σ}H=\{\sigma\}, then we use Kσ{{K}^{\sigma}} instead of K{σ}{{K}^{\{\sigma\}}} for brevity.

Remark.

K=K,KS{{K}^{\emptyset}}=K,\forall K\subseteq S.

Proposition 2.5.2.

Let SS be a TT-space and let HH be a subset of EndT(S)\operatorname{End}_{T}(S). Then SHqS{S^{H}}\leq_{q}S. Moreover, if IdT\operatorname{Id}\in T or SHSHT{{S}^{H}}\subseteq{{\left\langle{{S}^{H}}\right\rangle}_{T}}, then SHS{{S}^{H}}\leq S.

Proof.

Let aSHa\in{{S}^{H}} and fTf\in T. Then σH\forall\sigma\in H, σ(f(a))=f(σ(a))=f(a)\sigma(f(a))=f(\sigma(a))=f(a) (by Definition 2.5.1), and so f(a)SHf(a)\in{{S}^{H}}. Thus aTSH{{\left\langle a\right\rangle}_{T}}\subseteq{{S}^{H}}, and so SHTSH{{\left\langle{{S}^{H}}\right\rangle}_{T}}\subseteq{{S}^{H}}. Hence SH{{S}^{H}} is a quasi-TT-space by Definition 2.2.8, and thus SHqS{S^{H}}\leq_{q}S.

Moreover, if IdT\operatorname{Id}\in T or SHSHT{{S}^{H}}\subseteq{{\left\langle{{S}^{H}}\right\rangle}_{T}}, then by Proposition 2.2.10, SH{{S}^{H}} is a TT-space, and hence SHS{{S}^{H}}\leq S. ∎

However, in Proposition 2.5.2, if SHSHT{{S}^{H}}\nsubseteq{{\left\langle{{S}^{H}}\right\rangle}_{T}}, then it is possible that SH{{S}^{H}} is not a TT-space. We shall prove this claim right after Example 7.1.10.

The following is obvious.

Proposition 2.5.3.

Let SS be a TT-space and let HGEndT(S)H\subseteq G\subseteq\operatorname{End}_{T}(S). Then SHSG{{S}^{H}}\supseteq{{S}^{G}}.

2.6. Galois TT-monoids and Galois TT-groups

The concept of Galois group in the classical Galois theory is generalized to two notions as follows.

Definition 2.6.1.

Let SS be a TT-space and let BSB\subseteq S.

The Galois TT-monoid of SS over BB, denoted by GMnT(S/B)\operatorname{GMn}_{T}(S/B), is the set {σEndT(S)|bB,σ(b)=b}\{\sigma\in\operatorname{End}_{T}(S)\,|\,\forall b\in B,\sigma(b)=b\}.

The Galois TT-group of SS over BB, denoted by GGrT(S/B)\operatorname{GGr}_{T}(S/B), is the set {σAutT(S)|bB,σ(b)=b}\{\sigma\in\operatorname{Aut}_{T}(S)\,|\,\forall b\in B,\sigma(b)=b\}.

Remark.

Trivially, GMnT(/)=GGrT(/)={Id}\operatorname{GMn}_{T}(\emptyset/\emptyset)=\operatorname{GGr}_{T}(\emptyset/\emptyset)=\{\text{I}{{\text{d}}_{\emptyset}}\} because we regard the empty function \emptyset\to\emptyset as the identity function Id\text{I}{{\text{d}}_{\emptyset}}.

To justify Definition 2.6.1, the following two results are needed, of which the first one generalizes Proposition 2.3.8.

Proposition 2.6.2.

Let SS be a TT-space and let BSB\subseteq S. Then GMnT(S/B)\operatorname{GMn}_{T}(S/B) is a submonoid of EndT(S)\operatorname{End}_{T}(S) with composition of functions as the binary operation.

Proof.

Because IdGMnT(S/B)EndT(S)\operatorname{Id}\in\operatorname{GMn}_{T}(S/B)\subseteq\operatorname{End}_{T}(S), by Proposition 2.3.8, we only need to show that GMnT(S/B)\operatorname{GMn}_{T}(S/B) is closed under composition of functions.

Let σ1,σ2GMnT(S/B){{\sigma}_{1}},{{\sigma}_{2}}\in\operatorname{GMn}_{T}(S/B). By Definition 2.6.1, bB\forall b\in B, σ1(σ2(b))=σ1(b)=b{{\sigma}_{1}}({{\sigma}_{2}}(b))={{\sigma}_{1}}(b)=b. Hence σ1σ2GMnT(S/B){{\sigma}_{1}}\circ{{\sigma}_{2}}\in\operatorname{GMn}_{T}(S/B) (since σ1σ2EndT(S){{\sigma}_{1}}\circ{{\sigma}_{2}}\in\operatorname{End}_{T}(S) by Proposition 2.3.8), as desired. ∎

Analogously, Proposition 2.4.3 is generalized as follows, where the proof is omitted.

Proposition 2.6.3.

Let SS be a TT-space and let BSB\subseteq S. Then GGrT(S/B)\operatorname{GGr}_{T}(S/B) is a subgroup of AutT(S)\operatorname{Aut}_{T}(S) with composition of functions as the binary operation.

We will employ the following three propositions later, of which the proofs are obvious.

Proposition 2.6.4.

Let SS be a TT-space. Then BS,\forall B\subseteq S, BSGMnT(S/B)SGGrT(S/B)B\subseteq{{S}^{\operatorname{GMn}_{T}(S/B)}}\subseteq{{S}^{\operatorname{GGr}_{T}(S/B)}}.

Proposition 2.6.5.

Let SS be a TT-space. Then HEndT(S)\forall H\subseteq\operatorname{End}_{T}(S), HGMnT(S/SH)H\subseteq\operatorname{GMn}_{T}(S/{{S}^{H}}), and HAutT(S)\forall H\subseteq\operatorname{Aut}_{T}(S), HGGrT(S/SH)H\subseteq\operatorname{GGr}_{T}(S/{{S}^{H}}).

Proposition 2.6.6.

Let SS be a TT-space. Then BKS\forall B\subseteq K\subseteq S, GMnT(S/B)GMnT(S/K)\operatorname{GMn}_{T}(S/B)\supseteq\operatorname{GMn}_{T}(S/K) and GGrT(S/B)GGrT(S/K)\operatorname{GGr}_{T}(S/B)\supseteq\operatorname{GGr}_{T}(S/K).

2.7. Generated submonoids of EndT(S)\operatorname{End}_{T}(S) and subgroups of AutT(S)\operatorname{Aut}_{T}(S)

We will use the following notions and results later.

Definition 2.7.1.

Let SS be a TT-space.

If XEndT(S)X\subseteq\operatorname{End}_{T}(S), then the intersection of all submonoids of EndT(S)\operatorname{End}_{T}(S) containing XX, denoted by XEnd(S){{\left\langle X\right\rangle}_{\operatorname{End}(S)}}, is called the submonoid of EndT(S)\operatorname{End}_{T}(S) generated by X.

If XAutT(S)X\subseteq\operatorname{Aut}_{T}(S), then the intersection of all subgroups of AutT(S)\operatorname{Aut}_{T}(S) containing XX, denoted by XAut(S){{\left\langle X\right\rangle}_{\operatorname{Aut}(S)}}, is called the subgroup of AutT(S)\operatorname{Aut}_{T}(S) generated by XX.

Remark.

By Proposition 2.3.9, XEnd(S){{\left\langle X\right\rangle}_{\operatorname{End}(S)}} is the smallest submonoid of EndT(S)\operatorname{End}_{T}(S) which contains XX. And by Proposition 2.4.4, XAut(S){{\left\langle X\right\rangle}_{\operatorname{Aut}(S)}} is the smallest subgroup of AutT(S)\operatorname{Aut}_{T}(S) which contains XX.

Proposition 2.7.2.

Let SS be a TT-space and let XAutT(S)X\subseteq\operatorname{Aut}_{T}(S). Suppose that xX\forall x\in X, x1X{{x}^{-1}}\in X. Then XAut(S)=XEnd(S){{\left\langle X\right\rangle}_{\operatorname{Aut}(S)}}={{\left\langle X\right\rangle}_{\operatorname{End}(S)}}.

Proof.

Since XAutT(S)X\subseteq\operatorname{Aut}_{T}(S) and every subgroup of AutT(S)\operatorname{Aut}_{T}(S) is a submonoid of EndT(S)\operatorname{End}_{T}(S), the set of all subgroups of AutT(S)\operatorname{Aut}_{T}(S) containing XX is a subset of the set of all submonoids of EndT(S)\operatorname{End}_{T}(S) containing XX. Then by Definition 2.7.1 we can tell that XAut(S)XEnd(S){{\left\langle X\right\rangle}_{\operatorname{Aut}(S)}}\supseteq{{\left\langle X\right\rangle}_{\operatorname{End}(S)}}. So it is sufficient to prove that XAut(S)XEnd(S){{\left\langle X\right\rangle}_{\operatorname{Aut}(S)}}\subseteq{{\left\langle X\right\rangle}_{\operatorname{End}(S)}}.

If XX is empty, then XAut(S)={{{\left\langle X\right\rangle}_{\operatorname{Aut}(S)}}=\{Id on S}=XEnd(S)S\}={{\left\langle X\right\rangle}_{\operatorname{End}(S)}}, as desired.

Suppose that XX is not empty. Let σXAut(S)\sigma\in{{\left\langle X\right\rangle}_{\operatorname{Aut}(S)}}. Since XAut(S){\left\langle X\right\rangle}_{\operatorname{Aut}(S)} is the group generated by XX, n+\exists n\in\mathbb{Z}^{+}, x1,,xnXx_{1},\cdots,x_{n}\in X and e1,,ene_{1},\cdots,e_{n}\in\mathbb{Z} such that σ=x1e1xnen\sigma=x_{1}^{{{e}_{1}}}\cdots x_{n}^{{{e}_{n}}} (which is a product of group elements). Because it is assumed that xX\forall x\in X, x1X{{x}^{-1}}\in X, we can tell that any submonoid of EndT(S)\operatorname{End}_{T}(S) containing XX also contains x1e1xnen=σx_{1}^{{{e}_{1}}}\cdots x_{n}^{{{e}_{n}}}=\sigma. Hence XAut(S)XEnd(S){{\left\langle X\right\rangle}_{\operatorname{Aut}(S)}}\subseteq{{\left\langle X\right\rangle}_{\operatorname{End}(S)}}, as desired. ∎

3. Galois correspondences

In this section, we shall develop two main results on Galois correspondences. Roughly speaking, the first one shows the Galois correspondence between the Galois TT-monoids (Definition 2.6.1) of a TT-space SS and the fixed subsets of SS under actions of TT-endomorphisms of SS, and the second one shows the Galois correspondence between the Galois TT-groups (Definition 2.6.1) of a TT-space SS and the fixed subsets of SS under actions of TT-automorphisms of SS.

For a TT-space SS and HEndT(S)H\subseteq\operatorname{End}_{T}(S), by Proposition 2.5.2, SHS^{H} is a quasi-TT-subspace of SS, rather than a TT-subspace of SS unless IdT\operatorname{Id}\in T or SHSHT{{S}^{H}}\subseteq{{\left\langle{{S}^{H}}\right\rangle}_{T}}. This fact implies that, quasi-TT-subspaces, rather than TT-subspaces, of a TT-space may play a major role in establishing Galois correspondences, though Proposition 2.2.10 tells us that in the case such as IdT\operatorname{Id}\in T, quasi-TT-space and TT-space are actually the same notion.

Notation 3.0.1.

Let SS be a TT-space and let BSB\subseteq S. We denote by SubT(S)\operatorname{Sub}_{T}(S) the set of all quasi-TT-subspaces of SS, and we denote by IntT(S/B)\operatorname{Int}_{T}(S/B) the set of all intermediate quasi-TT-spaces between BB and SS, i.e. the set of all quasi-TT-spaces KK with SKBS\supseteq K\supseteq B.

Remark.
  1. (1)

    By Proposition 2.2.10, both SubT(S)\operatorname{Sub}_{T}(S) and IntT(S/B)\operatorname{Int}_{T}(S/B) are sets of TT-spaces if IdT\operatorname{Id}\in T.

  2. (2)

    SubT()=IntT(/)={}\operatorname{Sub}_{T}(\emptyset)=\operatorname{Int}_{T}(\emptyset/\emptyset)=\{\emptyset\}.

Notation 3.0.2.

We denote by SGr(G)\operatorname{SGr}(G) the set of all subgroups of a group GG, and we denote by SMn(M)\operatorname{SMn}(M) the set of all submonoids of a monoid MM.

3.1. Two examples

Let SS be a TT-space and let GG be a subset of AutT(S)\operatorname{Aut}_{T}(S). Then by Definition 2.5.1, SS is a Galois TT-extension of SG{S^{G}}. Let’s check whether there always exists a correspondence between SGr(GGrT(S/SG))\operatorname{SGr}(\operatorname{GGr}_{T}(S/{{S}^{G}})) and IntT(S/SG)\operatorname{Int}_{T}(S/{{S}^{G}}) as in the classical Galois field theory (see Definition 2.6.1 for the notation GGrT(S/SG)\operatorname{GGr}_{T}(S/{{S}^{G}})).

Example 3.1.1.

Let T={f:T=\{f:\mathbb{R}\to\mathbb{R} given by xx+a|a}x\mapsto x+a\,|\,a\in\mathbb{R}\}. Then TT is an operator semigroup on \mathbb{R}, IdT\operatorname{Id}\in T and r,rT=\forall r\in\mathbb{R},{{\left\langle r\right\rangle}_{T}}=\mathbb{R}. Therefore, \mathbb{R} is a TT-space and the only (quasi-)TT-subspaces of \mathbb{R} are \emptyset and \mathbb{R}.

Let σEndT()\sigma\in\operatorname{End}_{T}(\mathbb{R}). Then fT\forall f\in T, σ\sigma commutes with ff on \mathbb{R}. Thus for aa\in\mathbb{R} and f:f:\mathbb{R}\to\mathbb{R} given by xx+ax\mapsto x+a,

σ(x)+a=f(σ(x))=σ(f(x))=σ(x+a),x,\sigma(x)+a=f(\sigma(x))=\sigma(f(x))=\sigma(x+a),\forall x\in\mathbb{R},

which implies that σ\sigma is a translation along the real axis. Hence σT\sigma\in T and EndTT\operatorname{End}_{T}\mathbb{R}\subseteq T. On the other hand, we can tell TAutT()EndT()T\subseteq\operatorname{Aut}_{T}(\mathbb{R})\subseteq\operatorname{End}_{T}(\mathbb{R}). Thus, EndT()=AutT()=T\operatorname{End}_{T}(\mathbb{R})=\operatorname{Aut}_{T}(\mathbb{R})=T. Then \mathbb{R} is a Galois TT-extension of T={{\mathbb{R}}^{T}}=\emptyset and IntT(/T)={,}\operatorname{Int}_{T}(\mathbb{R}/{{\mathbb{R}}^{T}})=\{\emptyset,\mathbb{R}\} (since the only (quasi-)TT-subspaces of \mathbb{R} are \emptyset and \mathbb{R}).

However, GGrT(/T)(=GGrT(/)=AutT()=T)\operatorname{GGr}_{T}(\mathbb{R}/{{\mathbb{R}}^{T}})(=\operatorname{GGr}_{T}(\mathbb{R}/\emptyset)=\operatorname{Aut}_{T}(\mathbb{R})=T) has an infinite number of subgroups, so

|IntT(/T)|<|SGr(GGrT(/T))|.\left|\operatorname{Int}_{T}(\mathbb{R}/{{\mathbb{R}}^{T}})\right|<\left|\operatorname{SGr}(\operatorname{GGr}_{T}(\mathbb{R}/{{\mathbb{R}}^{T}}))\right|.

The preceding example shows that the number of all intermediate (quasi-)TT-spaces of a Galois TT-extension may be strictly smaller than that of all subgroups of the corresponding Galois TT-group. The converse is also possible, as shown below.

Example 3.1.2.

Let T={T=\{Id}\}. Then TT is an operator semigroup on any set DD and any subset of DD is a TT-space. Obviously any map from DD to DD belongs to EndT(D)\operatorname{End}_{T}(D), and so any bijective map from DD to DD lies in AutT(D)\operatorname{Aut}_{T}(D).

Suppose |D|=2\left|D\right|=2. Then AutT(D)=S2\operatorname{Aut}_{T}(D)={{S}_{2}}, which is the symmetric group on 22 letters. Then DD is a Galois TT-extension of DS2={D^{S_{2}}}=\emptyset, IntT(D/DS2)\operatorname{Int}_{T}(D/{{D}^{S_{2}}}) is the power set of DD, and

SGr(GGrT(D/DS2))=SGr(AutT(D))=SGr(S2).\operatorname{SGr}(\operatorname{GGr}_{T}(D/{{D}^{S_{2}}}))=\operatorname{SGr}(\operatorname{Aut}_{T}(D))=\operatorname{SGr}({{S}_{2}}).

Then

|IntT(D/DS2)|=4>2=|SGr(GGrT(D/DS2))|.\left|\operatorname{Int}_{T}(D/{{D}^{S_{2}}})\right|=4>2=\left|\operatorname{SGr}(\operatorname{GGr}_{T}(D/{{D}^{S_{2}}}))\right|.

Example 3.1.2 shows that the number of all intermediate (quasi-)TT-spaces of a Galois TT-extension may be strictly larger than that of all subgroups of the corresponding Galois TT-group.

However, if we focus on the fixed subsets of a TT-space SS under actions of TT-automorphisms of SS and on the subgroups (of AutT(S)\operatorname{Aut}_{T}(S)) which are Galois TT-groups, then we will see a Galois correspondence. Moreover, we shall find a Galois correspondence for Galois TT-monoids as well.

3.2. Galois correspondences

First, we need to define the fixed subsets of a TT-space SS under actions of TT-endomorphisms (or TT-automorphisms) of SS as follows.

Definition 3.2.1.

Let SS be a TT-space and let BSB\subseteq S. Then

IntTEnd(S/B):={SH|BSH,HEndT(S)},\operatorname{Int}_{T}^{\operatorname{End}}(S/B):=\{{{S}^{H}}\,|\,B\subseteq{{S}^{H}},H\subseteq\operatorname{End}_{T}(S)\},

and

IntTAut(S/B):={SH|BSH,HAutT(S)}.\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B):=\{{{S}^{H}}\,|\,B\subseteq{{S}^{H}},H\subseteq\operatorname{Aut}_{T}(S)\}.
Remark.
  1. (1)

    KIntTEnd(S/B)\forall K\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B), SS is a Galois TT-extension of KK (by Definition 2.5.1).

  2. (2)

    Since AutT(S)EndT(S)\operatorname{Aut}_{T}(S)\subseteq\operatorname{End}_{T}(S), IntTAut(S/B)IntTEnd(S/B)IntT(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)\subseteq\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\subseteq\operatorname{Int}_{T}(S/B) (by Proposition 2.5.2).

  3. (3)

    IntTAut(/)=IntTEnd(/)={}\operatorname{Int}_{T}^{\operatorname{Aut}}(\emptyset/\emptyset)=\operatorname{Int}_{T}^{\operatorname{End}}(\emptyset/\emptyset)=\{\emptyset\}.

The following characterizes IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) and IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B).

Lemma 3.2.2.

Let SS be a TT-space and let BKSB\subseteq K\subseteq S. Then

KIntTEnd(S/B)K=SGMnT(S/K)K\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\Leftrightarrow K={{S}^{\operatorname{GMn}_{T}(S/K)}}

and

KIntTAut(S/B)K=SGGrT(S/K).K\in\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)\Leftrightarrow K={{S}^{\operatorname{GGr}_{T}(S/K)}}.
Proof.

1. Sufficiency

If K=SGMnT(S/K)K={{S}^{\operatorname{GMn}_{T}(S/K)}}, then KIntTEnd(S/B)K\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B) by Definition 3.2.1.

If K=SGGrT(S/K)K={{S}^{\operatorname{GGr}_{T}(S/K)}}, then KIntTAut(S/B)K\in\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) by Definition 3.2.1.

2. Necessity

Suppose KIntTEnd(S/B)K\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B). Then HEndT(S)\exists H\subseteq\operatorname{End}_{T}(S) such that K=SHK={{S}^{H}}, and so by Proposition 2.6.5, HGMnT(S/SH)=GMnT(S/K)H\subseteq\operatorname{GMn}_{T}(S/{{S}^{H}})=\operatorname{GMn}_{T}(S/K). Hence K=SHSGMnT(S/K)K={{S}^{H}}\supseteq{{S}^{\operatorname{GMn}_{T}(S/K)}} by Proposition 2.5.3.

On the other hand, KSGMnT(S/K)K\subseteq{{S}^{\operatorname{GMn}_{T}(S/K)}} by Proposition 2.6.4.

Therefore, K=SGMnT(S/K)K={{S}^{\operatorname{GMn}_{T}(S/K)}}.

If KIntTAut(S/B)K\in\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B), analogously we can show K=SGGrT(S/K)K={{S}^{\operatorname{GGr}_{T}(S/K)}}. ∎

Second, we define

Definition 3.2.3.

Let SS be a TT-space and let BSB\subseteq S. Then

GSMnT(S/B):={GMnT(S/K)|BKS},\operatorname{GSMn}_{T}(S/B):=\{\operatorname{GMn}_{T}(S/K)\,|\,B\subseteq K\subseteq S\},

and

GSGrT(S/B):={GGrT(S/K)|BKS}.\operatorname{GSGr}_{T}(S/B):=\{\operatorname{GGr}_{T}(S/K)\,|\,B\subseteq K\subseteq S\}.
Remark.
  1. (1)

    GSMn stands for “Galois submonoids” and GSGr stands for “Galois subgroups”.

  2. (2)

    Clearly,

    GSMnT(S/B)SMn(GMnT(S/B))\operatorname{GSMn}_{T}(S/B)\subseteq\operatorname{SMn}(\operatorname{GMn}_{T}(S/B))

    (by Notation 3.0.2 and Proposition 2.6.2),

    GSGrT(S/B)SGr(GGrT(S/B))\operatorname{GSGr}_{T}(S/B)\subseteq\operatorname{SGr}(\operatorname{GGr}_{T}(S/B))

    (by Notation 3.0.2 and Proposition 2.6.3),

    GSGrT(S/B)GSMnT(S/B),\operatorname{GSGr}_{T}(S/B)\subseteq\operatorname{GSMn}_{T}(S/B),

    and

    GSGrT(/)=GSMnT(/)={{Id}}.\operatorname{GSGr}_{T}(\emptyset/\emptyset)=\operatorname{GSMn}_{T}(\emptyset/\emptyset)=\{\{\text{I}{{\text{d}}_{\emptyset}}\}\}.

The following characterizes GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) and GSGrT(S/B)\operatorname{GSGr}_{T}(S/B).

Lemma 3.2.4.

Let SS be a TT-space, let HEndT(S)H\subseteq\operatorname{End}_{T}(S), and let BSHSB\subseteq{{S}^{H}}\subseteq S. Then

HGSMnT(S/B)GMnT(S/SH)=HH\in\operatorname{GSMn}_{T}(S/B)\Leftrightarrow\operatorname{GMn}_{T}(S/{{S}^{H}})=H

and

HGSGrT(S/B)GGrT(S/SH)=H.H\in\operatorname{GSGr}_{T}(S/B)\Leftrightarrow\operatorname{GGr}_{T}(S/{{S}^{H}})=H.
Proof.

1. Sufficiency

If GMnT(S/SH)=H\operatorname{GMn}_{T}(S/{{S}^{H}})=H, then HGSMnT(S/B)H\in\operatorname{GSMn}_{T}(S/B) by Definition 3.2.3.

If GGrT(S/SH)=H\operatorname{GGr}_{T}(S/{{S}^{H}})=H, then HGSGrT(S/B)H\in\operatorname{GSGr}_{T}(S/B) by Definition 3.2.3.

2. Necessity

Suppose HGSMnT(S/B)H\in\operatorname{GSMn}_{T}(S/B). Then K\exists K such that BKSB\subseteq K\subseteq S and H=GMnT(S/K)H=\operatorname{GMn}_{T}(S/K), and thus by Proposition 2.6.4, KSGMnT(S/K)=SHK\subseteq{{S}^{\operatorname{GMn}_{T}(S/K)}}={{S}^{H}}. Hence H=GMnT(S/K)GMnT(S/SH)H=\operatorname{GMn}_{T}(S/K)\supseteq\operatorname{GMn}_{T}(S/{{S}^{H}}) by Proposition 2.6.6.

On the other hand, HGMnT(S/SH)H\subseteq\operatorname{GMn}_{T}(S/{{S}^{H}}) by Proposition 2.6.5.

Therefore, H=GMnT(S/SH)H=\operatorname{GMn}_{T}(S/{{S}^{H}}).

Analogously, if HGSGrT(S/B)H\in\operatorname{GSGr}_{T}(S/B), we can show H=GGrT(S/SH)H=\operatorname{GGr}_{T}(S/{{S}^{H}}). ∎

Now two main results of this section follow. Roughly speaking, the first one shows the Galois correspondence between the Galois TT-monoids of a TT-space SS and the fixed subsets of SS under actions of TT-endomorphisms of SS.

Corollary 3.2.5.

Let SS be a TT-space and let BSB\subseteq S. Then the correspondences

γ:HSH\gamma:H\mapsto{{S}^{H}} and δ:KGMnT(S/K)\delta:K\mapsto\operatorname{GMn}_{T}(S/K)

define inclusion-inverting mutually inverse bijective maps ((i.e. γ\gamma is an inclusion-reversing bijective map whose inverse δ\delta is also an inclusion-reversing bijective map)) between GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) and IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B).

Proof.

γ\gamma is inclusion-reversing by Proposition 2.5.3, and δ\delta is inclusion-reversing by Proposition 2.6.6.

Lemmas 3.2.2 and 3.2.4 imply that γδ\gamma\circ\delta and δγ\delta\circ\gamma are identity functions, respectively. Hence both δ\delta and γ\gamma are bijective maps and δ1=γ{{\delta}^{-1}}=\gamma. ∎

Analogously, we have the following, which, roughly speaking, shows the Galois correspondence between the Galois TT-groups of a TT-space SS and the fixed subsets of SS under actions of TT-automorphisms of SS.

Corollary 3.2.6.

Let SS be a TT-space and let BSB\subseteq S. Then the correspondences

γ:HSH\gamma:H\mapsto{{S}^{H}} and δ:KGGrT(S/K)\delta:K\mapsto\operatorname{GGr}_{T}(S/K)

define inclusion-inverting mutually inverse bijective maps between GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) and IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B).

Proof.

The same as that of Corollary 3.2.5. ∎

As a matter of fact, Corollaries 3.2.5 and 3.2.6 tell us, essentially, what Galois correspondences are: They tell us where to find Galois correspondences in any specific case. In fact, with examples we gave (e.g. Proposition 2.3.5) or will give for TT-morphisms, we can tell from Corollaries 3.2.5 and 3.2.6 that Galois correspondences are ubiquitous.

In Section 11, we shall prove that Corollaries 3.2.5 and 3.2.6 still hold for the case where elements of TT are allowed to be functions in more than one variable and/or even partial.

As is well known, in the Galois theory for infinite algebraic field extensions, Krull topology is put on Galois groups, and in differential Galois theory, differential Galois groups are endowed with Zariski topology. Then the fundamental theorem in each of the two theories tells us that there exists a Galois correspondence between the set of all closed subgroups of the Galois group of a Galois field extension (or a Picard-Vessiot extension) and the set of all intermediate (differential) fields.

Now for Corollary 3.2.5, can we define a topology on SS and a topology on EndT(S)\operatorname{End}_{T}(S) such that IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) is the set of all closed quasi-TT-subspaces (of SS) containing BB and GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) is the set of all closed submonoids of GMnT(S/B)\operatorname{GMn}_{T}(S/B)? If such topologies exist, then we can characterize the Galois correspondence between IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) and GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) in terms of closed quasi-TT-subspaces of SS and closed submonoids of GMnT(S/B)\operatorname{GMn}_{T}(S/B).

And for Corollary 3.2.6, can we define a topology on SS and a topology on AutT(S)\operatorname{Aut}_{T}(S) with IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) being the set of all closed quasi-TT-subspaces (of SS) containing BB and GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) being the set of all closed subgroups of GGrT(S/B)\operatorname{GGr}_{T}(S/B)? If such topologies exist, then we can characterize the Galois correspondence between IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) and GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) in terms of closed quasi-TT-subspaces of SS and closed subgroups of GGrT(S/B)\operatorname{GGr}_{T}(S/B).

Before we can answer the above questions in Section 5, we need to address a related issue in Section 4, which is about the lattice structures of the sets involved in Corollaries 3.2.5 and 3.2.6.

There is one more thing which we have not talked about for Galois correspondences. It is normally the last part of the fundamental theorem of a Galois theory, i.e. the correspondence between the normal subgroups of a Galois group and the Galois extensions of the base field. We will address this issue in Section 20.

4. Lattice structures of objects arising in Galois correspondences on a TT-space SS

In this section, we shall study lattice structures of objects which arise in Corollaries 3.2.5 and 3.2.6. The value of this section, though being a preparation for the next section, asserts itself.

Recall that a lattice is a (nonempty) partially ordered set LL in which every pair of elements a,bLa,b\in L has a meet aba\wedge b (greatest lower bound) and a join aba\vee b (least upper bound), and LL is a complete lattice if any subset of it has a greatest lower bound and a least upper bound. Note that if lattice LL is complete, then =L\wedge\emptyset=\vee L and =L\vee\emptyset=\wedge L (see e.g. [2]).

Notation 4.0.1.

To abbreviate our notations, by {Ai}\{{{A}_{i}}\}, Ai\bigcup{{{A}_{i}}} and Ai\bigcap{{{A}_{i}}} we mean {Ai|iI}\{{{A}_{i}}|i\in I\}, iIAi\bigcup\nolimits_{i\in I}{{{A}_{i}}} and iIAi\bigcap\nolimits_{i\in I}{{{A}_{i}}}, respectively.

4.1. SubT(S)\operatorname{Sub}_{T}(S), SMn(EndT(S))\operatorname{SMn}(\operatorname{End}_{T}(S)) and SGr(AutT(S))\operatorname{SGr}(\operatorname{Aut}_{T}(S))

Proposition 4.1.1.

Let SS be a TT-space. Then SubT(S)\operatorname{Sub}_{T}(S) ((Notation 3.0.1)) is a lattice with inclusion as the binary relation if S1,S2SubT(S)\forall{{S}_{1}},{{S}_{2}}\in\operatorname{Sub}_{T}(S), S1S2:=S1S2{{S}_{1}}\wedge{{S}_{2}}:={{S}_{1}}\bigcap{{S}_{2}} and S1S2:=S1S2{{S}_{1}}\vee{{S}_{2}}:={{S}_{1}}\bigcup{{S}_{2}}. Moreover, SubT(S)\operatorname{Sub}_{T}(S) is a complete lattice if let A=Si\wedge A=\bigcap{{{S}_{i}}} and let A=Si\vee A=\bigcup{{{S}_{i}}}, A:={Si}SubT(S)\forall A:=\{{{S}_{i}}\}\subseteq\operatorname{Sub}_{T}(S).

Proof.

Straightforward consequences of Propositions 2.2.12 and 2.2.13. Note that SubT()={}\operatorname{Sub}_{T}(\emptyset)=\{\emptyset\} is a complete lattice. ∎

Proposition 4.1.2.

Let SS be a TT-space. Then SMn(EndT(S))\operatorname{SMn}(\operatorname{End}_{T}(S)) ((Notation 3.0.2)) is a lattice with inclusion as the binary relation if M1M2:=M1M2{{M}_{1}}\wedge{{M}_{2}}:={{M}_{1}}\bigcap{{M}_{2}} and M1M2:=M1M2End(S){{M}_{1}}\vee{{M}_{2}}:={{\left\langle{{M}_{1}}\bigcup{{M}_{2}}\right\rangle}_{\operatorname{End}(S)}} ((Definition 2.7.1)), M1,M2SMn(EndT(S))\forall{{M}_{1}},{{M}_{2}}\in\operatorname{SMn}(\operatorname{End}_{T}(S)). Moreover, if let A=Mi\wedge A=\bigcap{{{M}_{i}}} and let A=MiEnd(S)\vee A={{\left\langle\bigcup{{{M}_{i}}}\right\rangle}_{\operatorname{End}(S)}}, A:={Mi}SMn(EndT(S))\forall A:=\{{{M}_{i}}\}\subseteq\operatorname{SMn}(\operatorname{End}_{T}(S)), then SMn(EndT(S))\operatorname{SMn}(\operatorname{End}_{T}(S)) is a complete lattice.

Proof.

Obvious (cf. Proposition 2.3.9 and the remark right after Definition 2.7.1). ∎

For SGr(AutT(S))\operatorname{SGr}(\operatorname{Aut}_{T}(S)) (Notation 3.0.2), we also have

Proposition 4.1.3.

Let SS be a TT-space. Then SGr(AutT(S))\operatorname{SGr}(\operatorname{Aut}_{T}(S)) is a lattice with inclusion as the binary relation if G1G2:=G1G2{{G}_{1}}\wedge{{G}_{2}}:={{G}_{1}}\bigcap{{G}_{2}} and G1G2:=G1G2Aut(S){{G}_{1}}\vee{{G}_{2}}:={{\left\langle{{G}_{1}}\bigcup{{G}_{2}}\right\rangle}_{\operatorname{Aut}(S)}} ((Definition 2.7.1)), G1,G2SGr(AutT(S))\forall{{G}_{1}},{{G}_{2}}\in\operatorname{SGr}(\operatorname{Aut}_{T}(S)). Moreover, if let A=Gi\wedge A=\bigcap{{{G}_{i}}} and let A=GiAut(S)\vee A={{\left\langle\bigcup{{{G}_{i}}}\right\rangle}_{\operatorname{Aut}(S)}}, A:={Gi}SGr(AutT(S))\forall A:=\{{{G}_{i}}\}\subseteq\operatorname{SGr}(\operatorname{Aut}_{T}(S)), then SGr(AutT(S))\operatorname{SGr}(\operatorname{Aut}_{T}(S)) is a complete lattice.

Proof.

Obvious by Proposition 2.4.4 and the remark right after Definition 2.7.1. ∎

Moreover, we have

Proposition 4.1.4.

Let SS be a TT-space. Then SGr(AutT(S))\operatorname{SGr}(\operatorname{Aut}_{T}(S)) is a complete sublattice of the complete lattice SMn(EndT(S))\operatorname{SMn}(\operatorname{End}_{T}(S)) defined in Proposition 4.1.2.

Proof.

It suffices to show that for any A:={Gi}SGr(AutT(S))A:=\{{{G}_{i}}\}\subseteq\operatorname{SGr}(\operatorname{Aut}_{T}(S)),

A=GiAut(S)=GiEnd(S).\vee A={{\left\langle\bigcup{{{G}_{i}}}\right\rangle}_{\operatorname{Aut}(S)}}={{\left\langle\bigcup{{{G}_{i}}}\right\rangle}_{\operatorname{End}(S)}}.

From Proposition 2.7.2, we can tell that the equations hold. ∎

4.2. IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) and IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)

We are interested in the two sets because they arise in Corollary 3.2.5 or 3.2.6.

Lemma 4.2.1.

Let SS be a TT-space and let {Hi}\{{{H}_{i}}\} be a set of subsets of EndT(S)\operatorname{End}_{T}(S). Then SHi=SHi\bigcap{{{S}^{{{H}_{i}}}}}={{S}^{\bigcup{{{H}_{i}}}}}.

Proof.
SHi\displaystyle\bigcap{{{S}^{{{H}_{i}}}}} ={aS|σHi,σ(a)=a}\displaystyle=\bigcap{\{a\in S\,|\,\forall\sigma\in{{H}_{i}},\sigma(a)=a\}}
={aS|σHi,σ(a)=a}\displaystyle=\{a\in S\,|\,\forall\sigma\in\bigcup{{{H}_{i}}},\sigma(a)=a\}
=SHi.\displaystyle={{S}^{\bigcup{{{H}_{i}}}}}.

Lemma 4.2.1 implies

Proposition 4.2.2.

IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) is a complete \wedge-sublattice of the complete lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 4.1.1.

Proof.

By Definition 3.2.1, each element of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) has the form SHi(B){{S}^{{{H}_{i}}}}(\supseteq B), where HiEndT(S){{H}_{i}}\subseteq\operatorname{End}_{T}(S). Then by Lemma 4.2.1, SHi=SHiIntTEnd(S/B)\bigcap{{{S}^{{{H}_{i}}}}}={{S}^{\bigcup{{{H}_{i}}}}}\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B).

Moreover, by Proposition 2.5.2, IntTEnd(S/B)SubT(S)\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\subseteq\operatorname{Sub}_{T}(S).

Therefore, IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) is a complete \wedge-sublattice of the complete lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 4.1.1. ∎

However, IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) is not necessarily a \vee-sublattice of the lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 4.1.1. In fact, it is possible that IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) is not even a \vee-semilattice if S1,S2IntTEnd(S/B)\forall{{S}_{1}},{{S}_{2}}\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B), S1S2:=S1S2{{S}_{1}}\vee{{S}_{2}}:={{S}_{1}}\bigcup{{S}_{2}}, as shown in the following example.

Example 4.2.3.

Let TT be the operator semigroup on \mathbb{C} induced by [x]\mathbb{Q}[x] as defined in Example 2.1.2. That is, let

T={f: given by af(a)|f(x)[x]}.T=\{{{f}^{*}}:\mathbb{C}\to\mathbb{C}\text{ given by }a\mapsto f(a)\,|\,f(x)\in\mathbb{Q}[x]\}.

Let a TT-space S=2+3T=(2+3)S={{\left\langle\sqrt{2}+\sqrt{3}\right\rangle}_{T}}=\mathbb{Q}(\sqrt{2}+\sqrt{3}). Then S=(2,3)S=\mathbb{Q}(\sqrt{2},\sqrt{3}) and S/S/\mathbb{Q} is a finite Galois field extension (see e.g. [7]).

By Corollary 2.3.4, a map σ:SS\sigma:S\to S is a TT-morphism if and only if it is a field automorphism with \mathbb{Q} fixed pointwisely, and so by Definition 3.2.1 and the fundamental theorem of the classical Galois theory,

IntTEnd(S/)=IntTAut(S/)={FS|F is a field},\operatorname{Int}_{T}^{\operatorname{End}}(S/\mathbb{Q})=\operatorname{Int}_{T}^{\operatorname{Aut}}(S/\mathbb{Q})=\{\mathbb{Q}\subseteq F\subseteq S\,|\,F\text{ is a field}\},

and it contains both (2)\mathbb{Q}(\sqrt{2}) and (3)\mathbb{Q}(\sqrt{3}).

Let S=(2)(3){S}^{\prime}=\mathbb{Q}(\sqrt{2})\bigcup\mathbb{Q}(\sqrt{3}). Then SIntTEnd(S/){S}^{\prime}\notin\operatorname{Int}_{T}^{\operatorname{End}}(S/\mathbb{Q}) because S{S}^{\prime} is not a field. But S=(2)(3){S}^{\prime}=\mathbb{Q}(\sqrt{2})\vee\mathbb{Q}(\sqrt{3}) as defined in Proposition 4.1.1. Therefore, IntTEnd(S/)\operatorname{Int}_{T}^{\operatorname{End}}(S/\mathbb{Q}) is not a \vee-sublattice of the lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 4.1.1. In fact, IntTEnd(S/)\operatorname{Int}_{T}^{\operatorname{End}}(S/\mathbb{Q}) is not even a \vee-semilattice if S1,S2IntTEnd(S/)\forall{{S}_{1}},{{S}_{2}}\in\operatorname{Int}_{T}^{\operatorname{End}}(S/\mathbb{Q}), S1S2:=S1S2{{S}_{1}}\vee{{S}_{2}}:={{S}_{1}}\bigcup{{S}_{2}}.

By the way, note that in the preceding example, S(SubT(S)){S}^{\prime}(\in\operatorname{Sub}_{T}(S)) is not a subfield of SS. This shows that the notion of TT-space is not equivalent to the notion of field in this case. Basically this is because all elements of TT defined in Example 2.1.2 are unary functions. We will solve this problem in Section 9 (cf. Example 9.2.2).

Because AutT(S)EndT(S)\operatorname{Aut}_{T}(S)\subseteq\operatorname{End}_{T}(S), the following is a straightforward consequence of Lemma 4.2.1.

Lemma 4.2.4.

Let SS be a TT-space and let {Hi}\{{{H}_{i}}\} be a set of subsets of AutT(S)\operatorname{Aut}_{T}(S). Then SHi=SHi\bigcap{{{S}^{{{H}_{i}}}}}={{S}^{\bigcup{{{H}_{i}}}}}.

Analogously, Lemma 4.2.4 implies

Proposition 4.2.5.

IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) is a complete \wedge-sublattice of the complete lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 4.1.1.

Proof.

By Definition 3.2.1, each element of IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) has the form SHi(B){{S}^{{{H}_{i}}}}(\supseteq B), where HiAutT(S){{H}_{i}}\subseteq\operatorname{Aut}_{T}(S). Then by Lemma 4.2.4, SHi=SHiIntTAut(S/B)\bigcap{{{S}^{{{H}_{i}}}}}={{S}^{\bigcup{{{H}_{i}}}}}\in\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B).

Moreover, by Proposition 2.5.2, IntTAut(S/B)SubT(S)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)\subseteq\operatorname{Sub}_{T}(S).

Therefore, IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) is a complete \wedge-sublattice of the complete lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 4.1.1. ∎

However, recall that in Example 4.2.3, IntTAut(S/)=IntTEnd(S/)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/\mathbb{Q})=\operatorname{Int}_{T}^{\operatorname{End}}(S/\mathbb{Q}). So Example 4.2.3 also tells us that IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) is not always a \vee-sublattice of the lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 4.1.1.

4.3. GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) and GSGrT(S/B)\operatorname{GSGr}_{T}(S/B)

GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) and GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) are important because they arise in Corollary 3.2.5 or 3.2.6.

Lemma 4.3.1.

Let SS be a TT-space and let {Si}\{{{S}_{i}}\} be a set of subsets of SS. Then

GMnT(S/Si)=GMnT(S/Si).\bigcap{\operatorname{GMn}_{T}(S/{{S}_{i}})}=\operatorname{GMn}_{T}(S/\bigcup{{{S}_{i}}}).
Proof.

By Definition 2.6.1,

GMnT(S/Si)\displaystyle\bigcap{\operatorname{GMn}_{T}(S/{{S}_{i}})} ={σEndT(S)|σ(a)=a,aSi}\displaystyle=\bigcap{\{\sigma\in\operatorname{End}_{T}(S)\,|\,\sigma(a)=a,\forall a\in{{S}_{i}}\}}
={σEndT(S)|σ(a)=a,aSi}\displaystyle=\{\sigma\in\operatorname{End}_{T}(S)\,|\,\sigma(a)=a,\forall a\in\bigcup{{{S}_{i}}}\}
=GMnT(S/Si).\displaystyle=\operatorname{GMn}_{T}(S/\bigcup{{{S}_{i}}}).

Lemma 4.3.1 implies

Proposition 4.3.2.

GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) is a complete \wedge-sublattice of the complete lattice SMn(EndT(S))\operatorname{SMn}(\operatorname{End}_{T}(S)) defined in Proposition 4.1.2.

Proof.

By Definition 3.2.3, each element of GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) has the form GMnT(S/Si)\operatorname{GMn}_{T}(S/{{S}_{i}}), where BSiSB\subseteq{{S}_{i}}\subseteq S. Then by Lemma 4.3.1,

GMnT(S/Si)=GMnT(S/Si)GSMnT(S/B).\bigcap{\operatorname{GMn}_{T}(S/{{S}_{i}})}=\operatorname{GMn}_{T}(S/\bigcup{{{S}_{i}}})\in\operatorname{GSMn}_{T}(S/B).

Moreover, by Proposition 2.6.2, GSMnT(S/B)SMn(EndT(S))\operatorname{GSMn}_{T}(S/B)\subseteq\operatorname{SMn}(\operatorname{End}_{T}(S)).

Therefore, GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) is a complete \wedge-sublattice of the complete lattice SMn(EndT(S))\operatorname{SMn}(\operatorname{End}_{T}(S)) defined in Proposition 4.1.2. ∎

Nevertheless, the following example tells us that GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) is not necessarily a \vee-sublattice of the lattice SMn(EndT(S))\operatorname{SMn}(\operatorname{End}_{T}(S)) defined in Proposition 4.1.2. Indeed, if M1,M2GSMnT(S/B)\forall{{M}_{1}},{{M}_{2}}\in\operatorname{GSMn}_{T}(S/B), M1M2:=M1M2End(S){{M}_{1}}\vee{{M}_{2}}:={{\left\langle{{M}_{1}}\bigcup{{M}_{2}}\right\rangle}_{\operatorname{End}(S)}}, then it is possible that GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) is not even a \vee-semilattice.

Example 4.3.3.

Let D={a,b}D=\{a,b\} and let T=T={Id on DD}. Then DD is a TT-space and EndT(D)\operatorname{End}_{T}(D) is the set of all maps from DD to DD. Let

M1={all maps from D to D with a fixed}{{M}_{1}}=\{\text{all maps from $D$ to $D$ with $a$ fixed}\}

and

M2={all maps from D to D with b fixed}.{{M}_{2}}=\{\text{all maps from $D$ to $D$ with $b$ fixed}\}.

Then as monoids, M1,M2GSMnT(D/){{M}_{1}},{{M}_{2}}\in\operatorname{GSMn}_{T}(D/\emptyset) by Definition 3.2.3. And it is not hard to see that M1M2SMn(EndT(D)){{M}_{1}}\bigcup{{M}_{2}}\in\operatorname{SMn}(\operatorname{End}_{T}(D)).

Thus by Proposition 4.1.2, M1M2=M1M2End(D)=M1M2{{M}_{1}}\vee{{M}_{2}}={{\left\langle{{M}_{1}}\bigcup{{M}_{2}}\right\rangle}_{End(D)}}={{M}_{1}}\bigcup{{M}_{2}}.

However, since the transposition of aa and bb does not lie in M1M2{{M}_{1}}\bigcup{{M}_{2}},

GMnT(D/DM1M2)=GMnT(D/)={all maps from D to D}M1M2.\operatorname{GMn}_{T}(D/{{D}^{{{M}_{1}}\vee{{M}_{2}}}})=\operatorname{GMn}_{T}(D/\emptyset)=\{\text{all maps from $D$ to $D$}\}\neq{{M}_{1}}\vee{{M}_{2}}.

Hence by Lemma 3.2.4, M1M2GSMnT(D/){{M}_{1}}\vee{{M}_{2}}\notin\operatorname{GSMn}_{T}(D/\emptyset).

Therefore, GSMnT(D/)\operatorname{GSMn}_{T}(D/\emptyset) is not a \vee-sublattice of the lattice SMn(EndT(D))\operatorname{SMn}(\operatorname{End}_{T}(D)) defined in Proposition 4.1.2. In fact, GSMnT(D/)\operatorname{GSMn}_{T}(D/\emptyset) is not even a \vee-semilattice if M1,M2GSMnT(D/)\forall{{M}_{1}},{{M}_{2}}\in\operatorname{GSMn}_{T}(D/\emptyset), M1M2:=M1M2End(D){{M}_{1}}\vee{{M}_{2}}:={{\left\langle{{M}_{1}}\bigcup{{M}_{2}}\right\rangle}_{End(D)}}.

By a similar argument as in the proof of Lemma 4.3.1, we can show the following, whose proof is omitted.

Lemma 4.3.4.

Let SS be a TT-space and let {Si}\{{{S}_{i}}\} be a set of subsets of SS. Then GGrT(S/Si)=GGrT(S/Si)\bigcap{\operatorname{GGr}_{T}(S/{{S}_{i}})}=\operatorname{GGr}_{T}(S/\bigcup{{{S}_{i}}}).

Proposition 4.3.5.

GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) is a complete \wedge-sublattice of the complete lattice SGr(AutT(S))\operatorname{SGr}(\operatorname{Aut}_{T}(S)) defined in Proposition 4.1.3.

Proof.

By Definition 3.2.3, each element of GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) has the form GGrT(S/Si)\operatorname{GGr}_{T}(S/{{S}_{i}}), where BSiSB\subseteq{{S}_{i}}\subseteq S. Then by Lemma 4.3.4,

GGrT(S/Si)=GGrT(S/Si)GSGrT(S/B).\bigcap{\operatorname{GGr}_{T}(S/{{S}_{i}})}=\operatorname{GGr}_{T}(S/\bigcup{{{S}_{i}}})\in\operatorname{GSGr}_{T}(S/B).

Moreover, by Proposition 2.6.3, GSGrT(S/B)SGr(AutT(S))\operatorname{GSGr}_{T}(S/B)\subseteq\operatorname{SGr}(\operatorname{Aut}_{T}(S)).

Therefore, GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) is a complete \wedge-sublattice of the complete lattice SGr(AutT(S))\operatorname{SGr}(\operatorname{Aut}_{T}(S)) defined in Proposition 4.1.3. ∎

However, the following example shows that GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) is not necessarily a \vee-sublattice of the lattice SGr(AutT(S))\operatorname{SGr}(\operatorname{Aut}_{T}(S)) defined in Proposition 4.1.3. Indeed, if G1,G2GSGrT(S/B)\forall{{G}_{1}},{{G}_{2}}\in\operatorname{GSGr}_{T}(S/B), G1G2:=G1G2Aut(S){{G}_{1}}\vee{{G}_{2}}:={{\left\langle{{G}_{1}}\bigcup{{G}_{2}}\right\rangle}_{\operatorname{Aut}(S)}}, then it is possible that GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) is not even a \vee-semilattice.

Example 4.3.6.

Let D={1,2,3,4}D=\{1,2,3,4\} and let T={T=\{Id on D}D\}. Then DD is a TT-space and AutT(D)\operatorname{Aut}_{T}(D) is the set of all bijective maps from DD to DD. Let

G1={Id on D, the transposition of 1 and 2 (with 3 and 4 fixed)}{{G}_{1}}=\{\text{Id on $D$, the transposition of 1 and 2 (with 3 and 4 fixed)}\}

and

G2={Id on D, the transposition of 3 and 4 (with 1 and 2 fixed)}.{{G}_{2}}=\{\text{Id on $D$, the transposition of 3 and 4 (with 1 and 2 fixed)}\}.

Then we can tell that GGrT(D/DG1)=G1\operatorname{GGr}_{T}(D/{{D}^{{{G}_{1}}}})={{G}_{1}} and GGrT(D/DG2)=G2\operatorname{GGr}_{T}(D/{{D}^{{{G}_{2}}}})={{G}_{2}}, and so by Lemma 3.2.4, G1,G2GSGrT(D/){{G}_{1}},{{G}_{2}}\in\operatorname{GSGr}_{T}(D/\emptyset).

By Proposition 4.1.3, G1G2=G1G2Aut(D)=G1×G2{{G}_{1}}\vee{{G}_{2}}={{\left\langle{{G}_{1}}\bigcup{{G}_{2}}\right\rangle}_{\operatorname{Aut}(D)}}={{G}_{1}}\times{{G}_{2}}.

However, GGrT(D/DG1×G2)=GGrT(D/)=S4\operatorname{GGr}_{T}(D/{{D}^{{{G}_{1}}\times{{G}_{2}}}})=\operatorname{GGr}_{T}(D/\emptyset)={{S}_{4}}, which is the symmetric group on four letters. Thus GGrT(D/DG1G2)G1G2\operatorname{GGr}_{T}(D/{{D}^{{{G}_{1}}\vee{{G}_{2}}}})\neq{{G}_{1}}\vee{{G}_{2}}, and so by Lemma 3.2.4, G1G2GSGrT(D/){{G}_{1}}\vee{{G}_{2}}\notin\operatorname{GSGr}_{T}(D/\emptyset). Hence GSGrT(D/)\operatorname{GSGr}_{T}(D/\emptyset) is not a \vee-sublattice of the lattice SGr(AutT(D))\operatorname{SGr}(\operatorname{Aut}_{T}(D)) defined in Proposition 4.1.3. Indeed, GSGrT(D/)\operatorname{GSGr}_{T}(D/\emptyset) is not even a \vee-semilattice if G1,G2GSGrT(D/)\forall{{G}_{1}},{{G}_{2}}\in\operatorname{GSGr}_{T}(D/\emptyset), G1G2:=G1G2Aut(D){{G}_{1}}\vee{{G}_{2}}:={{\left\langle{{G}_{1}}\bigcup{{G}_{2}}\right\rangle}_{\operatorname{Aut}(D)}}.

5. Topologies employed to construct Galois correspondences

As was promised near the end of Section 3, now we are going to answer the following questions.

To construct the Galois correspondence described in Corollary 3.2.5 without resort to Definition 3.2.1 or 3.2.3, can we define a topology on SS and a topology on EndT(S)\operatorname{End}_{T}(S) such that IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) is the set of all closed quasi-TT-subspaces (of SS) containing BB and GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) is the set of all closed submonoids of GMnT(S/B)\operatorname{GMn}_{T}(S/B)? This question will be answered in Subsections 5.1 and 5.3.

And for Corollary 3.2.6, can we define a topology on SS and a topology on AutT(S)\operatorname{Aut}_{T}(S) with IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) being the set of all closed quasi-TT-subspaces (of SS) containing BB and GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) being the set of all closed subgroups of GGrT(S/B)\operatorname{GGr}_{T}(S/B)? This question will be answered in Subsections 5.2 and 5.4.

5.1. Topologies on a TT-space SS for IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B)

Lemma 5.1.1.

Let SS be a TT-space with BSB\subseteq S. Then the topology 𝒯1{{\mathcal{T}}_{1}} on SS generated by a subbasis β1:={S\A|AIntTEnd(S/B)}{S}{{\beta}_{1}}:=\{S\backslash A\,|\,A\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\}\bigcup\{S\} is the smallest ((coarsest)) topology on SS such that the collection of all closed sets contains IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B).

Proof.

Since the union of all elements of β1{{\beta}_{1}} is SS, β1{{\beta}_{1}} qualifies as a subbasis for a topology on SS. Hence the topology 𝒯1{{\mathcal{T}}_{1}} generated by β1{{\beta}_{1}} equals the intersection of all topologies on SS that contain β1{{\beta}_{1}} (see e.g. [6]), and so 𝒯1{{\mathcal{T}}_{1}} is the smallest topology on SS containing β1{{\beta}_{1}}. Thus by the definition of β1{{\beta}_{1}}, 𝒯1{{\mathcal{T}}_{1}} is the smallest topology on SS such that the collection of all closed sets contains IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B). ∎

For a TT-space SS with BSB\subseteq S, to characterize the Galois correspondence depicted in Corollary 3.2.5 in terms of closed quasi-TT-subspaces of SS, i.e. to replace IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) in Corollary 3.2.5 with the set of all closed quasi-TT-subspaces (of SS) containing BB, we need a topology on SS such that the following equation is satisfied.

(5.1) IntTEnd(S/B)\{}={BSqS|S is closed and nonempty}\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\}=\{B\subseteq S^{\prime}\leq_{q}S\,|\,S^{\prime}\text{ is closed and nonempty}\}
Remark.

The reason why we require the above equality rather than the following one is explained as follows.

(5.2) IntTEnd(S/B)={BSqS|S is closed}\operatorname{Int}_{T}^{\operatorname{End}}(S/B)=\{B\subseteq{S}^{\prime}\leq_{q}S\,|\,{S}^{\prime}\text{ is closed\}}

Suppose B=S=B={S}^{\prime}=\emptyset. Then BSqSB\subseteq S^{\prime}\leq_{q}S and S{S}^{\prime} is closed. Hence \emptyset lies in the right side of (5.2). But it is possible that S(=)IntTEnd(S/B){S}^{\prime}(=\emptyset)\notin\operatorname{Int}_{T}^{\operatorname{End}}(S/B) (e.g. when SS\neq\emptyset and EndT(S)={Id}\operatorname{End}_{T}(S)=\{\operatorname{Id}\}), and thus in this case we cannot replace IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) with the right side of (5.2).

The following is to determine whether there exists a topology on SS such that Equation ((5.1)) is satisfied: If P1P_{1} defined as follows is nonempty, then we can replace IntTEnd(S/B)\{}\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\} with the set of all nonempty closed quasi-TT-subspaces (of SS) containing BB.

Theorem 5.1.2.

Let SS be a TT-space and let BSB\subseteq S. Let

P1={all topologies on S such that Equation (5.1) is satisfied}{{P}_{1}}=\{\text{all topologies on $S$ such that Equation $($\ref{5.1}$)$ is satisfied}\}

and let

Q1={all topologies on S which are finer (larger) than 𝒯1},{{Q}_{1}}=\{\text{all topologies on $S$ which are finer $($larger$)$ than ${{\mathcal{T}}_{1}}$}\},

where 𝒯1{{\mathcal{T}}_{1}} is defined in Lemma 5.1.1. Then P1Q1{{P}_{1}}\subseteq{{Q}_{1}} and the following statements are equivalent:

  1. (i)

    P1{P_{1}}\neq\emptyset.

  2. (ii)

    S1,S2IntTEnd(S/B)\forall{{S}_{1}},{{S}_{2}}\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B), S1S2IntTEnd(S/B){{S}_{1}}\bigcup{{S}_{2}}\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B). ((Or equivalently, β1{{\beta}_{1}} defined in Lemma 5.1.1 is a basis for 𝒯1({{\mathcal{T}}_{1}}\ (see e.g. [6])))).

  3. (iii)

    IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) is a \vee-sublattice of the lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 4.1.1.

  4. (iv)

    𝒯1P1(Q1){{\mathcal{T}}_{1}}\in{{P}_{1}}(\subseteq{{Q}_{1}}); that is, 𝒯1{{\mathcal{T}}_{1}} is the coarsest ((smallest)) topology on SS such that Equation ((5.1)) is satisfied.

Proof.

By Lemma 5.1.1, 𝒯1{\mathcal{T}}_{1} is the smallest topology on SS such that the collection of all closed sets contains IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B). Suppose 𝒯P1\mathcal{T}\in{{P}_{1}}. Then all elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) are closed in 𝒯\mathcal{T}, and so 𝒯1𝒯{\mathcal{T}}_{1}\subseteq\mathcal{T}. Hence 𝒯Q1\mathcal{T}\in{{Q}_{1}}. Thus P1Q1{{P}_{1}}\subseteq{{Q}_{1}}.

(i) \Rightarrow (ii): Let 𝒯P1\mathcal{T}\in{{P}_{1}}. To prove (ii), we only need to consider the case where S1,S2IntTEnd(S/B)\{}{{S}_{1}},{{S}_{2}}\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\}. Then by the definition of P1{P_{1}},

S1,S2{BSqS|S is nonempty and closed in 𝒯}.{{S}_{1}},{{S}_{2}}\in\{B\subseteq{S}^{\prime}\leq_{q}S\,|\,{S}^{\prime}\text{ is nonempty and closed in }\mathcal{T}\}.

By Proposition 2.2.13, S1S2qS{{S}_{1}}\bigcup{{S}_{2}}\leq_{q}S. Hence

S1S2{BSqS|S is nonempty and closed in 𝒯},{{S}_{1}}\bigcup{{S}_{2}}\in\{B\subseteq{S}^{\prime}\leq_{q}S\,|\,{S}^{\prime}\text{ is nonempty and closed in }\mathcal{T}\},

which equals IntTEnd(S/B)\{}\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\} by (5.1). Therefore, S1S2IntTEnd(S/B){{S}_{1}}\bigcup{{S}_{2}}\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B), and so (ii) is true.

(ii) \Rightarrow (iii): By Definition 3.2.1 and Proposition 2.5.2, IntTEnd(S/B)SubT(S)\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\subseteq\operatorname{Sub}_{T}(S). Then from Proposition 4.1.1, we can tell that (iii) is true if (ii) is true.

(iii) \Rightarrow (iv): By the definition of 𝒯1{{\mathcal{T}}_{1}} in Lemma 5.1.1, all elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) are closed in 𝒯1{{\mathcal{T}}_{1}}, and so by Definition 3.2.1 and Proposition 2.5.2,

IntTEnd(S/B)\{}{BSqS|S is nonempty and closed in 𝒯1}.\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\}\subseteq\{B\subseteq S^{\prime}\leq_{q}S\,|\,S^{\prime}\text{ is nonempty and closed in }{{\mathcal{T}}_{1}}\}.

On the other hand, we show

IntTEnd(S/B)\{}{BSqS|S is nonempty and closed in 𝒯1}\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\}\supseteq\{B\subseteq S^{\prime}\leq_{q}S\,|\,S^{\prime}\text{ is nonempty and closed in }{{\mathcal{T}}_{1}}\}

as follows.

Let BSqSB\subseteq{S}^{\prime}\leq_{q}S such that S{S}^{\prime} is nonempty and closed in 𝒯1{{\mathcal{T}}_{1}}. Then we need to show SIntTEnd(S/B){S}^{\prime}\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B).

Since 𝒯1{{\mathcal{T}}_{1}} is the collection of all unions of finite intersections of elements of β1={S\A|AIntTEnd(S/B)}{S}{{\beta}_{1}}=\{S\backslash A\,|\,A\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\}\bigcup\{S\}, by DeMorgan’s Laws, the collection of all closed sets in 𝒯1{{\mathcal{T}}_{1}} is the family of all intersections of finite unions of elements of IntTEnd(S/B){}\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\bigcup\{\emptyset\}. Then S{S}^{\prime} is an intersection of finite unions of elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B). Because (iii) is supposed to be true, any finite union of elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) belongs to IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B). So S{S}^{\prime} is an intersection of elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B). Then by Lemma 4.2.1, SIntTEnd(S/B){S}^{\prime}\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B), as desired. Hence

IntTEnd(S/B)\{}{BSqS|S is nonempty and closed in 𝒯1}.\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\}\supseteq\{B\subseteq S^{\prime}\leq_{q}S\,|\,S^{\prime}\text{ is nonempty and closed in }{{\mathcal{T}}_{1}}\}.

Therefore,

IntTEnd(S/B)\{}={BSqS|S is nonempty and closed in 𝒯1}.\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\}=\{B\subseteq S^{\prime}\leq_{q}S\,|\,S^{\prime}\text{ is nonempty and closed in }{{\mathcal{T}}_{1}}\}.

Thus 𝒯1P1{{\mathcal{T}}_{1}}\in{{P}_{1}}. We showed P1Q1{{P}_{1}}\subseteq{{Q}_{1}} at the beginning of our proof, and so by the definition of Q1{{Q}_{1}}, 𝒯1{{\mathcal{T}}_{1}} is the coarsest topology in P1P_{1}. Then by the definition of P1{{P}_{1}}, 𝒯1{{\mathcal{T}}_{1}} is the coarsest topology on SS such that Equation ((5.1)) is satisfied.

(iv) \Rightarrow (i): Obvious. ∎

By Theorem 5.1.2, when P1{{P}_{1}}\neq\emptyset, we can put a topology such as 𝒯1{{\mathcal{T}}_{1}} on SS such that Equation (5.1) is satisfied. Then IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) in Corollary 3.2.5 can be replaced by {BSqS|SB\subseteq{S}^{\prime}\leq_{q}S\,|\,{S}^{\prime} is closed and nonempty} if IntTEnd(S/B)\emptyset\notin\operatorname{Int}_{T}^{\operatorname{End}}(S/B), and it can be replaced by {BSqS|SB\subseteq{S}^{\prime}\leq_{q}S\,|\,{S}^{\prime} is closed} if IntTEnd(S/B)\emptyset\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B).

However, Example 4.2.3 tells us that statement (ii) (and (iii)) in Theorem 5.1.2 is not always true. Hence it is possible that P1={{P}_{1}}=\emptyset. When this occurs, we cannot endow SS with any topology such that Equation (5.1) is satisfied. Then in this case, there is no topology on SS to characterize the Galois correspondence depicted in Corollary 3.2.5 in terms of closed subsets of SS.

5.2. Topologies on a TT-space SS for IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)

To replace IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) in Corollary 3.2.6, we have

Lemma 5.2.1.

Let SS be a TT-space with BSB\subseteq S. Then the topology 𝒯2{{\mathcal{T}}_{2}} on SS generated by a subbasis β2:={S\A|AIntTAut(S/B)}{S}{{\beta}_{2}}:=\{S\backslash A\,|\,A\in\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)\}\bigcup\{S\} is the smallest topology on SS such that the collection of all closed sets contains IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B).

Proof.

Almost the same as the proof of Lemma 5.1.1 except that IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B), 𝒯1{{\mathcal{T}}_{1}} and β1{{\beta}_{1}} are replaced by IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B), 𝒯2{{\mathcal{T}}_{2}} and β2{{\beta}_{2}}, respectively. ∎

For a TT-space SS with BSB\subseteq S, to characterize the Galois correspondence depicted in Corollary 3.2.6 in terms of closed quasi-TT-subspaces of SS, we need a topology on SS such that the following equation is satisfied.

(5.3) IntTAut(S/B)\{}={BSqS|S is closed and nonempty}\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)\backslash\{\emptyset\}=\{B\subseteq S^{\prime}\leq_{q}S\,|\,S^{\prime}\text{ is closed and nonempty}\}
Theorem 5.2.2.

Let SS be a TT-space and let BSB\subseteq S. Let

P2={all topologies on S such that Equation (5.3) is satisfied}{{P}_{2}}=\{\text{all topologies on $S$ such that Equation $($\ref{5.3}$)$ is satisfied}\}

and let

Q2={all topologies on S which are finer than 𝒯2},{{Q}_{2}}=\{\text{all topologies on $S$ which are finer than ${{\mathcal{T}}_{2}}$}\},

where 𝒯2{{\mathcal{T}}_{2}} is defined in Lemma 5.2.1. Then P2Q2{{P}_{2}}\subseteq{{Q}_{2}} and the following statements are equivalent:

  1. (i)

    P2{P_{2}}\neq\emptyset.

  2. (ii)

    S1,S2IntTAut(S/B)\forall{{S}_{1}},{{S}_{2}}\in\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B), S1S2IntTAut(S/B){{S}_{1}}\bigcup{{S}_{2}}\in\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B). ((Or equivalently, β2{{\beta}_{2}} defined in Lemma 5.2.1 is a basis for 𝒯2){{\mathcal{T}}_{2}}).

  3. (iii)

    IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) is a \vee-sublattice of the lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 4.1.1.

  4. (iv)

    𝒯2P2(Q2){{\mathcal{T}}_{2}}\in{{P}_{2}}(\subseteq{{Q}_{2}}); that is, 𝒯2{{\mathcal{T}}_{2}} is the coarsest topology on SS such that Equation ((5.3)) is satisfied.

Proof.

Almost the same as the proof of Theorem 5.1.2 except that IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B), 𝒯1{{\mathcal{T}}_{1}}, β1{{\beta}_{1}}, P1{{P}_{1}} and Q1{{Q}_{1}} are replaced by IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B), 𝒯2{{\mathcal{T}}_{2}}, β2{{\beta}_{2}}, P2{{P}_{2}} and Q2{{Q}_{2}}, respectively, and accordingly, Lemmas 4.2.1 and 5.1.1 are replaced by Lemmas 4.2.4 and 5.2.1, respectively. ∎

It follows that when P2{{P}_{2}}\neq\emptyset, we can give a topology such as 𝒯2{{\mathcal{T}}_{2}} on SS such that Equation (5.3) is satisfied. Then IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) in Corollary 3.2.6 can be substituted with {BSqS|SB\subseteq{S}^{\prime}\leq_{q}S\,|\,{S}^{\prime} is closed and nonempty} if IntTAut(S/B)\emptyset\notin\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B), and it can be substituted with {BSqS|SB\subseteq{S}^{\prime}\leq_{q}S\,|\,{S}^{\prime} is closed} if IntTAut(S/B)\emptyset\in\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B).

Recall that at the end of Subsection 5.1, we use Example 4.2.3 to show that statement (ii) (and (iii)) in Theorem 5.1.2 is not always true. Because in Example 4.2.3, IntTEnd(S/)=IntTAut(S/)\operatorname{Int}_{T}^{\operatorname{End}}(S/\mathbb{Q})=\operatorname{Int}_{T}^{\operatorname{Aut}}(S/\mathbb{Q}), statement (ii) (and (iii)) in Theorem 5.2.2 is not necessarily true, either. Hence it is possible that P2={{P}_{2}}=\emptyset. When this occurs, we cannot give any topology on SS such that Equation (5.3) is satisfied. Then in this case, there does not exist any topology on SS such that the Galois correspondence depicted in Corollary 3.2.6 can be characterized in terms of closed subsets of SS.

5.3. Topologies on EndT(S)\operatorname{End}_{T}(S)

To replace GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) in Corollary 3.2.5, we need

Lemma 5.3.1.

Let SS be a TT-space with BSB\subseteq S. Then the topology 𝒯3{{\mathcal{T}}_{3}} on EndT(S)\operatorname{End}_{T}(S) generated by a subbasis

β3:={EndT(S)\H|HGSMnT(S/B)}{EndT(S)}{{\beta}_{3}}:=\{\operatorname{End}_{T}(S)\backslash H\,|\,H\in\operatorname{GSMn}_{T}(S/B)\}\bigcup\{\operatorname{End}_{T}(S)\}

is the smallest topology on EndT(S)\operatorname{End}_{T}(S) such that the collection of all closed sets contains GSMnT(S/B)\operatorname{GSMn}_{T}(S/B).

Proof.

Since the union of all elements of β3{{\beta}_{3}} is EndT(S)\operatorname{End}_{T}(S), β3{{\beta}_{3}} qualifies as a subbasis for a topology on EndT(S)\operatorname{End}_{T}(S). Hence the topology 𝒯3{{\mathcal{T}}_{3}} generated by β3{{\beta}_{3}} equals the intersection of all topologies on EndT(S)\operatorname{End}_{T}(S) that contain β3{{\beta}_{3}} (see e.g. [6]), and so 𝒯3{{\mathcal{T}}_{3}} is the smallest topology on EndT(S)\operatorname{End}_{T}(S) containing β3{{\beta}_{3}}. Thus by the definition of β3{{\beta}_{3}}, 𝒯3{{\mathcal{T}}_{3}} is the smallest topology on EndT(S)\operatorname{End}_{T}(S) such that the collection of all closed sets contains GSMnT(S/B)\operatorname{GSMn}_{T}(S/B). ∎

For a TT-space SS with BSB\subseteq S, to characterize the Galois correspondence depicted in Corollary 3.2.5 in terms of closed submonoids of GMnT(S/B)\operatorname{GMn}_{T}(S/B), we need a topology on EndT(S)\operatorname{End}_{T}(S) such that the following equation is satisfied.

(5.4) GSMnT(S/B)={MSMn(GMnT(S/B))|M is closed}\operatorname{GSMn}_{T}(S/B)=\{M\in\operatorname{SMn}(\operatorname{GMn}_{T}(S/B))\,|\,M\text{ is closed}\}
Remark.

See Notation 3.0.2 for SMn\operatorname{SMn}.

Theorem 5.3.2.

Let SS be a TT-space and let BSB\subseteq S. Let

P3={all topologies on EndT(S) such that Equation (5.4) is satisfied}{{P}_{3}}=\{\text{all topologies on $\operatorname{End}_{T}(S)$ such that Equation $($\ref{5.4}$)$ is satisfied\}}

and let

Q3={all topologies on EndT(S) which are finer than 𝒯3},\text{${{Q}_{3}}=$\{all topologies on $\operatorname{End}_{T}(S)$ which are finer than }{{\mathcal{T}}_{3}}\},

where 𝒯3{{\mathcal{T}}_{3}} is defined in Lemma 5.3.1. Then P3Q3{{P}_{3}}\subseteq{{Q}_{3}} and the following statements are equivalent:

  1. (i)

    P3{P_{3}}\neq\emptyset.

  2. (ii)

    For any intersection MM of finite unions of elements of GSMnT(S/B)\operatorname{GSMn}_{T}(S/B), MSMn(EndT(S))M\in\operatorname{SMn}(\operatorname{End}_{T}(S)) implies MGSMnT(S/B)M\in\operatorname{GSMn}_{T}(S/B).

  3. (iii)

    𝒯3P3(Q3){{\mathcal{T}}_{3}}\in{{P}_{3}}(\subseteq{{Q}_{3}}); that is, 𝒯3{{\mathcal{T}}_{3}} is the coarsest topology on EndT(S)\operatorname{End}_{T}(S) such that Equation ((5.4)) is satisfied.

Moreover, if GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) is a complete \vee-sublattice of the complete lattice SMn(EndT(S))\operatorname{SMn}(\operatorname{End}_{T}(S)) defined in Proposition 4.1.2, then the above three statements are true.

Proof.

By Lemma 5.3.1, 𝒯3{{\mathcal{T}}_{3}} is the smallest topology on EndT(S)\operatorname{End}_{T}(S) such that the collection of all closed sets contains GSMnT(S/B)\operatorname{GSMn}_{T}(S/B). Suppose 𝒯P3\mathcal{T}\in{{P}_{3}}. Then all elements of GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) are closed in 𝒯\mathcal{T}, and so 𝒯3𝒯{\mathcal{T}}_{3}\subseteq\mathcal{T}. Hence 𝒯Q3\mathcal{T}\in{{Q}_{3}}. Thus P3Q3{{P}_{3}}\subseteq{{Q}_{3}}.

(i)\Rightarrow(ii): For any intersection MM given in (ii), from Definitions 3.2.3 and 2.6.1, we can tell MGMnT(S/B)M\subseteq\operatorname{GMn}_{T}(S/B). Suppose MSMn(EndT(S))M\in\operatorname{SMn}(\operatorname{End}_{T}(S)). Then MSMn(GMnT(S/B))M\in\operatorname{SMn}(\operatorname{GMn}_{T}(S/B)). Since P3{P_{3}}\neq\emptyset, let 𝒯P3\mathcal{T}\in{{P}_{3}}. By the definition of P3{{P}_{3}}, every element of GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) is closed in 𝒯\mathcal{T}, and so is the intersection MM (of finite unions of elements of GSMnT(S/B)\operatorname{GSMn}_{T}(S/B)). Again by the definition of P3{{P}_{3}}, MGSMnT(S/B)M\in\operatorname{GSMn}_{T}(S/B).

(ii)\Rightarrow(iii):

By the definition of 𝒯3{{\mathcal{T}}_{3}} in Lemma 5.3.1, all elements of GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) are closed in 𝒯3{{\mathcal{T}}_{3}}, and so by Definition 3.2.3 and Proposition 2.6.2,

GSMnT(S/B){MSMn(GMnT(S/B))|M is closed in 𝒯3}.\operatorname{GSMn}_{T}(S/B)\subseteq\{M\in\operatorname{SMn}(\operatorname{GMn}_{T}(S/B))\,|\,M\text{ is closed in }{{\mathcal{T}}_{3}}\}.

On the other hand, we prove

GSMnT(S/B){MSMn(GMnT(S/B))|M is closed in 𝒯3}\operatorname{GSMn}_{T}(S/B)\supseteq\{M\in\operatorname{SMn}(\operatorname{GMn}_{T}(S/B))\,|\,M\text{ is closed in }{{\mathcal{T}}_{3}}\}

as follows.

Let MSMn(GMnT(S/B))M\in\operatorname{SMn}(\operatorname{GMn}_{T}(S/B)) such that MM is closed in 𝒯3{{\mathcal{T}}_{3}}. We need to show MGSMnT(S/B)M\in\operatorname{GSMn}_{T}(S/B).

Because 𝒯3{{\mathcal{T}}_{3}} is generated by subbasis β3{{\beta}_{3}}, 𝒯3{{\mathcal{T}}_{3}} is the collection of all unions of finite intersections of elements of

β3={EndT(S)\H|HGSMnT(S/B)}{EndT(S)}.{{\beta}_{3}}=\{\operatorname{End}_{T}(S)\backslash H\,|\,H\in\operatorname{GSMn}_{T}(S/B)\}\bigcup\{\operatorname{End}_{T}(S)\}.

Since MM is closed in 𝒯3{{\mathcal{T}}_{3}}, EndT(S)\M\operatorname{End}_{T}(S)\backslash M is a union of finite intersections of elements of β3{{\beta}_{3}}. It follows from DeMorgan’s Laws that MM is an intersection of finite unions of elements of GSMnT(S/B){}\operatorname{GSMn}_{T}(S/B)\bigcup\{\emptyset\}. Because (Id)M\text{(Id}\in)M\neq\emptyset, MM is an intersection of finite unions of elements of GSMnT(S/B)\operatorname{GSMn}_{T}(S/B). Since (ii) is assumed to be true and MSMn(GMnT(S/B))SMn(EndT(S))M\in\operatorname{SMn}(\operatorname{GMn}_{T}(S/B))\subseteq\operatorname{SMn}(\operatorname{End}_{T}(S)), MGSMnT(S/B)M\in\operatorname{GSMn}_{T}(S/B), as desired. Hence

GSMnT(S/B){MSMn(GMnT(S/B))|M is closed in 𝒯3}\operatorname{GSMn}_{T}(S/B)\supseteq\{M\in\operatorname{SMn}(\operatorname{GMn}_{T}(S/B))\,|\,M\text{ is closed in }{{\mathcal{T}}_{3}}\}

Therefore,

GSMnT(S/B)={MSMn(GMnT(S/B))|M is closed in 𝒯3}\operatorname{GSMn}_{T}(S/B)=\{M\in\operatorname{SMn}(\operatorname{GMn}_{T}(S/B))\,|\,M\text{ is closed in }{{\mathcal{T}}_{3}}\}

Thus 𝒯3P3{{\mathcal{T}}_{3}}\in{{P}_{3}}. We showed P3Q3{{P}_{3}}\subseteq{{Q}_{3}} at the beginning of our proof. So by the definition of Q3{{Q}_{3}}, 𝒯3{{\mathcal{T}}_{3}} is the coarsest topology in P3P_{3}. Then by the definition of P3{{P}_{3}}, 𝒯3{{\mathcal{T}}_{3}} is the coarsest topology on EndT(S)\operatorname{End}_{T}(S) such that Equation ((5.4)) is satisfied.

(iii)\Rightarrow(i): Obvious.

By Proposition 4.3.2, any intersection of elements of GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) still lies in GSMnT(S/B)\operatorname{GSMn}_{T}(S/B). Then as a result of the first distributive law of set operations, any intersection MM of finite unions of elements of GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) is a union of elements of GSMnT(S/B)\operatorname{GSMn}_{T}(S/B). Suppose that GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) is a complete \vee-sublattice of the complete lattice SMn(EndT(S))\operatorname{SMn}(\operatorname{End}_{T}(S)) defined in Proposition 4.1.2. Then MSMn(EndT(S))M\in\operatorname{SMn}(\operatorname{End}_{T}(S)) implies MGSMnT(S/B)M\in\operatorname{GSMn}_{T}(S/B), and so (ii) is true (in this case). ∎

When P3{{P}_{3}}\neq\emptyset, by Theorem 5.3.2, we can endow EndT(S)\operatorname{End}_{T}(S) with a topology such as 𝒯3{{\mathcal{T}}_{3}} such that Equation ((5.4)) is satisfied. Then GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) in Corollary 3.2.5 can be replaced by the set of all closed submonoids of GMnT(S/B)\operatorname{GMn}_{T}(S/B).

Nevertheless, it is possible that P3={{P}_{3}}=\emptyset. Recall that in Example 4.3.3, M1,M2GSMnT(D/){{M}_{1}},{{M}_{2}}\in\operatorname{GSMn}_{T}(D/\emptyset) and M1M2SMn(EndT(D)){{M}_{1}}\bigcup{{M}_{2}}\in\operatorname{SMn}(\operatorname{End}_{T}(D)), but M1M2GSMnT(D/){{M}_{1}}\bigcup{{M}_{2}}\notin\operatorname{GSMn}_{T}(D/\emptyset). Thus for Example 4.3.3, statement (ii) in Theorem 5.3.2 is not true, and hence P3={{P}_{3}}=\emptyset. Therefore, it is possible that we cannot endow EndT(S)\operatorname{End}_{T}(S) with any topology such that the Galois correspondence depicted in Corollary 3.2.5 can be characterized in terms of closed submonoids of GMnT(S/B)\operatorname{GMn}_{T}(S/B).

5.4. Topologies on AutT(S)\operatorname{Aut}_{T}(S)

To replace GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) in Corollary 3.2.6, we need

Lemma 5.4.1.

Let SS be a TT-space and let BSB\subseteq S. Then the topology 𝒯4{{\mathcal{T}}_{4}} on AutT(S)\operatorname{Aut}_{T}(S) generated by a subbasis

β4:={AutT(S)\H|HGSGrT(S/B)}{AutT(S)}{{\beta}_{4}}:=\{\operatorname{Aut}_{T}(S)\backslash H\,|\,H\in\operatorname{GSGr}_{T}(S/B)\}\bigcup\{\operatorname{Aut}_{T}(S)\}

is the smallest topology on AutT(S)\operatorname{Aut}_{T}(S) such that the collection of all closed sets contains GSGrT(S/B)\operatorname{GSGr}_{T}(S/B).

Proof.

Almost the same as the proof of Lemma 5.3.1 except that GSMnT(S/B)\operatorname{GSMn}_{T}(S/B), EndT(S)\operatorname{End}_{T}(S), 𝒯3{{\mathcal{T}}_{3}} and β3{{\beta}_{3}} are replaced by GSGrT(S/B)\operatorname{GSGr}_{T}(S/B), AutT(S)\operatorname{Aut}_{T}(S), 𝒯4{{\mathcal{T}}_{4}} and β4{{\beta}_{4}}, respectively. ∎

For a TT-space SS with BSB\subseteq S, to characterize the Galois correspondence depicted in Corollary 3.2.6 in terms of closed subgroups of GGrT(S/B)\operatorname{GGr}_{T}(S/B), we need a topology on AutT(S)\operatorname{Aut}_{T}(S) such that the following equation is satisfied.

(5.5) GSGrT(S/B)={MSGr(GGrT(S/B))|M is closed}\operatorname{GSGr}_{T}(S/B)=\{M\in\operatorname{SGr}(\operatorname{GGr}_{T}(S/B))\,|\,M\text{ is closed}\}
Remark.

See Notation 3.0.2 for SGr\operatorname{SGr}.

Theorem 5.4.2.

Let SS be a TT-space and let BSB\subseteq S. Let

P4={all topologies on AutT(S) such that Equation (5.5) is satisfied}{{P}_{4}}=\{\text{all topologies on $\operatorname{Aut}_{T}(S)$ such that Equation $($\ref{5.5}$)$ is satisfied\}}

and let

Q4={all topologies on AutT(S) which are finer than 𝒯4},\text{${{Q}_{4}}=$\{all topologies on $\operatorname{Aut}_{T}(S)$ which are finer than }{{\mathcal{T}}_{4}}\},

where 𝒯4{{\mathcal{T}}_{4}} is defined in Lemma 5.4.1. Then P4Q4{{P}_{4}}\subseteq{{Q}_{4}} and the following statements are equivalent:

  1. (i)

    P4{P_{4}}\neq\emptyset.

  2. (ii)

    For any intersection GG of finite unions of elements of GSGrT(S/B)\operatorname{GSGr}_{T}(S/B), GSGr(AutT(S))G\in\operatorname{SGr}(\operatorname{Aut}_{T}(S)) implies GGSGrT(S/B)G\in\operatorname{GSGr}_{T}(S/B).

  3. (iii)

    𝒯4P4(Q4){{\mathcal{T}}_{4}}\in{{P}_{4}}(\subseteq{{Q}_{4}}); that is, 𝒯4{{\mathcal{T}}_{4}} is the coarsest topology on AutT(S)\operatorname{Aut}_{T}(S) such that Equation ((5.5)) is satisfied.

Moreover, if GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) is a complete \vee-sublattice of the complete lattice SGr(AutT(S))\operatorname{SGr}(\operatorname{Aut}_{T}(S)) defined in Proposition 4.1.3, then the above three statements are true.

Proof.

Almost the same as the proof of Theorem 5.3.2 except that EndT(S)\operatorname{End}_{T}(S), GSMnT(S/B)\operatorname{GSMn}_{T}(S/B), 𝒯3{{\mathcal{T}}_{3}}, β3{{\beta}_{3}}, P3{{P}_{3}}, Q3{{Q}_{3}}, and MM are replaced by AutT(S)\operatorname{Aut}_{T}(S), GSGrT(S/B)\operatorname{GSGr}_{T}(S/B), 𝒯4{{\mathcal{T}}_{4}}, β4{{\beta}_{4}}, P4{{P}_{4}}, Q4{{Q}_{4}}, and GG, respectively, and accordingly, Lemma 5.3.1 and Propositions 2.6.2, 4.1.2 and 4.3.2 are replaced by Lemma 5.4.1 and Propositions 2.6.3, 4.1.3 and 4.3.5, respectively. ∎

Nevertheless, we do not know whether the three equivalent statements in Theorem 5.4.2 are always true, and hence we do not know whether there always exists a topology on AutT(S)\operatorname{Aut}_{T}(S) such that Equation ((5.5)) is satisfied. In other words, for any TT-space SS with BSB\subseteq S, we do not know whether we can always endow AutT(S)\operatorname{Aut}_{T}(S) with a topology such that GSGrT(S/B)\operatorname{GSGr}_{T}(S/B) in Corollary 3.2.6 can be replaced by the set of all closed subgroups of GGrT(S/B)\operatorname{GGr}_{T}(S/B).

6. Generalized morphisms and isomorphisms from a T1T_{1}-space to a T2T_{2}-space

In this section, we shall define generalized morphisms and isomorphisms from a T1T_{1}-space to a T2T_{2}-space, where both T1T_{1} and T2T_{2} are operator semigroups.

6.1. θ\theta-morphisms

In algebra, ring homomorphisms from the ring B[x]B[x] to the ring B[x]{B}^{\prime}[x], where both BB and B{B}^{\prime} are commutative rings, are defined. However, Definition 2.3.1 does not cover this kind of morphisms because, by Example 2.1.2, B[x]B[x] and B[x]{B}^{\prime}[x] may induce different operator semigroups. So we generalize Definition 2.3.1 as follows.

Definition 6.1.1.

Let T1T_{1} and T2T_{2} be operator semigroups, let θ\theta be a map from T1T_{1} to T2T_{2}, and let ϕ\phi be a map from a T1T_{1}-space SS to a T2T_{2}-space. If fT1\forall f\in{{T}_{1}} and aSa\in S, ϕ(f(a))=θ(f)(ϕ(a))\phi(f(a))=\theta(f)(\phi(a)), then ϕ\phi is called a θ\theta-morphism.

Remark.

If T1=T2{{T}_{1}}={{T}_{2}} and θ\theta is the identity map, then ϕ\phi is also a T1{{T}_{1}}-morphism defined by Definition 2.3.1.

Let us see two examples as follows.

Proposition 6.1.2.

Let AA and RR be differential rings, let TA{{T}_{A}} ((resp. TR){{T}_{R}}) be the operator semigroup on AA ((resp. R)R) defined as in Example 2.1.5, and let a map θ:TATR\theta:{{T}_{A}}\to{{T}_{R}} be given by AnRn\partial_{A}^{n}\mapsto\partial_{R}^{n}, n0\forall n\in{{\mathbb{N}}_{0}}, where A{{\partial}_{A}} and R{{\partial}_{R}} are derivations of AA and RR, respectively. Let ϕ:AR\phi:A\to R be a ring homomorphism. Then ϕ\phi is a differential homomorphism if and only if ϕ\phi is a θ\theta-morphism.

Proof.

Recall that a ring homomorphism ϕ:AR\phi:A\to R is a differential homomorphism if and only if it satisfies ϕ(A(a))=R(ϕ(a)),aA\phi({{\partial}_{A}}(a))={{\partial}_{R}}(\phi(a)),\forall a\in A (see e.g. [1]). Thus we only need to show that ϕ\phi is a θ\theta-morphism if and only if ϕ\phi satisfies ϕ(A(a))=R(ϕ(a)),aA\phi({{\partial}_{A}}(a))={{\partial}_{R}}(\phi(a)),\forall a\in A. Since n0\forall n\in{{\mathbb{N}}_{0}}, θ(An)=Rn\theta(\partial_{A}^{n})=\partial_{R}^{n}, the latter equivalence is obvious by Definition 6.1.1. ∎

Now Proposition 2.3.5 can be generalized to continuous functions from a topological space to another.

Proposition 6.1.3.

Let XXand YY be topological spaces. Let 𝒫(X)\mathcal{P}(X) and 𝒫(Y)\mathcal{P}(Y) denote the power sets of XX and YY, respectively. Let

TX={IdX,ClX:𝒫(X)𝒫(X) given by AXAX¯},{{T}_{X}}=\{{\operatorname{Id}_{X}},{\operatorname{Cl}_{X}}:\mathcal{P}(X)\to\mathcal{P}(X)\text{ given by }{{A}_{X}}\mapsto\overline{{{A}_{X}}}\},

and

TY={IdY,ClY:𝒫(Y)𝒫(Y) given by AYAY¯},{{T}_{Y}}=\{{\operatorname{Id}_{Y}},{\operatorname{Cl}_{Y}}:\mathcal{P}(Y)\to\mathcal{P}(Y)\text{ given by }{{A}_{Y}}\mapsto\overline{{{A}_{Y}}}\},

where IdX{\operatorname{Id}_{X}} and IdY{\operatorname{Id}_{Y}} denote the identity functions on 𝒫(X)\mathcal{P}(X) and 𝒫(Y)\mathcal{P}(Y), respectively. Let a map p:XYp:X\to Y induce a map p:𝒫(X)𝒫(Y)p^{*}:\mathcal{P}(X)\to\mathcal{P}(Y) as follows.

AX𝒫(X)\forall{{A}_{X}}\in\mathcal{P}(X) that is closed in XX, let p(AX)=p(AX)¯p^{*}({{A}_{X}})=\overline{p({{A}_{X}})}, where p(AX)p({{A}_{X}}) denotes the set {p(x)|xAX}\{p(x)\,|\,x\in{{A}_{X}}\} for convenience. And AX𝒫(X)\forall{{A}_{X}}\in\mathcal{P}(X) which is not closed in XX, let p(AX)=p(AX)p^{*}({{A}_{X}})=p({{A}_{X}}).

Let θ\theta be the map from TX{{T}_{X}} to TY{{T}_{Y}} given by θ(IdX)=IdY\theta({\operatorname{Id}_{X}})={\operatorname{Id}_{Y}} and θ(ClX)=ClY\theta({\operatorname{Cl}_{X}})={\operatorname{Cl}_{Y}}. Then pp is continuous if and only if pp^{*} is a θ\theta-morphism.

Proof.

Our proof is analogous to that of Proposition 2.3.5.

Both TX{{T}_{X}} and TY{{T}_{Y}} are operator semigroups. 𝒫(X)TX=𝒫(X){{\left\langle\mathcal{P}(X)\right\rangle}_{{{T}_{X}}}}=\mathcal{P}(X) and 𝒫(Y)TY=𝒫(Y){{\left\langle\mathcal{P}(Y)\right\rangle}_{{{T}_{Y}}}}=\mathcal{P}(Y) are TX{{T}_{X}}-space and TY{{T}_{Y}}-space, respectively.

1. Necessity Suppose that pp is continuous. To prove that pp^{*} is a θ-\theta\text{-}morphism, it is sufficient to show that AX𝒫(X)\forall{{A}_{X}}\in\mathcal{P}(X) and fTXf\in{{T}_{X}}, p(f(AX))=θ(f)(p(AX))p^{*}(f({{A}_{X}}))=\theta(f)(p^{*}({{A}_{X}})). The equation holds when f=IdXf=\text{I}{{\text{d}}_{X}}. So it suffices to show that

AX𝒫(X)\forall{{A}_{X}}\in\mathcal{P}(X), p(ClX(AX))=ClY(p(AX))(=θ(ClX)(p(AX)))p^{*}(\text{C}{{\text{l}}_{X}}({{A}_{X}}))=\text{C}{{\text{l}}_{Y}}(p^{*}({{A}_{X}}))(=\theta(\text{C}{{\text{l}}_{X}})(p^{*}({{A}_{X}}))).

If AX=AX¯{{A}_{X}}=\overline{{{A}_{X}}}, by the definitions of pp^{*}, ClX\text{C}{{\text{l}}_{X}} and ClY\text{C}{{\text{l}}_{Y}},

ClY(p(AX))=ClY(p(AX)¯)=p(AX)¯=p(AX)=p(ClX(AX))\text{C}{{\text{l}}_{Y}}(p^{*}({{A}_{X}}))=\text{C}{{\text{l}}_{Y}}(\overline{p({{A}_{X}})})=\overline{p({{A}_{X}})}=p^{*}({{A}_{X}})=p^{*}(\text{C}{{\text{l}}_{X}}({{A}_{X}})),

as desired.

Since pp is continuous, p(AX¯)p(AX)¯p(\overline{{{A}_{X}}})\subseteq\overline{p({{A}_{X}})} (see e.g. [6]). Hence p(AX¯)¯p(AX)¯\overline{p(\overline{{{A}_{X}}})}\subseteq\overline{p({{A}_{X}})}. On the other hand, p(AX¯)¯p(AX)¯\overline{p(\overline{{{A}_{X}}})}\supseteq\overline{p({{A}_{X}})} because p(AX¯)p(AX)p(\overline{{{A}_{X}}})\supseteq p({{A}_{X}}). So p(AX¯)¯=p(AX)¯\overline{p(\overline{{{A}_{X}}})}=\overline{p({{A}_{X}})}. Thus if AXAX¯{{A}_{X}}\neq\overline{{{A}_{X}}}, then

ClY(p(AX))=ClY(p(AX))=p(AX)¯=p(AX¯)¯=p(AX¯)=p(ClX(AX))\text{C}{{\text{l}}_{Y}}(p^{*}({{A}_{X}}))=\text{C}{{\text{l}}_{Y}}(p({{A}_{X}}))=\overline{p({{A}_{X}})}=\overline{p(\overline{{{A}_{X}}})}=p^{*}(\overline{{{A}_{X}}})=p^{*}(\text{C}{{\text{l}}_{X}}({{A}_{X}})),

as desired.

2. Sufficiency Suppose that pp^{*} is a θ\theta-morphism. To prove that pp is continuous, we only need to show that p1(AY)=p1(AY)¯{{p}^{-1}}({{A}_{Y}})=\overline{{{p}^{-1}}({{A}_{Y}})}, AY𝒫(Y)\forall{{A}_{Y}}\in\mathcal{P}(Y) with AY=AY¯{{A}_{Y}}=\overline{{{A}_{Y}}}.

Assume AY𝒫(Y)\exists{{A}_{Y}}\in\mathcal{P}(Y) with AY=AY¯{{A}_{Y}}=\overline{{{A}_{Y}}} and p1(AY)p1(AY)¯{{p}^{-1}}({{A}_{Y}})\neq\overline{{{p}^{-1}}({{A}_{Y}})}. Let B=p1(AY)B={{p}^{-1}}({{A}_{Y}}). Then BB¯B\neq\overline{B} and p(B¯)AYp(\overline{B})\nsubseteq{{A}_{Y}} (because otherwise B=B¯B=\overline{B}). Hence

p(ClX(B))=p(B¯)=p(B¯)¯p(B¯)AYp^{*}(\text{C}{{\text{l}}_{X}}(B))=p^{*}(\overline{B})=\overline{p(\overline{B})}\supseteq p(\overline{B})\nsubseteq{{A}_{Y}},

but

ClY(p(B))=ClY(p(B))=p(B)¯AY¯=AY\text{C}{{\text{l}}_{Y}}(p^{*}(B))=\text{C}{{\text{l}}_{Y}}(p(B))=\overline{p(B)}\subseteq\overline{{{A}_{Y}}}={{A}_{Y}}

because p(B)AYp(B)\subseteq{{A}_{Y}}. Therefore, p(ClX(B))ClY(p(B))(=θ(ClX)(p(B)))p^{*}(\text{C}{{\text{l}}_{X}}(B))\neq\text{C}{{\text{l}}_{Y}}(p^{*}(B))(=\theta(\text{C}{{\text{l}}_{X}})(p^{*}(B))), which is contrary to the assumption that pp^{*} is a θ\theta-morphism. ∎

However, when we tried to generalize Proposition 2.3.2 to the case of θ\theta-morphisms, we found that in Definition 6.1.1, the condition θ\theta being a map is too strong for ring homomorphisms. Basically this is because in Example 2.1.2, the map τ\tau which induces TT is not necessarily injective (and so θ\theta defined in Proposition 6.1.10 is not necessarily a map). Hence we generalize Definition 6.1.1 to Definition 6.1.8, where the condition θ\theta being a map is replaced by a weaker one.

First, we allow θ\theta to be a binary relation, not necessarily a function.

Notation 6.1.4.

Let D1D_{1} (resp. D2D_{2}) be a set, let M1M_{1} (resp. M2M_{2}) be a set of functions from D1D_{1} to D1D_{1} (resp. a set of functions from D2D_{2} to D2D_{2}), let binary relation θM1×M2\theta\subseteq{{M}_{1}}\times{{M}_{2}}, and let AD2A\subseteq{{D}_{2}}. We denote by θ|A\theta{{|}_{A}} the binary relation {(f,g|A)|(f,g)θ}\{(f,g{{|}_{A}})\,|\,(f,g)\in\theta\}, where g|Ag{{|}_{A}} is the restriction of gg to AA.

Definition 6.1.5.

Let θ\theta and AA be defined as in Notation 6.1.4. Let fDomθ(M1)f\in\operatorname{Dom}\theta(\subseteq M_{1}) and aAa\in A. If (f,g1),(f,g2)θ\forall(f,g_{1}),(f,g_{2})\in\theta, g1(a)=g2(a)g_{1}(a)=g_{2}(a), then we say that θ(f)(a)\theta(f)(a) is well-defined and let θ(f)(a)=g1(a)\theta(f)(a)=g_{1}(a).

Proposition 6.1.6.

Let θ\theta and AA be defined as in Notation 6.1.4. Then the following statements are equivalent:

  1. (i)

    θ|A\theta{{|}_{A}} is a map.

  2. (ii)

    θ(f)(a)\theta(f)(a) is well-defined, fDomθ\forall f\in\operatorname{Dom}\theta and aAa\in A.

Proof.

θ|A\theta{{|}_{A}} is a map;

\Leftrightarrow (f,g1),(f,g2)θ\forall(f,g_{1}),(f,g_{2})\in\theta, g1|A=g2|Ag_{1}{{|}_{A}}=g_{2}{{|}_{A}};

\Leftrightarrow (f,g1),(f,g2)θ\forall(f,g_{1}),(f,g_{2})\in\theta and aAa\in A, g1(a)=g2(a)g_{1}(a)=g_{2}(a);

\Leftrightarrow fDomθ\forall f\in\operatorname{Dom}\theta and aAa\in A, θ(f)(a)\theta(f)(a) is well-defined. ∎

Moreover, to simplify our descriptions, we need a convention as follows.

Convention 6.1.7.

In this article, unless otherwise specified, by an equation we always imply that both sides of the equation are well-defined.

Then Definition 6.1.1 is generalized to

Definition 6.1.8.

Let T1T_{1} and T2T_{2} be operator semigroups, let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}}, and let ϕ\phi be a map from a T1T_{1}-space SS to a T2T_{2}-space. If fDomθ\forall f\in\operatorname{Dom}\theta and aSa\in S, ϕ(f(a))=θ(f)(ϕ(a))\phi(f(a))=\theta(f)(\phi(a)), then ϕ\phi is called a θ\theta-morphism.

Corollary 6.1.9.

Let ϕ\phi be a map, not necessarily a θ\theta-morphism, given as in Definition 6.1.8. If ϕ\phi is a θ\theta-morphism, then θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map. Conversely, if θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map, then fDomθ\forall f\in\operatorname{Dom}\theta and aSa\in S, θ(f)(ϕ(a))\theta(f)(\phi(a)) is well-defined.

Proof.

If ϕ\phi is a θ\theta-morphism, then by Definition 6.1.8 and Convention 6.1.7, fDomθ\forall f\in\operatorname{Dom}\theta and aSa\in S, θ(f)(ϕ(a))\theta(f)(\phi(a)) is well-defined, and thus by Proposition 6.1.6, θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map. The converse also follows from Proposition 6.1.6. ∎

Remark.

By Corollary 6.1.9, to define a θ\theta-morphism, we may dismiss Convention 6.1.7 and put the condition θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} being a map into Definition 6.1.8. However, later we shall find that it is convenient to use Convention 6.1.7 when things become complicated.

From Corollary 6.1.9, we can tell that the condition θ\theta being a map in Definition 6.1.1 is replaced by a weaker one (i.e. θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} being a map) in Definition 6.1.8. So Definition 6.1.1 is a special case of Definition 6.1.8.

In the remaining part of Section 6 and Section 7, a θ\theta-morphism is always defined by Definition 6.1.8 unless otherwise specified.

Now Proposition 2.3.2 can be generalized to

Proposition 6.1.10.

Let F/BF/B ((resp. F/B{F}^{\prime}/{B}^{\prime})) be a field extension, let TT ((resp. T{T}^{\prime})) be the corresponding operator semigroup defined as in Example 2.1.2, let φ:BB\varphi:B\to{B}^{\prime} be a field isomorphism, and let ϑ:B[x]B[x]\vartheta:B[x]\to{B}^{\prime}[x] be the map given by fff\mapsto{f}^{\prime} where

f(x)=φ(a0)+φ(a1)x++φ(an)xn{f}^{\prime}(x)=\varphi({{a}_{0}})+\varphi({{a}_{1}})x+\cdots+\varphi({{a}_{n}}){{x}^{n}} for f(x)=a0+a1x++anxnf(x)={{a}_{0}}+{{a}_{1}}x+\cdots+{{a}_{n}}{{x}^{n}}.

Let θ={(τ(f),τ(ϑ(f))|fB[x]}\theta=\{(\tau(f),{\tau}^{\prime}(\vartheta(f))\,|\,f\in B[x]\}, where τ:B[x]T\tau:B[x]\to T and τ:B[x]T\tau^{\prime}:B^{\prime}[x]\to T^{\prime} are the maps which induce TT and TT^{\prime}, respectively, defined as in Example 2.1.2. Then uF\forall u\in F and uF{u}^{\prime}\in{F}^{\prime}, a map ϕ:uTuT\phi:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle{{u}^{\prime}}\right\rangle}_{{{T}^{\prime}}}} is a ring homomorphism extending φ\varphi if and only if ϕ\phi is a θ\theta-morphism.

Proof.

Obviously, θT×T\theta\subseteq T\times{T}^{\prime} with Domθ=T\operatorname{Dom}\theta=T, and uT{{\left\langle u\right\rangle}_{T}} and uT{{\left\langle{{u}^{\prime}}\right\rangle}_{{{T}^{\prime}}}} are the rings B[u]B[u] and B[u]{B}^{\prime}[{u}^{\prime}], respectively.

1. Sufficiency Suppose that ϕ\phi is a θ\theta-morphism. By Definition 6.1.8, ϕ(g(a))=θ(g)(ϕ(a))\phi(g(a))=\theta(g)(\phi(a)), gDomθ\forall g\in\operatorname{Dom}\theta and auTa\in{{\left\langle u\right\rangle}_{T}}.

Let gDomθ(=T)g\in\operatorname{Dom}\theta(=T) be a constant polynomial function given by ac(B),aFa\mapsto c(\in B),\forall a\in F. Then by the definitions of θ\theta and ϑ\vartheta, we can tell that θ(g)\theta(g) is the constant polynomial function given by aφ(c)(B),aFa^{\prime}\mapsto\varphi(c)(\in B^{\prime}),\forall a^{\prime}\in F^{\prime}. Thus, cB\forall c\in B, ϕ(c)(=ϕ(g(a))=θ(g)(ϕ(a)))=φ(c)\phi(c)(=\phi(g(a))=\theta(g)(\phi(a)))=\varphi(c), that is, ϕ\phi extends φ\varphi.

Let a,buTa,b\in{{\left\langle u\right\rangle}_{T}}. Then p(x),q(x)B[x]\exists p(x),q(x)\in B[x] with p(u)=ap(u)=a and q(u)=bq(u)=b. Hence

ϕ(a+b)\phi(a+b)

=ϕ(p(u)+q(u))=\phi(p(u)+q(u))

=ϕ((p+q)(u))=\phi((p+q)(u))

=ϕ(τ(p+q)(u))=\phi(\tau(p+q)(u))

=θ(τ(p+q))(ϕ(u))=\theta(\tau(p+q))(\phi(u)) (since ϕ\phi is a θ\theta-morphism)

=τ(ϑ(p+q))(ϕ(u))={\tau}^{\prime}(\vartheta(p+q))(\phi(u)) (by the definition of θ\theta)

=ϑ(p+q)(ϕ(u))=\vartheta(p+q)(\phi(u))

=ϑ(p)(ϕ(u))+ϑ(q)(ϕ(u))=\vartheta(p)(\phi(u))+\vartheta(q)(\phi(u)) (since φ:BB\varphi:B\to{B}^{\prime} is a field isomorphism)

=τ(ϑ(p))(ϕ(u))+τ(ϑ(q))(ϕ(u))={\tau}^{\prime}(\vartheta(p))(\phi(u))+{\tau}^{\prime}(\vartheta(q))(\phi(u))

=θ(τ(p))(ϕ(u))+θ(τ(q))(ϕ(u))=\theta(\tau(p))(\phi(u))+\theta(\tau(q))(\phi(u)) (since by Corollary 6.1.9, θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a function)

=ϕ(τ(p)(u))+ϕ(τ(q)(u))=\phi(\tau(p)(u))+\phi(\tau(q)(u)) (since ϕ\phi is a θ\theta-morphism)

=ϕ(p(u))+ϕ(q(u))=\phi(p(u))+\phi(q(u))

=ϕ(a)+ϕ(b)=\phi(a)+\phi(b).

And

ϕ(ab)\phi(ab)

=ϕ(p(u)q(u))=\phi(p(u)q(u))

=ϕ((pq)(u))=\phi((pq)(u))

=ϕ(τ(pq)(u))=\phi(\tau(pq)(u))

=θ(τ(pq))(ϕ(u))=\theta(\tau(pq))(\phi(u)) (since ϕ\phi is a θ\theta-morphism)

=τ(ϑ(pq))(ϕ(u))={\tau}^{\prime}(\vartheta(pq))(\phi(u)) (by the definition of θ\theta)

=ϑ(pq)(ϕ(u))=\vartheta(pq)(\phi(u))

=ϑ(p)(ϕ(u))ϑ(q)(ϕ(u))=\vartheta(p)(\phi(u))\vartheta(q)(\phi(u)) (since φ:BB\varphi:B\to{B}^{\prime} is a field isomorphism)

=τ(ϑ(p))(ϕ(u))τ(ϑ(q))(ϕ(u))={\tau}^{\prime}(\vartheta(p))(\phi(u)){\tau}^{\prime}(\vartheta(q))(\phi(u))

=θ(τ(p))(ϕ(u))θ(τ(q))(ϕ(u))=\theta(\tau(p))(\phi(u))\theta(\tau(q))(\phi(u)) (since by Corollary 6.1.9, θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a function)

=ϕ(τ(p)(u))ϕ(τ(q)(u))=\phi(\tau(p)(u))\phi(\tau(q)(u)) (since ϕ\phi is a θ\theta-morphism)

=ϕ(p(u))ϕ(q(u))=\phi(p(u))\phi(q(u))

=ϕ(a)ϕ(b)=\phi(a)\phi(b).

Therefore ϕ\phi is a ring homomorphism extending φ\varphi.

2.Necessity Suppose that ϕ\phi is a ring homomorphism extending φ\varphi. Then cB\forall c\in B, ϕ(c)=φ(c)\phi(c)=\varphi(c), and a,buT\forall a,b\in{{\left\langle u\right\rangle}_{T}}, ϕ(a+b)=ϕ(a)+ϕ(b)\phi(a+b)=\phi(a)+\phi(b) and ϕ(ab)=ϕ(a)ϕ(b)\phi(ab)=\phi(a)\phi(b).

Let zuTz\in{{\left\langle u\right\rangle}_{T}} and let f=a0+a1x++anxnB[x]f={{a}_{0}}+{{a}_{1}}x+\cdots+{{a}_{n}}{{x}^{n}}\in B[x]. Then

ϕ(f(z))\phi(f(z))

=ϕ(a0+a1z++anzn)=\phi({a_{0}}+{a_{1}}z+\cdots+{a_{n}}{z^{n}})

=φ(a0)+φ(a1)ϕ(z)++φ(an)(ϕ(z))n=\varphi({a_{0}})+\varphi({a_{1}})\phi(z)+\cdots+\varphi({a_{n}}){(\phi(z))^{n}}

=ϑ(f)(ϕ(z)),=\vartheta(f)(\phi(z)),

that is

(6.1) ϕ(f(z))=ϑ(f)(ϕ(z))\phi(f(z))=\vartheta(f)(\phi(z))

To show that ϕ\phi is a θ\theta-morphism, by Corollary 6.1.9, we need to show that θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map. Let (h,g|Imϕ),(h,g|Imϕ)θ|Imϕ(h,g{{|}_{\operatorname{Im}\phi}}),(h,{g}^{\prime}{{|}_{\operatorname{Im}\phi}})\in\theta{{|}_{\operatorname{Im}\phi}} with (h,g),(h,g)θ(h,g),(h,{g}^{\prime})\in\theta. By the definition of θ\theta, l,mB[x]\exists l,m\in B[x] such that (h,g)=(τ(l),τ(ϑ(l)))(h,g)=(\tau(l),{\tau}^{\prime}(\vartheta(l))) and (h,g)=(τ(m),τ(ϑ(m)))(h,{g}^{\prime})=(\tau(m),{\tau}^{\prime}(\vartheta(m))), and so τ(l)=τ(m)\tau(l)=\tau(m).

Assume g|Imϕg|Imϕg{{|}_{\operatorname{Im}\phi}}\neq{g}^{\prime}{{|}_{\operatorname{Im}\phi}}; that is, τ(ϑ(l))|Imϕτ(ϑ(m))|Imϕ{\tau}^{\prime}(\vartheta(l)){{|}_{\operatorname{Im}\phi}}\neq{\tau}^{\prime}(\vartheta(m)){{|}_{\operatorname{Im}\phi}}. Then zuT\exists z\in{{\left\langle u\right\rangle}_{T}} such that ϑ(l)(ϕ(z))ϑ(m)(ϕ(z))\vartheta(l)(\phi(z))\neq\vartheta(m)(\phi(z)). However, since τ(l)=τ(m)\tau(l)=\tau(m), ϕ(l(z))=ϕ(m(z))\phi(l(z))=\phi(m(z)), and so from Equation ((6.1)), we can tell that ϑ(l)(ϕ(z))=ϕ(l(z))=ϕ(m(z))=ϑ(m)(ϕ(z))\vartheta(l)(\phi(z))=\phi(l(z))=\phi(m(z))=\vartheta(m)(\phi(z)), a contradiction.

Hence θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} must be a map. Then zuT\forall z\in{{\left\langle u\right\rangle}_{T}} and fB[x]f\in B[x],

ϕ(τ(f)(z))\phi(\tau(f)(z))

=ϕ(f(z))=\phi(f(z))

=ϑ(f)(ϕ(z))=\vartheta(f)(\phi(z)) (by Equation ((6.1)))

=τ(ϑ(f))(ϕ(z))={\tau}^{\prime}(\vartheta(f))(\phi(z))

=θ(τ(f))(ϕ(z))=\theta(\tau(f))(\phi(z)) (by the definition of θ\theta and the fact that θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map).

So by Definition 6.1.8, ϕ\phi is a θ\theta-morphism. ∎

Note that up to now all polynomial rings involved are in one variable. We shall discuss ring homomorphisms involving polynomial rings in more than one variable in Sections 8 and 9.

For Imϕ\operatorname{Im}\phi of θ\theta-morphisms ϕ\phi, we generalize Proposition 2.3.6 as follows.

Proposition 6.1.11.

Let ϕ\phi be a θ\theta-morphism from a T1T_{1}-space S1S_{1} to a T2T_{2}-space S2S_{2}. Suppose Imθ=T2\operatorname{Im}\theta=T_{2}. Then ImϕqS2\operatorname{Im}\phi\leq_{q}{S_{2}}, i.e. Imϕ\operatorname{Im}\phi is a quasi-T2T_{2}-subspace of S2S_{2} (Definition 2.2.8). Moreover, if IdT2\operatorname{Id}\in T_{2} or more generally, ImϕImϕT2\operatorname{Im}\phi\subseteq{{\left\langle\operatorname{Im}\phi\right\rangle}_{T_{2}}}, then ImϕS2\operatorname{Im}\phi\leq{{S}_{2}}, i.e. Imϕ\operatorname{Im}\phi is a T2T_{2}-subspace of S2S_{2} (Definition 2.2.1).

Proof.

By Definition 6.1.8, aS1\forall a\in{{S}_{1}} and fDomθf\in\operatorname{Dom}\theta, θ(f)(ϕ(a))=ϕ(f(a))Imϕ\theta(f)(\phi(a))=\phi(f(a))\in\operatorname{Im}\phi, and thus ϕ(a)T2Imϕ{{\left\langle\phi(a)\right\rangle}_{T_{2}}}\subseteq\operatorname{Im}\phi because Imθ=T2\operatorname{Im}\theta=T_{2}. Hence ImϕT2Imϕ{{\left\langle\operatorname{Im}\phi\right\rangle}_{T_{2}}}\subseteq\operatorname{Im}\phi, and so by Definition 2.2.8, Imϕ\operatorname{Im}\phi is a quasi-T2T_{2}-subspace of S2S_{2}.

Moreover, if IdT2\operatorname{Id}\in T_{2} or ImϕImϕT2\operatorname{Im}\phi\subseteq{{\left\langle\operatorname{Im}\phi\right\rangle}_{T_{2}}}, then by Proposition 2.2.10, Imϕ\operatorname{Im}\phi is a T2T_{2}-space and hence ImϕS2\operatorname{Im}\phi\leq{{S}_{2}}. ∎

6.2. θ\theta-isomorphisms

The notion of TT-isomorphism is generalized as follows.

Definition 6.2.1.

Let ϕ\phi be a θ\theta-morphism from a T1T_{1}-space S1S_{1} to a T2T_{2}-space S2S_{2}. If ϕ\phi is bijective, then we call ϕ\phi a θ\theta-isomorphism. Moreover, we denote by Isoθ(S1,S2)\operatorname{Iso}_{\theta}({{S}_{1}},{{S}_{2}}) the set of all θ\theta-isomorphisms from S1{{S}_{1}} to S2{{S}_{2}}.

To justify the definition, we need to show that ϕ1\phi^{-1} is a θ1\theta^{-1}-morphism from S2S_{2} to S1S_{1}. For this purpose, we need

Lemma 6.2.2.

Let ϕ\phi be a θ\theta-isomorphism from a T1T_{1}-space S1S_{1} to a T2T_{2}-space S2S_{2}. Then θ1|S1\theta^{-1}{{|}_{S_{1}}} is a map, where θ1:={(g,f)|(f,g)θ}\theta^{-1}:=\{(g,f)\,|\,(f,g)\in\theta\}.

Proof.

Let (g,f1),(g,f2)θ1(g,f_{1}),(g,f_{2})\in\theta^{-1} and aS1a\in S_{1}. Then to show that θ1|S1\theta^{-1}{{|}_{S_{1}}} is a map, by Proposition 6.1.6 and Definition 6.1.5, it suffices to show f1(a)=f2(a)f_{1}(a)=f_{2}(a). Moreover, since ϕ\phi is injective, we only need to show ϕ(f1(a))=ϕ(f2(a))\phi(f_{1}(a))=\phi(f_{2}(a)):

ϕ(f1(a))\displaystyle\phi(f_{1}(a)) =θ(f1)(ϕ(a)) (by Definition 6.1.8)\displaystyle=\theta(f_{1})(\phi(a))\text{ (by Definition }\ref{5.3.5})
=g(ϕ(a)) (because (f1,g)θ and θ|Imϕ is a map by Corollary 6.1.9)\displaystyle=g(\phi(a))\text{ (because }(f_{1},g)\in\theta\text{ and }\theta{|}_{\operatorname{Im}\phi}\text{ is a map by Corollary }\ref{5.3.c})
=θ(f2)(ϕ(a)) (because (f2,g)θ and θ|Imϕ is a map)\displaystyle=\theta(f_{2})(\phi(a))\text{ (because }(f_{2},g)\in\theta\text{ and }\theta{|}_{\operatorname{Im}\phi}\text{ is a map})
=ϕ(f2(a)) (by Definition 6.1.8).\displaystyle=\phi(f_{2}(a))\text{ (by Definition }\ref{5.3.5}).

Proposition 6.2.3.

Let S1S_{1} and S2S_{2} be a T1T_{1}-space and a T2T_{2}-space, respectively, let ϕIsoθ(S1,S2)\phi\in\operatorname{Iso}_{\theta}({{S}_{1}},{{S}_{2}}) and let ϕ1{{\phi}^{-1}} be the inverse map of ϕ\phi. Then ϕ1Isoθ1(S2,S1){{\phi}^{-1}}\in\operatorname{Iso}_{\theta^{-1}}({{S}_{2}},{{S}_{1}}).

Proof.

By Definition 6.1.8, fDomθ\forall f\in\operatorname{Dom}\theta and aS1a\in{{S}_{1}}, ϕ(f(a))=θ(f)(ϕ(a))\phi(f(a))=\theta(f)(\phi(a)), and so f(a)=ϕ1(θ(f)(ϕ(a)))f(a)={{\phi}^{-1}}(\theta(f)(\phi(a))). Hence

(6.2) f(ϕ1(b))=ϕ1(θ(f)(b)),fDomθ and bS2.f({{\phi}^{-1}}(b))={{\phi}^{-1}}(\theta(f)(b)),\forall f\in\operatorname{Dom}\theta\text{ and }b\in{{S}_{2}}.

By Lemma 6.2.2, θ1|S1\theta^{-1}{{|}_{S_{1}}} is a map. Then by Proposition 6.1.6, θ1(g)(a)\theta^{-1}(g)(a) is well-defined, gDomθ1(=Imθ)\forall g\in\operatorname{Dom}\theta^{-1}(=\operatorname{Im}\theta) and aS1a\in S_{1}. Thus it follows from (6.2) that

θ1(g)(ϕ1(b))=ϕ1(g(b)),gDomθ1 and bS2.\theta^{-1}(g)({{\phi}^{-1}}(b))={{\phi}^{-1}}(g(b)),\forall g\in\operatorname{Dom}\theta^{-1}\text{ and }b\in{{S}_{2}}.

So by Definition 6.1.8, ϕ1\phi^{-1} is a θ1\theta^{-1}-morphism from S2S_{2} to S1S_{1}.

Moreover, since ϕ1{\phi}^{-1} is bijective, by Definition 6.2.1, ϕ1Isoθ1(S2,S1){{\phi}^{-1}}\in\operatorname{Iso}_{\theta^{-1}}({{S}_{2}},{{S}_{1}}). ∎

7. Constructions of the generalized morphisms and isomorphisms

For a TT-morphism or a θ\theta-morphism from a TT-space SS, Subsections 7.1, 7.2 and 7.3 discuss how to construct it by a map from a set which generates SS. Specifically, Subsection 7.1 is about a construction of a TT-morphism from uT{{\left\langle u\right\rangle}_{T}}, where uDu\in D, by a map from {u}\{u\}. And Subsection 7.2 generalizes the construction to a TT-morphism from UT{{\left\langle U\right\rangle}_{T}}, where UDU\subseteq D, by a map from UU. Then in Subsection 7.3, the construction is further generalized to θ\theta-morphisms. In Subsection 7.4, we shall introduce a way of constructing TT-morphisms in terms of topology.

Moreover, some notions and results in algebra are generalized in Section 7 as “byproducts” (see e.g. Propositions 7.1.2, 7.1.3 and 7.1.4).

7.1. A construction of TT-morphisms from uT(uD){{\left\langle u\right\rangle}_{T}}(u\in D)

Let F/BF/B be a field extension and let u,vFu,v\in F be algebraic over BB. If uu and vv share the same minimal polynomial over BB, then

p(u)=q(u)p(v)=q(v),p(x),q(x)B[x].p(u)=q(u)\Leftrightarrow p(v)=q(v),\forall p(x),q(x)\in B[x].

In light of this observation, we introduce the following.

Notation 7.1.1.

Let DD be a set, let MM be a set of maps from DD to DD, and let u,vDu,v\in D. Then by uMvu\xrightarrow{M}v we mean that

f,gM\forall f,g\in M, if f(u)=g(u)f(u)=g(u), then f(v)=g(v)f(v)=g(v).

Moreover, by uMvu\overset{M}{\longleftrightarrow}v we mean that

uMvu\xrightarrow{M}v and vMuv\xrightarrow{M}u,

or equivalently, by uMvu\overset{M}{\longleftrightarrow}v we mean that

f(u)=g(u)f(v)=g(v)f(u)=g(u)\Leftrightarrow f(v)=g(v), f,gM\forall f,g\in M.

Besides, we denote by [u)M{{[u)}_{M}} and [u]M{[u]_{M}} the set {wD|uMw}\{w\in D\,|\,u\xrightarrow{M}w\} and the set {wD|uMw}\{w\in D\,|\,u\overset{M}{\longleftrightarrow}w\}, respectively.

Remark.

Apparently, uMvu\overset{M}{\longleftrightarrow}v defines an equivalence relation on DD.

The following serves as an example.

Proposition 7.1.2.

Let BB be a subfield of a field FF. Let TT be the operator semigroup defined in Example 2.1.2.

  1. (a)

    If uFu\in F is algebraic over BB, then

    [u]T=[u)T={{[u]}_{T}}={{[u)}_{T}}={all roots of min(B,u)\min(B,u)},

    where min(B,u)\min(B,u) is the minimal polynomial of uu over BB.

  2. (b)

    If uFu\in F is transcendental over BB, then

    [u)T=F{{[u)}_{T}}=F and [u]T={{[u]}_{T}}={vF|vv\in F\,|\,v is transcendental over BB}.

Proof.

For (a): Let uFu\in F be algebraic over BB.

Let w[u)Tw\in{{[u)}_{T}}. By Notation 7.1.1, f,gT\forall f^{*},g^{*}\in T (cf. Example 2.1.2), f(u)=g(u)f^{*}(u)=g^{*}(u) implies f(w)=g(w)f^{*}(w)=g^{*}(w). So f(x),g(x)B[x]\forall f(x),g(x)\in B[x], f(u)=g(u)f(u)=g(u) implies f(w)=g(w)f(w)=g(w). Thus ww is a root of min(B,u)\min(B,u). Hence

[u)T{{[u)}_{T}}\subseteq{all roots of min(B,u)\min(B,u)}.

Moreover, let vFv\in F be a root of min(B,u)\min(B,u). Then f(x),g(x)B[x]\forall f(x),g(x)\in B[x], f(u)=g(u)f(v)=g(v)f(u)=g(u)\Leftrightarrow f(v)=g(v). It follows that f,gT\forall f^{*},g^{*}\in T, f(u)=g(u)f(v)=g(v)f^{*}(u)=g^{*}(u)\Leftrightarrow f^{*}(v)=g^{*}(v); that is, uTvu\overset{T}{\longleftrightarrow}v, and so v[u]Tv\in{{[u]}_{T}}. Hence

([u)T)({{[u)}_{T}}\subseteq){all roots of min(B,u)\min(B,u)}[u]T\subseteq{{[u]}_{T}}.

On the other hand, [u)T[u]T{{[u)}_{T}}\supseteq{{[u]}_{T}} by Notation 7.1.1. Therefore,

[u)T={{[u)}_{T}}={all roots of min(B,u)\min(B,u)}=[u]T={{[u]}_{T}}.

For (b): Let uFu\in F be transcendental over BB.

Then f(x),g(x)B[x]\forall f(x),g(x)\in B[x], f(u)=g(u)f(u)=g(u) implies f=gf=g. So f,gT\forall f^{*},g^{*}\in T, f(u)=g(u)f^{*}(u)=g^{*}(u) implies f=gf^{*}=g^{*}. Then vF\forall v\in F, uTvu\xrightarrow{T}v, and so [u)T=F{{[u)}_{T}}=F.

Hence [u]T={vF|vTu}={{[u]}_{T}}=\{v\in F\,|\,v\overset{T}{\longleftrightarrow}u\}={vF|vTuv\in F\,|\,v\xrightarrow{T}u}.

vF\forall v\in F, if vv is also transcendental over BB, then as was just shown for uu, vTuv\xrightarrow{T}u. So

[u]T={vF|vTu}{{[u]}_{T}}=\{v\in F\,|\,v\xrightarrow{T}u\}\supseteq{vF|vv\in F\,|\,v is transcendental over BB}.

On the other hand, if vFv\in F is algebraic over BB, then by (a), u[v)Tu\notin{{[v)}_{T}}, i.e. vTuv\xrightarrow{T}u is false; That is, if vTuv\xrightarrow{T}u is true, then vv is transcendental over BB. So

[u]T={vF|vTu}{{[u]}_{T}}=\{v\in F\,|\,v\xrightarrow{T}u\}\subseteq{vF|vv\in F\,|\,v is transcendental over BB}.

Therefore, [u]T={{[u]}_{T}}={vF|vv\in F\,|\,v is transcendental over BB}. ∎

We will talk more about the notion of transcendental elements in Section 21.

Recall that in the classical Galois theory, each element of the Galois group of a polynomial permutes its roots. From Proposition 7.1.2, we can tell that the following two results generalize this well-known fact.

Proposition 7.1.3.

Let σ\sigma be a TT-morphism from a TT-space SS (to a TT-space). Then aS\forall a\in S, aTσ(a)a\xrightarrow{T}\sigma(a) ((and so σ(a)[a)T)\sigma(a)\in{{[a)}_{T}}).

Proof.

For aSa\in S, let f,gTf,g\in T such that f(a)=g(a)f(a)=g(a). Then by Definition 2.3.1, f(σ(a))=σ(f(a))=σ(g(a))=g(σ(a))f(\sigma(a))=\sigma(f(a))=\sigma(g(a))=g(\sigma(a)). Hence by Notation 7.1.1, aTσ(a)a\xrightarrow{T}\sigma(a). ∎

Proposition 7.1.4.

Let σ\sigma be a TT-morphism from a TT-space SS (to a TT-space) such that σ\sigma is injective. Then aS\forall a\in S, aTσ(a)a\overset{T}{\longleftrightarrow}\sigma(a) ((and so σ(a)[a]T)\sigma(a)\in{{[a]}_{T}}).

Proof.

Let aSa\in S. By Proposition 7.1.3, we only need to show σ(a)Ta\sigma(a)\xrightarrow{T}a. Let f,gTf,g\in T such that f(σ(a))=g(σ(a))f(\sigma(a))=g(\sigma(a)). Then by Definition 2.3.1, σ(f(a))=f(σ(a))=g(σ(a))=σ(g(a))\sigma(f(a))=f(\sigma(a))=g(\sigma(a))=\sigma(g(a)). Hence f(a)=g(a)f(a)=g(a) because σ\sigma is injective, as desired. ∎

The converse of Proposition 7.1.3 may imply a potential way to construct TT-morphisms. However, the converse of Proposition 7.1.3 is not true. That is, for a map σ\sigma from a TT-space SS, the condition “aS\forall a\in S, aTσ(a)a\xrightarrow{T}\sigma(a)” is not sufficient for σ\sigma to be a TT-morphism, as shown below.

Example 7.1.5.

Let RR be a differential ring with a derivation \partial such that {(a)|aR}=R\{\partial(a)\,|\,a\in R\}=R (it is easy to find such differential ring), let TT be the operator semigroup on RR given by {n|n+}\{{{\partial}^{n}}\,|\,n\in{{\mathbb{Z}}^{+}}\} (note that unlike the operator semigroup defined in Example 2.1.5, now IdT\operatorname{Id}\notin T), let CC be the constant ring of RR, and let a map σ:RR\sigma:R\to R be given by aa+ca\mapsto a+c, where cCc\in C and c0c\neq 0. Then f,gT\forall f,g\in T and aRa\in R, f(a)=g(a)f(a)=g(a) implies f(σ(a))=g(σ(a))f(\sigma(a))=g(\sigma(a)). Hence aR\forall a\in R, aTσ(a)a\xrightarrow{T}\sigma(a). But obviously fT\exists f\in T and aRa\in R such that f(σ(a))σ(f(a))f(\sigma(a))\neq\sigma(f(a)). So σ\sigma, a map from the TT-space RT=R{{\left\langle R\right\rangle}_{T}}=R to itself, is not a TT-morphism.

Thus, to make a map σ\sigma from a TT-space SS be a TT-morphism, we need to impose a condition on σ\sigma which is stronger than “aS\forall a\in S, aTσ(a)a\xrightarrow{T}\sigma(a)”. Before we do this by Proposition 7.1.7, let’s see a fact as follows.

Proposition 7.1.6.

Let u,vDu,v\in D. Then uTvu\xrightarrow{T}v implies f(u)Tf(v)f(u)\xrightarrow{T}f(v), fT\forall f\in T.

Remark.

So, if uTvu\xrightarrow{T}v and there exists a map σ:uTvT\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}} given by f(u)f(v),fTf(u)\mapsto f(v),\forall f\in T, then auT\forall a\in{{\left\langle u\right\rangle}_{T}}, aTσ(a)a\xrightarrow{T}\sigma(a) (since f(u)Tf(v)f(u)\xrightarrow{T}f(v), fT\forall f\in T). We shall see in Proposition 7.1.7 that the map σ\sigma is a TT-morphism.

Proof.

Suppose uTvu\xrightarrow{T}v. Let f,g,hTf,g,h\in T such that g(f(u))=h(f(u))g(f(u))=h(f(u)). Then by Notation 7.1.1, it suffices to show g(f(v))=h(f(v))g(f(v))=h(f(v)).

By Definition 2.1.1, gfTg\circ f\in T and hfTh\circ f\in T. Since uTvu\xrightarrow{T}v and g(f(u))=h(f(u))g(f(u))=h(f(u)), by Notation 7.1.1, we have g(f(v))=h(f(v))g(f(v))=h(f(v)), as desired. ∎

Proposition 7.1.7.

Let u,vDu,v\in D. Then

  1. (a)

    uTvu\xrightarrow{T}v if and only if there exists a map σ:uTvT\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}} given by f(u)f(v),fTf(u)\mapsto f(v),\forall f\in T.

  2. (b)

    The map σ\sigma given in (a) is a TT-morphism.

Proof.

For (a):

uTvu\xrightarrow{T}v;

\Leftrightarrow f,gT\forall f,g\in T, f(u)=g(u)f(u)=g(u) implies f(v)=g(v)f(v)=g(v);

\Leftrightarrow σ:uTvT\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}} given by f(u)f(v),fTf(u)\mapsto f(v),\forall f\in T, is a well-defined map.

For (b): Let auTa\in{{\left\langle u\right\rangle}_{T}}. Then fT\exists f\in T such that f(u)=af(u)=a. Let gTg\in T. Then σ(g(a))=σ(g(f(u)))\sigma(g(a))=\sigma(g(f(u))). By Definition 2.1.1, gfTg\circ f\in T, and so by the definition of σ\sigma in (a), σ(g(f(u)))=g(f(v))\sigma(g(f(u)))=g(f(v)). Besides, g(σ(a))=g(σ(f(u)))=g(f(v))g(\sigma(a))=g(\sigma(f(u)))=g(f(v)) also by the definition of σ\sigma. Combining the above equations, we have g(σ(a))=σ(g(a))g(\sigma(a))=\sigma(g(a)). Therefore, σ\sigma is a TT-morphism. ∎

For TT-isomorphisms (Definition 2.4.1), we have the following straightforward result of Proposition 7.1.7.

Corollary 7.1.8.

Let u,vDu,v\in D.

  1. (a)

    uTvu\overset{T}{\longleftrightarrow}v if and only if there exists a bijective map σ:uTvT\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}} given by f(u)f(v),fTf(u)\mapsto f(v),\forall f\in T.

  2. (b)

    The bijective map σ\sigma given in (a) is a TT-isomorphism.

The converse of statement (b) in Proposition 7.1.7 raises a question as follows.

Question 7.1.9.

For uDu\in D and a TT-morphism σ\sigma from uT{{\left\langle u\right\rangle}_{T}}, does there exist any vDv\in D such that σ(f(u))=f(v),fT\sigma(f(u))=f(v),\forall f\in T?

The significance of this question is that, in the cases where the answer is yes, we can construct a TT-morphism σ\sigma by letting σ(f(u))=f(v),fT\sigma(f(u))=f(v),\forall f\in T. Proposition 7.1.7 implies that there are cases where the answer to the question is positive, and we shall talk about these cases in Subsection 7.2. It seems not easy to find a case which gives negative answer to the question, but we found one as follows.

Example 7.1.10.

Define h:h:\mathbb{R}\to\mathbb{R} by xx+1x\mapsto x+1 and g:g:\mathbb{R}\to\mathbb{R} by xπxx\mapsto\pi x. Let T={h,g}T=\left\langle\{h,g\}\right\rangle (Definition 2.1.6) and S=1TS={{\left\langle 1\right\rangle}_{T}}.

Then each fTf\in T can be expressed as a finite composite of alternating “powers” of hh or gg as follows, where each hki{{h}^{{{k}_{i}}}} or gki{{g}^{{{k}_{i}}}} denotes the composite of ki{{k}_{i}} copies of hh or gg, respectively.

(7.1) f=hkngkn1hk2gk1,f={h^{{k_{n}}}}{g^{{k_{n-1}}}}\cdots{h^{{k_{2}}}}{g^{{k_{1}}}},
(7.2) f=gknhkn1hk2gk1,f={g^{{k_{n}}}}{h^{{k_{n-1}}}}\cdots{h^{{k_{2}}}}{g^{{k_{1}}}},
(7.3) f=hkngkn1gk2hk1,f={h^{{k_{n}}}}g^{{k_{n-1}}}\cdots{g^{{k_{2}}}}h^{{k_{1}}},

or

(7.4) f=gknhkn1gk2hk1,f={g^{{k_{n}}}}h^{{k_{n-1}}}\cdots{g^{{k_{2}}}}h^{{k_{1}}},

where nn is an even number for ((7.1)) and ((7.4)), nn is an odd number for ((7.2)) and ((7.3)), and ki+,i=1,,n{{k}_{i}}\in{{\mathbb{Z}}^{+}},\forall i=1,\cdots,n. In particular, hh has the form (7.3), where n=1n=1, and gg has the form (7.2), where n=1n=1.

Correspondingly, aS=1T\forall a\in S={{\left\langle 1\right\rangle}_{T}}, a=f(1)a=f(1) has the form

(7.5) a=((πk1+k2)πk3++kn2)πkn1+kn,a=(\cdots({\pi^{{k_{1}}}}+{k_{2}}){\pi^{{k_{3}}}}+\cdots+{k_{n-2}}){\pi^{{k_{n-1}}}}+{k_{n}},
(7.6) a=((πk1+k2)πk3++kn1)πkn,a=(\cdots({\pi^{{k_{1}}}}+{k_{2}}){\pi^{{k_{3}}}}+\cdots+{k_{n-1}}){\pi^{{k_{n}}}},
(7.7) a=(((1+k1)πk2+k3)πk4++kn2)πkn1+kn,a=(\cdots((1+{k_{1}}){\pi^{{k_{2}}}}+{k_{3}}){\pi^{{k_{4}}}}+\cdots+{k_{n-2}}){\pi^{{k_{n-1}}}}+{k_{n}},

or

(7.8) a=(((1+k1)πk2+k3)πk4++kn1)πkn,a=(\cdots((1+{k_{1}}){\pi^{{k_{2}}}}+{k_{3}}){\pi^{{k_{4}}}}+\cdots+{k_{n-1}}){\pi^{{k_{n}}}},

where nn is an even number for ((7.5)) and ((7.8)), nn is an odd number for ((7.6)) and ((7.7)), and ki+,i=1,,n{{k}_{i}}\in{{\mathbb{Z}}^{+}},\forall i=1,\cdots,n.

Let aSa\in S. Since π\pi is transcendental over \mathbb{Q}, the expression for aa in the above forms is unique. It follows that there is only one fTf\in T in the form of ((7.1)), ((7.2)), ((7.3)), or ((7.4)) such that a=f(1)a=f(1). Hence we can define a map σ:SS\sigma:S\to S as follows.

For all aSa\in S, if fTf\in T has the form ((7.1)) or ((7.2)) such that a=f(1)a=f(1), let σ(a)=f(π)\sigma(a)=f(\pi), which clearly belongs to SS, and if fTf\in T has the form ((7.3)) or ((7.4)) such that a=f(1)a=f(1), let σ(a)=a\sigma(a)=a.

Now let’s show σEndT(S)\sigma\in\operatorname{End}_{T}(S). Let lTl\in T.

Suppose that a=f(1)a=f(1) has the form ((7.5)) or ((7.6)). Then ff has the form ((7.1)) or ((7.2)), and so lfl\circ f also has the form ((7.1)) or ((7.2)). Hence by the definition of σ\sigma, σ(l(a))=σ(l(f(1)))=l(f(π))=l(σ(a))\sigma(l(a))=\sigma(l(f(1)))=l(f(\pi))=l(\sigma(a)), as desired.

Suppose that a=f(1)a=f(1) has the form ((7.7)) or ((7.8)). Then ff has the form ((7.3)) or ((7.4)), and so lfl\circ f also has the form ((7.3)) or ((7.4)). Hence by the definition of σ\sigma, σ(l(a))=σ(l(f(1)))=l(f(1))=l(a)=l(σ(a))\sigma(l(a))=\sigma(l(f(1)))=l(f(1))=l(a)=l(\sigma(a)), as desired.

Therefore, σ(l(a))=l(σ(a))\sigma(l(a))=l(\sigma(a)), lT\forall l\in T and aSa\in S. So σEndT(S)\sigma\in\operatorname{End}_{T}(S).

For 1 and σ\sigma, assume that the answer to Question 7.1.9 is positive, that is, there exists vv\in\mathbb{R} such that σ(f(1))=f(v),fT\sigma(f(1))=f(v),\forall f\in T. Recall the definitions of hh and gg at the beginning of the example. Since hh has the form (7.3), h(v)=σ(h(1))=h(1)=2h(v)=\sigma(h(1))=h(1)=2, and so v=1v=1; but since gg has the form (7.2), g(v)=σ(g(1))=g(π)=π2g(v)=\sigma(g(1))=g(\pi)={{\pi}^{2}}, and so v=πv=\pi, a contradiction.

Therefore, for 1 and σ\sigma, the answer to Question 7.1.9 is negative.

Recall that right after Proposition 2.5.2, we claimed that for a TT-space SS and HEndT(S)H\subseteq\operatorname{End}_{T}(S), it is possible that SH{{S}^{H}} is not a TT-space. Now we show this claim as follows.

Proof.

Let TT, SS and σ\sigma be the same as in Example 7.1.10 and let D=S{1}D=S\bigcup\{1\}. Let us restrict the domain of TT to DD. We can tell that the “new” TT, which we still denote by TT for convenience, is an operator semigroup on DD, and all the argument in Example 7.1.10 still holds. In particular, σEndT(S)\sigma\in\operatorname{End}_{T}(S), and 2Sσ2\in{{S}^{\sigma}} because 2 has the form ((7.7)). Suppose that Sσ{{S}^{\sigma}} is a TT-space. Then UD\exists U\subseteq D such that UT=Sσ{{\left\langle U\right\rangle}_{T}}={{S}^{\sigma}}. Since 2Sσ2\in{{S}^{\sigma}}, uUD=S{1}\exists u\in U\subseteq D=S\bigcup\{1\} such that 2uT2\in{{\left\langle u\right\rangle}_{T}}. It follows that uu must be 1, and so Sσ1T=S{{S}^{\sigma}}\supseteq{{\left\langle 1\right\rangle}_{T}}=S. Then Sσ=S{S}^{\sigma}=S because SσS{S}^{\sigma}\subseteq S, and hence every element of SS is fixed under the action of σ\sigma. However, πS\pi\in S and σ(π)=σ(g(1))=g(π)=π2π\sigma(\pi)=\sigma(g(1))=g(\pi)={{\pi}^{2}}\neq\pi, a contradiction. ∎

7.2. A construction of TT-morphisms from UT(UD){{\left\langle U\right\rangle}_{T}}(U\subseteq D)

TT-morphisms discussed in Subsection 7.1 are from TT-spaces generated by a single element of DD. To study TT-morphisms from other TT-spaces, we generalize Notation 7.1.1 as follows.

Notation 7.2.1.

Let DD be a set, let MM be a set of maps from DD to DD and let α:U(D)V(D)\alpha:U(\subseteq D)\to V(\subseteq D) be a map. Then by UM,αVU\xrightarrow{M,\alpha}V we mean that

f,gM\forall f,g\in M and u,wUu,w\in U, if f(u)=g(w)f(u)=g(w), then f(α(u))=g(α(w))f(\alpha(u))=g(\alpha(w)).

Moreover, by UM,αVU\overset{M,\alpha}{\longleftrightarrow}V we mean that

α\alpha is bijective, UM,αVU\xrightarrow{M,\alpha}V and VM,α1UV\xrightarrow{M,{{\alpha}^{-1}}}U,

where α1:VU\alpha^{-1}:V\to U denotes the inverse of α\alpha;

or equivalently, by UM,αVU\overset{M,\alpha}{\longleftrightarrow}V we mean that

α\alpha is bijective, and f,gM\forall f,g\in M and u,wUu,w\in U, f(u)=g(w)f(α(u))=g(α(w))f(u)=g(w)\Leftrightarrow f(\alpha(u))=g(\alpha(w)).

Now Proposition 7.1.3 can be generalized to

Proposition 7.2.2.

Let σ\sigma be a TT-morphism from a TT-space S1S_{1} to a TT-space S2S_{2}. Then AS1\forall A\subseteq{{S}_{1}}, AT,αS2A\xrightarrow{T,\alpha}{S}_{2}, where α:=σ|A\alpha:=\sigma{{|}_{A}}. In particular, S1T,σS2{{S}_{1}}\xrightarrow{T,\sigma}{S}_{2}.

Proof.

Let f,gTf,g\in T and let a,bAS1a,b\in A\subseteq{{S}_{1}}. If f(a)=g(b)f(a)=g(b), then

f(α(a))=f(σ(a))=σ(f(a))=σ(g(b))=g(σ(b))=g(α(b)).f(\alpha(a))=f(\sigma(a))=\sigma(f(a))=\sigma(g(b))=g(\sigma(b))=g(\alpha(b)).

Hence by Notation 7.2.1, AT,αS2A\xrightarrow{T,\alpha}{S}_{2}. ∎

However, the converse of Proposition 7.2.2 is not true, as shown below.

Example 7.2.3.

Let RR, TT and σ\sigma be defined as in Example 7.1.5. Let ARA\subseteq R and let α=σ|A\alpha=\sigma{{|}_{A}}. Then a,bA\forall a,b\in A and f,gTf,g\in T, f(a)=g(b)f(a)=g(b) implies f(α(a))=g(α(b))f(\alpha(a))=g(\alpha(b)). Hence AT,αRA\xrightarrow{T,\alpha}R. But σ\sigma is not a TT-morphism, as explained in Example 7.1.5.

To construct TT-morphisms, we generalize Proposition 7.1.7 as follows.

Proposition 7.2.4.

Let α:U(D)V(D)\alpha:U(\subseteq D)\to V(\subseteq D) be a map.

  1. (a)

    UT,αVU\xrightarrow{T,\alpha}V if and only if there exists a map σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} given by f(u)f(α(u))f(u)\mapsto f(\alpha(u)), uU\forall u\in U and fTf\in T.

  2. (b)

    The map σ\sigma given in (a) is a TT-morphism.

Proof.

For (a):

UT,αVU\xrightarrow{T,\alpha}V;

\Leftrightarrow f(u)=g(w)f(u)=g(w) implies f(α(u))=g(α(w))f(\alpha(u))=g(\alpha(w)), f,gT\forall f,g\in T and u,wUu,w\in U; (by Notation 7.2.1)

\Leftrightarrow σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} given by f(u)f(α(u))f(u)\mapsto f(\alpha(u)), uU\forall u\in U and fTf\in T, is a well-defined map.

For (b): Let aUTa\in{{\left\langle U\right\rangle}_{T}}. Then uU\exists u\in U and gTg\in T such that a=g(u)a=g(u). Let fTf\in T. Then σ(f(a))=σ((fg)(u))\sigma(f(a))=\sigma((f\circ g)(u)). By Definition 2.1.1, fgTf\circ g\in T, and so by the definition of σ\sigma, σ((fg)(u))=(fg)(α(u))\sigma((f\circ g)(u))=(f\circ g)(\alpha(u)). Besides, g(α(u))=σ(g(u))g(\alpha(u))=\sigma(g(u)) also by the definition of σ\sigma. Combining the above equations, we have

σ(f(a))=f(g(α(u)))=f(σ(g(u)))=f(σ(a)).\sigma(f(a))=f(g(\alpha(u)))=f(\sigma(g(u)))=f(\sigma(a)).

Thus σ\sigma is a TT-morphism. ∎

To construct TT-isomorphisms, Corollary 7.1.8 is generalized as follows.

Proposition 7.2.5.

Let α:U(D)V(D)\alpha:U(\subseteq D)\to V(\subseteq D) be a map.

  1. (a)

    The following statements are equivalent:

    1. (i)

      UT,αVU\overset{T,\alpha}{\longleftrightarrow}V.

    2. (ii)

      α\alpha is bijective and there exists a bijective map σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} given by f(u)f(α(u))f(u)\mapsto f(\alpha(u)), uU\forall u\in U and fTf\in T.

  2. (b)

    The map σ\sigma given in (ii) is a TT-isomorphism.

Proof.

For (a):

UT,αVU\overset{T,\alpha}{\longleftrightarrow}V;

\Leftrightarrow α\alpha is bijective, and f,gT\forall f,g\in T and u,wUu,w\in U, f(u)=g(w)f(u)=g(w) if and only if f(α(u))=g(α(w))f(\alpha(u))=g(\alpha(w)); (by Notation 7.2.1)

\Leftrightarrow α\alpha is bijective and σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} given by f(u)f(α(u))f(u)\mapsto f(\alpha(u)), uU\forall u\in U and fTf\in T, is a well-defined bijective map.

For (b): By Proposition 7.2.4, σ\sigma is a TT-morphism. Since σ\sigma is bijective, by Definition 2.4.1, it is a TT-isomorphism. ∎

The converse of statement (b) in Proposition 7.2.4 raises a question as follows, which generalizes Question 7.1.9.

Question 7.2.6.

For UDU\subseteq D and a TT-morphism σ\sigma from UT{{\left\langle U\right\rangle}_{T}}, is there any map α:UD\alpha:U\to D such that σ(f(u))=f(α(u))\sigma(f(u))=f(\alpha(u)), uU\forall u\in U and fTf\in T?

The significance of this question is that, in the cases where the answer is yes, we can construct a TT-morphism σ\sigma by letting σ(f(u))=f(α(u))\sigma(f(u))=f(\alpha(u)), uU\forall u\in U and fTf\in T. Example 7.1.10 shows a case where the answer to Question 7.2.6 is negative, but soon we shall see that there are cases where the answer is positive. Thus the property in question deserves a name.

Definition 7.2.7.

Let TT be an operator semigroup on DD, let UDU\subseteq D, and let σ\sigma be a TT-morphism from UT{{\left\langle U\right\rangle}_{T}}. If there exists a map α:UD\alpha:U\to D such that σ(f(u))=f(α(u))\sigma(f(u))=f(\alpha(u)), uU\forall u\in U and fTf\in T, then we say that σ\sigma is constructible by α\alpha, or just say that σ\sigma is constructible for brevity.

Recall that it is possible that UT{{\left\langle U\right\rangle}_{T}} does not contain UU (Proposition 2.2.6). But if the identity function lies in TT, which is true in the cases such as Examples 2.1.2, 2.1.4 and 2.1.5, then UUTU\subseteq{{\left\langle U\right\rangle}_{T}}, where we can get a positive answer to Question 7.2.6 as shown below.

Corollary 7.2.8.

Let UDU\subseteq D. If UUTU\subseteq{{\left\langle U\right\rangle}_{T}}, then any TT-morphism σ\sigma from UT{{\left\langle U\right\rangle}_{T}} is constructible by α:=σ|U\alpha:=\sigma{{|}_{U}}.

Proof.

Since UUTU\subseteq{{\left\langle U\right\rangle}_{T}}, by Definition 2.3.1, σ(f(u))=f(σ(u))=f(α(u))\sigma(f(u))=f(\sigma(u))=f(\alpha(u)), uU\forall u\in U and fTf\in T. Thus σ\sigma is constructible by α\alpha. ∎

Moreover, if the operator semigroup TT can be generated by a single element, then we also have a positive answer to Question 7.2.6 as shown below.

Theorem 7.2.9.

Let σ\sigma be a TT-morphism from UT{{\left\langle U\right\rangle}_{T}} to VT{{\left\langle V\right\rangle}_{T}}. If T=gT=\left\langle g\right\rangle (Definition 2.1.6), where gg is a function from DD to DD, then there exists a map α:UVVT\alpha:U\to V\bigcup{{\left\langle V\right\rangle}_{T}} such that σ\sigma is constructible by α\alpha.

Proof.

If σ\sigma is the empty map \emptyset\to\emptyset, then by Definition 7.2.7, σ\sigma is constructible by the empty map trivially. Hence it suffices to show the case where U(D)U(\subseteq D) and V(D)V(\subseteq D) are nonempty.

Let A=VVTA=V\bigcup\left\langle V\right\rangle_{T}. xD\forall x\in D and fTf\in T, set

(7.9) fA1(x)={aA|f(a)=x}.{f_{A}^{-1}}(x)=\{a\in A\,|\,f(a)=x\}.

Let uUu\in U. Let Au=fTfA1(σ(f(u))){{A}_{u}}=\bigcap\nolimits_{f\in T}{f_{A}^{-1}}(\sigma(f(u))). Suppose Au{{A}_{u}}\neq\emptyset. Then by Equation ((7.9)), f(a)=σ(f(u))f(a)=\sigma(f(u)), aAu\forall a\in{{A}_{u}} and fTf\in T. To define a map α:UVVT\alpha:U\to V\bigcup{{\left\langle V\right\rangle}_{T}} such that σ\sigma is constructible by α\alpha, let α(u)\alpha(u) be any aAu(A=VVT)a\in A_{u}(\subseteq A=V\bigcup{\left\langle V\right\rangle}_{T}). Then fT\forall f\in T, σ(f(u))=f(a)=f(α(u))\sigma(f(u))=f(a)=f(\alpha(u)), as desired for σ\sigma to be constructible by α\alpha.

Therefore, it suffices to show Au{{A}_{u}}\neq\emptyset.

To show Au{{A}_{u}}\neq\emptyset, we shall first show gA1(σ(g(u)))Au{g_{A}^{-1}}(\sigma(g(u)))\subseteq A_{u}. Then we shall prove gA1(σ(g(u))){g_{A}^{-1}}(\sigma(g(u)))\neq\emptyset.

Since T=gT=\left\langle g\right\rangle, Au=fTfA1(σ(f(u)))=k+(gk)A1(σ(gk(u))){{A}_{u}}=\bigcap\nolimits_{f\in T}{f_{A}^{-1}}(\sigma(f(u)))=\bigcap\nolimits_{k\in{{\mathbb{Z}}^{+}}}{{(g^{k})_{A}^{-1}}(\sigma({{g}^{k}}(u))}), where gk{{g}^{k}} denotes the composite of kk copies of gg.

Let k(+)>1k(\in{{\mathbb{Z}}^{+}})>1. Because g(u)UTg(u)\in{{\left\langle U\right\rangle}_{T}}, by Definition 2.3.1,

σ(gk(u))=σ(gk1(g(u)))=gk1(σ(g(u))).\sigma({{g}^{k}}(u))=\sigma({{g}^{k-1}}(g(u)))=g^{k-1}(\sigma(g(u))).

Applying (gk)A1{(g^{k})}_{A}^{-1} to the left and right sides of the above equations, we have

(7.10) (gk)A1(σ(gk(u)))=(gk)A1(gk1(σ(g(u)))).{(g^{k})}_{A}^{-1}(\sigma({{g}^{k}}(u)))={(g^{k})}_{A}^{-1}({g^{k-1}}(\sigma(g(u)))).

To show gA1(σ(g(u)))Au{g_{A}^{-1}}(\sigma(g(u)))\subseteq A_{u}, suppose agA1(σ(g(u)))\exists a\in{g_{A}^{-1}}(\sigma(g(u))).

Then by ((7.9)) (with f:=gf:=g and x:=σ(g(u))x:=\sigma(g(u))), g(a)=σ(g(u))g(a)=\sigma(g(u)). Hence gk(a)=gk1(g(a))=gk1(σ(g(u))){g^{k}}(a)={g^{k-1}}(g(a))={g^{k-1}}(\sigma(g(u))). Then by ((7.9)) (with f:=gkf:=g^{k} and x:=gk1(σ(g(u)))x:={g^{k-1}}(\sigma(g(u)))), a(gk)A1(gk1(σ(g(u))))a\in{{({g}^{k})}_{A}^{-1}}({g^{k-1}}(\sigma(g(u)))). Thus,

(7.11) gA1(σ(g(u)))(gk)A1(gk1(σ(g(u)))).{g_{A}^{-1}}(\sigma(g(u)))\subseteq{{({g}^{k})}_{A}^{-1}}({g^{k-1}}(\sigma(g(u)))).

Combining ((7.10)) with ((7.11)), we have

(gk)A1(σ(gk(u)))gA1(σ(g(u))).{(g^{k})}_{A}^{-1}(\sigma({{g}^{k}}(u)))\supseteq{g_{A}^{-1}}(\sigma(g(u))).

Hence k1,(gk)A1(σ(gk(u)))gA1(σ(g(u)))\forall k\geq 1,{(g^{k})}_{A}^{-1}(\sigma({{g}^{k}}(u)))\supseteq{g_{A}^{-1}}(\sigma(g(u))), and so

Au(=k+(gk)A1(σ(gk(u))))gA1(σ(g(u))).{{A}_{u}}(=\bigcap\nolimits_{k\in{{\mathbb{Z}}^{+}}}{{(g^{k})_{A}^{-1}}(\sigma({{g}^{k}}(u))}))\supseteq{g_{A}^{-1}}(\sigma(g(u))).

Therefore, to show Au{{A}_{u}}\neq\emptyset, it suffices to prove gA1(σ(g(u))){g_{A}^{-1}}(\sigma(g(u)))\neq\emptyset.

Since σ(g(u))VT\sigma(g(u))\in{\left\langle V\right\rangle}_{T}, vV\exists v\in V and h+h\in{{\mathbb{Z}}^{+}} such that σ(g(u))=gh(v)\sigma(g(u))=g^{h}(v). If h=1h=1, then by (7.9)(\ref{6.10a}) (with f:=gf:=g and x:=σ(g(u))x:=\sigma(g(u))), vgA1(σ(g(u)))v\in{g_{A}^{-1}}(\sigma(g(u))). And if h>1h>1, then σ(g(u))=g(gh1(v))\sigma(g(u))=g(g^{h-1}(v)), and thus by (7.9)(\ref{6.10a}) (with f:=gf:=g, x:=σ(g(u))x:=\sigma(g(u)) and the fact that gh1(v)VTAg^{h-1}(v)\in{{\left\langle V\right\rangle}_{T}}\subseteq A), gh1(v)gA1(σ(g(u)))g^{h-1}(v)\in g_{A}^{-1}(\sigma(g(u))). In both cases of hh, gA1(σ(g(u)))g_{A}^{-1}(\sigma(g(u)))\neq\emptyset, as desired. Therefore, Au{{A}_{u}}\neq\emptyset, as desired.

In summary, to define a map α:UVVT\alpha:U\to V\bigcup{{\left\langle V\right\rangle}_{T}}, for each uUu\in U, let α(u)\alpha(u) be any element of Au{A}_{u}. Then α\alpha satisfies σ(f(u))=f(α(u))\sigma(f(u))=f(\alpha(u)), uU\forall u\in U and fTf\in T. Hence by Definition 7.2.7, σ\sigma is constructible by α\alpha. ∎

7.3. A construction of θ\theta-morphisms

For θ\theta-morphisms, we generalize Notation 7.2.1 to

Notation 7.3.1.

Let D1D_{1} (resp. D2D_{2}) be a set, let M1M_{1} (resp. M2M_{2}) be a set of maps from D1D_{1} to D1D_{1} (resp. a set of maps from D2D_{2} to D2D_{2}), let θM1×M2\theta\subseteq{{M}_{1}}\times{{M}_{2}}, and let α:U(D1)V(D2)\alpha:U(\subseteq{D_{1}})\to V(\subseteq{{D}_{2}}) be a map.

By Uθ,αVU\xrightarrow{\theta,\alpha}V we mean that f,gDomθ\forall f,g\in\operatorname{Dom}\theta and u,wUu,w\in U,

f(u)=g(w)f(u)=g(w) implies θ(f)(α(u))=θ(g)(α(w))\theta(f)(\alpha(u))=\theta(g)(\alpha(w)).

Moreover, by Uθ,αVU\overset{\theta,\alpha}{\longleftrightarrow}V we mean that

α\alpha is bijective, Uθ,αVU\xrightarrow{\theta,\alpha}V and Vθ1,α1UV\xrightarrow{{{\theta}^{-1}},{{\alpha}^{-1}}}U,

where θ1:={(h,f)|(f,h)θ}{{\theta}^{-1}}:=\{(h,f)\,|\,(f,h)\in\theta\}.

Remark.
  1. (1)

    fDomθ\forall f\in\operatorname{Dom}\theta and uUu\in U, f(u)=f(u)f(u)=f(u), and thus Uθ,αVU\xrightarrow{\theta,\alpha}V implies that fDomθ\forall f\in\operatorname{Dom}\theta and uUu\in U, θ(f)(α(u))=θ(f)(α(u))\theta(f)(\alpha(u))=\theta(f)(\alpha(u)). Hence by Convention 6.1.7, Uθ,αVU\xrightarrow{\theta,\alpha}V implies that fDomθ\forall f\in\operatorname{Dom}\theta and uUu\in U, θ(f)(α(u))\theta(f)(\alpha(u)) is well-defined. Thus by Proposition 6.1.6, Uθ,αVU\xrightarrow{\theta,\alpha}V implies that θ|Imα\theta{{|}_{\operatorname{Im}\alpha}} is a map.

  2. (2)

    If M1=M2=:M{{M}_{1}}={{M}_{2}}=:M and θ\theta is the identity map on MM, then Uθ,αVU\xrightarrow{\theta,\alpha}V is equivalent to UM,αVU\xrightarrow{M,\alpha}V (defined in Notation 7.2.1).

The following generalizes Proposition 7.2.2.

Proposition 7.3.2.

Let ϕ\phi be a θ\theta-morphism from a T1T_{1}-space S1S_{1} to a T2T_{2}-space S2S_{2}. Then AS1\forall A\subseteq S_{1}, Aθ,αS2A\xrightarrow{\theta,\alpha}S_{2}, where α:=ϕ|A\alpha:=\phi{{|}_{A}}. In particular, S1θ,ϕS2S_{1}\xrightarrow{\theta,\phi}S_{2}.

Proof.

By Corollary 6.1.9, θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map, and so is θ|Imα\theta{{|}_{\operatorname{Im}\alpha}} (since ImαImϕ\operatorname{Im}\alpha\subseteq\operatorname{Im}\phi and by Proposition 6.1.6).

Let f,gDomθf,g\in\operatorname{Dom}\theta and let a,bAS1a,b\in A\subseteq S_{1}. If f(a)=g(b)f(a)=g(b), then by Definition 6.1.8,

θ(f)(α(a))=θ(f)(ϕ(a))=ϕ(f(a))=ϕ(g(b))=θ(g)(ϕ(b))=θ(g)(α(b))\theta(f)(\alpha(a))=\theta(f)(\phi(a))=\phi(f(a))=\phi(g(b))=\theta(g)(\phi(b))=\theta(g)(\alpha(b)).

Hence by Notation 7.3.1, Aθ,αS2A\xrightarrow{\theta,\alpha}S_{2}. ∎

For some reasons which will be clear soon, we are more interested when θ\theta has the following property.

Definition 7.3.3.

Let T1T_{1} and T2T_{2} be operator semigroups on D1D_{1} and D2D_{2}, respectively, let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}}, and let AD2A\subseteq{{D}_{2}}. If θ(fg)(a)=(θ(f)θ(g))(a)\theta(f\circ g)(a)=(\theta(f)\circ\theta(g))(a), f,gDomθ\forall f,g\in\operatorname{Dom}\theta and aAa\in A, then we say that θ\theta is distributive over AA.

Remark.

By Convention 6.1.7, the condition

θ(fg)(a)=(θ(f)θ(g))(a)\theta(f\circ g)(a)=(\theta(f)\circ\theta(g))(a), f,gDomθ\forall f,g\in\operatorname{Dom}\theta and aAa\in A

implies that f,gDomθ\forall f,g\in\operatorname{Dom}\theta and aAa\in A, both θ(fg)(a)\theta(f\circ g)(a) and (θ(f)θ(g))(a)(\theta(f)\circ\theta(g))(a) are well-defined, and hence the condition implies that Domθ\operatorname{Dom}\theta is an operator semigroup on D1D_{1} and θ|A\theta{{|}_{A}} is a map (by Proposition 6.1.6).

Then to construct θ\theta-morphisms, we generalize Proposition 7.2.4 as follows.

Proposition 7.3.4.

Let T1T_{1} and T2T_{2} be operator semigroups on D1D_{1} and D2D_{2} respectively, let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}} with Domθ=T1\operatorname{Dom}\theta={{T}_{1}}, and let α:U(D1)V(D2)\alpha:U(\subseteq{D_{1}})\to V(\subseteq{{D}_{2}}) be a map.

  1. (a)

    Uθ,αVU\xrightarrow{\theta,\alpha}V if and only if there exists a map ϕ:UT1VT2\phi:{{\left\langle U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}} given by f(u)θ(f)(α(u))f(u)\mapsto\theta(f)(\alpha(u)), uU\forall u\in U and fT1f\in{{T}_{1}}.

  2. (b)

    The map ϕ\phi given in (a) is a θ\theta-morphism if θ\theta is distributive over VV.

Proof.

For (a):

Uθ,αVU\xrightarrow{\theta,\alpha}V;

\Leftrightarrow f,gDomθ(=T1)\forall f,g\in\operatorname{Dom}\theta(={{T}_{1}}) and a,bUa,b\in U, f(a)=g(b)f(a)=g(b) implies that θ(f)(α(a))=θ(g)(α(b))\theta(f)(\alpha(a))=\theta(g)(\alpha(b)); (by Notation 7.3.1)

\Leftrightarrow ϕ:UT1VT2\phi:{{\left\langle U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}} given by f(u)θ(f)(α(u))f(u)\mapsto\theta(f)(\alpha(u)), uU\forall u\in U and fT1f\in{{T}_{1}}, is a well-defined map.

For (b): Suppose that θ\theta is distributive over VV.

Let aUT1a\in{{\left\langle U\right\rangle}_{{{T}_{1}}}}. Then uU\exists u\in U and gT1g\in{{T}_{1}} such that a=g(u)a=g(u). Let fT1f\in{{T}_{1}}. Then fgT1f\circ g\in{{T}_{1}} by Definition 2.1.1. So ϕ(f(a))=ϕ((fg)(u))=θ(fg)(α(u))\phi(f(a))=\phi((f\circ g)(u))=\theta(f\circ g)(\alpha(u)) by the definition of ϕ\phi. Then because θ\theta is distributive over VV, by Definition 7.3.3, θ(fg)(α(u))=θ(f)(θ(g)(α(u)))\theta(f\circ g)(\alpha(u))=\theta(f)(\theta(g)(\alpha(u))). And θ(g)(α(u))=ϕ(g(u))\theta(g)(\alpha(u))=\phi(g(u)) by the definition of ϕ\phi. Combining the above equations, we have

ϕ(f(a))=θ(f)(ϕ(g(u)))=θ(f)(ϕ(a)).\phi(f(a))=\theta(f)(\phi(g(u)))=\theta(f)(\phi(a)).

Thus by Definition 6.1.8, ϕ\phi is a θ\theta-morphism from UT1{{\left\langle U\right\rangle}_{{{T}_{1}}}} to VT2{{\left\langle V\right\rangle}_{{{T}_{2}}}}. ∎

Then Proposition 7.2.5 is generalized as follows.

Proposition 7.3.5.

Let T1T_{1}, T2T_{2}, UU and α\alpha be defined as in Proposition 7.3.4. Let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}} such that Domθ=T1\operatorname{Dom}\theta={{T}_{1}} and Imθ=T2\operatorname{Im}\theta={{T}_{2}}. Then

  1. (a)

    The following statements are equivalent:

    1. (i)

      Uθ,αVU\overset{\theta,\alpha}{\longleftrightarrow}V.

    2. (ii)

      α\alpha is bijective, there exists a bijective map ϕ:UT1VT2\phi:{{\left\langle U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}} given by f(u)θ(f)(α(u))f(u)\mapsto\theta(f)(\alpha(u)), uU\forall u\in U and fT1f\in{{T}_{1}}, and its inverse ϕ1:VT2UT1\phi^{-1}:{{\left\langle V\right\rangle}_{{{T}_{2}}}}\to{{\left\langle U\right\rangle}_{{{T}_{1}}}} can be given by g(v)θ1(g)(α1(v))g(v)\mapsto\theta^{-1}(g)(\alpha^{-1}(v)), vV\forall v\in V and gT2g\in{{T}_{2}}.

  2. (b)

    Suppose that θ\theta is distributive over VV. Then the bijective map ϕ\phi given in (ii) is a θ\theta-isomorphism.

Proof.

For (a):

Uθ,αVU\overset{\theta,\alpha}{\longleftrightarrow}V;

\Leftrightarrow α\alpha is bijective, Uθ,αVU\xrightarrow{\theta,\alpha}V and Vθ1,α1UV\xrightarrow{{{\theta}^{-1}},{{\alpha}^{-1}}}U (by Notation 7.3.1);

\Leftrightarrow α\alpha is bijective, ϕ:UT1VT2\phi:{{\left\langle U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}} given by f(u)θ(f)(α(u))f(u)\mapsto\theta(f)(\alpha(u)), uU\forall u\in U and fT1f\in{{T}_{1}}, is a well-defined map, and ϕ:VT2UT1\phi^{\prime}:{{\left\langle V\right\rangle}_{{{T}_{2}}}}\to{{\left\langle U\right\rangle}_{{{T}_{1}}}} given by g(v)θ1(g)(α1(v))g(v)\mapsto\theta^{-1}(g)(\alpha^{-1}(v)), vV\forall v\in V and gT2g\in{{T}_{2}} is also a well-defined map (by Proposition 7.3.4);

\Leftrightarrow α\alpha is bijective, ϕ:UT1VT2\phi:{{\left\langle U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}} given by f(u)θ(f)(α(u))f(u)\mapsto\theta(f)(\alpha(u)), uU\forall u\in U and fT1f\in{{T}_{1}}, is a well-defined bijective map, and its inverse ϕ1:VT2UT1\phi^{-1}:{{\left\langle V\right\rangle}_{{{T}_{2}}}}\to{{\left\langle U\right\rangle}_{{{T}_{1}}}} can be given by g(v)θ1(g)(α1(v))g(v)\mapsto\theta^{-1}(g)(\alpha^{-1}(v)), vV\forall v\in V and gT2g\in{{T}_{2}}.

The third equivalence relation is explained as follows. The sufficiency (\Leftarrow) is obvious, so we only show the necessity (\Rightarrow). For this purpose, we show that both ϕϕ\phi^{\prime}\circ\phi and ϕϕ\phi\circ\phi^{\prime} are the identity map.

Let (f,g)θ(f,g)\in\theta and uUu\in U. Then

ϕ(ϕ(f(u)))\phi^{\prime}(\phi(f(u)))

=ϕ(θ(f)(α(u)))=\phi^{\prime}(\theta(f)(\alpha(u))) (by the definition of ϕ\phi)

=ϕ(g(α(u)))=\phi^{\prime}(g(\alpha(u))) (by Definition 6.1.5)

=θ1(g)(u)=\theta^{-1}(g)(u) (by the definition of ϕ\phi^{\prime})

=f(u)=f(u) (by (the θ1\theta^{-1} version of) Definition 6.1.5).

Then ϕϕ\phi^{\prime}\circ\phi is the identity map on UT1{{\left\langle U\right\rangle}_{{{T}_{1}}}} (because Domθ=T1\operatorname{Dom}\theta={{T}_{1}}).

Analogously, we can show that ϕϕ\phi\circ\phi^{\prime} is the identity map on VT2{{\left\langle V\right\rangle}_{{{T}_{2}}}}. Therefore, both ϕ\phi and ϕ\phi^{\prime} are bijective and ϕ\phi^{\prime} is the inverse of ϕ\phi.

For (b):

By (b) in Proposition 7.3.4, the map ϕ\phi given in (ii) is a θ\theta-morphism. Since ϕ\phi is bijective, by Definition 6.2.1, ϕ\phi is a θ\theta-isomorphism. ∎

Proposition 7.3.4 raises a question as follows, which generalizes Question 7.2.6.

Question 7.3.6.

Let T1T_{1} and T2T_{2} be operator semigroups on D1D_{1} and D2D_{2}, respectively, let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}}, let UD1U\subseteq D_{1} and let ϕ\phi be a θ\theta-morphism from UT1{{\left\langle U\right\rangle}_{{{T}_{1}}}} (to a T2T_{2}-space). If Domθ=T1\operatorname{Dom}\theta={{T}_{1}}, is there any map α:UD2\alpha:U\to{{D}_{2}} such that ϕ(f(u))=θ(f)(α(u))\phi(f(u))=\theta(f)(\alpha(u)), uU\forall u\in U and fT1f\in{{T}_{1}}?

The significance of this question is that, in the cases where the answer is yes, we can construct a θ\theta-morphism ϕ\phi by letting ϕ(f(u))=θ(f)(α(u))\phi(f(u))=\theta(f)(\alpha(u)), uU\forall u\in U and fT1f\in{{T}_{1}}. The property in question deserves a name as follows, which generalizes Definition 7.2.7.

Definition 7.3.7.

Let T1T_{1} and T2T_{2} be operator semigroups on D1D_{1} and D2D_{2}, respectively, let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}}, let UD1U\subseteq D_{1} and let ϕ\phi be a θ\theta-morphism from UT1{{\left\langle U\right\rangle}_{{{T}_{1}}}} (to a T2T_{2}-space). If Domθ=T1\operatorname{Dom}\theta={{T}_{1}} and there exists a map α:UD2\alpha:U\to{{D}_{2}} such that ϕ(f(u))=θ(f)(α(u))\phi(f(u))=\theta(f)(\alpha(u)), uU\forall u\in U and fT1f\in{{T}_{1}}, then we say that ϕ\phi is constructible by α\alpha, or just say that ϕ\phi is constructible for brevity.

Remark.

By Convention 6.1.7, the condition “ϕ(f(u))=θ(f)(α(u))\phi(f(u))=\theta(f)(\alpha(u)), uU\forall u\in U and fT1f\in{{T}_{1}}” implies that θ|Imα\theta{{|}_{\operatorname{Im}\alpha}} is a map.

Since TT-morphisms are special cases of θ\theta-morphisms, Example 7.1.10 shows that it is possible that a θ\theta-morphism ϕ\phi from UT1{{\left\langle U\right\rangle}_{{{T}_{1}}}} is not constructible by any map α:UD2\alpha:U\to{{D}_{2}} even if θ\theta is distributive over D2{{D}_{2}}. Now let’s show some cases where θ\theta-morphisms are constructible. The following generalizes Corollary 7.2.8.

Corollary 7.3.8.

Let T1T_{1} and T2T_{2} be operator semigroups on D1D_{1} and D2D_{2}, respectively, let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}} and let UD1U\subseteq{D_{1}}. If Domθ=T1\operatorname{Dom}\theta={{T}_{1}} and UUT1U\subseteq{{\left\langle U\right\rangle}_{{{T}_{1}}}}, then any θ\theta-morphism ϕ\phi from UT1{{\left\langle U\right\rangle}_{{{T}_{1}}}} is constructible by α:=ϕ|U\alpha:=\phi{{|}_{U}}.

Proof.

Since Domθ=T1\operatorname{Dom}\theta={{T}_{1}} and UUT1U\subseteq{{\left\langle U\right\rangle}_{{{T}_{1}}}}, by Definition 6.1.8, uU\forall u\in U and fT1f\in{{T}_{1}}, ϕ(f(u))=θ(f)(ϕ(u))=θ(f)(α(u))\phi(f(u))=\theta(f)(\phi(u))=\theta(f)(\alpha(u)). ∎

Theorem 7.2.9 is generalized for θ\theta-morphisms as follows.

Theorem 7.3.9.

Let T1T_{1} and T2T_{2} be operator semigroups on D1D_{1} and D2D_{2}, respectively, let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}}, and let ϕ\phi be a θ\theta-morphism from UT1{{\left\langle U\right\rangle}_{{{T}_{1}}}} to VT2{{\left\langle V\right\rangle}_{{{T}_{2}}}}. If Domθ=T1=g\operatorname{Dom}\theta={{T}_{1}}=\left\langle g\right\rangle, Imθ=T2\operatorname{Im}\theta={{T}_{2}}, and θ\theta is distributive over VVT2V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}}, then there exists a map α:UVVT2\alpha:U\to V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}} such that ϕ\phi is constructible by α\alpha.

Proof.

The proof is analogous to the one for Theorem 7.2.9.

If ϕ\phi is the empty map \emptyset\to\emptyset, then by Definition 7.3.7, ϕ\phi is constructible by the empty map trivially. Hence we only need to prove the case where U(D1)U(\subseteq{D_{1}}) and V(D2)V(\subseteq{{D}_{2}}) are nonempty.

By (the remark under) Definition 7.3.3, θ|A\theta{{|}_{A}} is a map, where A:=VVT2A:=V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}}, and so is θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} because ImϕVT2A\operatorname{Im}\phi\subseteq{{\left\langle V\right\rangle}_{{{T}_{2}}}}\subseteq A.

First, xD2\forall x\in{{D}_{2}} and fT1f\in{{T}_{1}}, set

(7.12) (θ(f))A1(x)={aA|θ(f)(a)=x}.{{(\theta(f))}_{A}^{-1}}(x)=\{a\in A\,|\,\theta(f)(a)=x\}.

It is well-defined because θ|A\theta{{|}_{A}} is a map.

Let uUu\in U. Let Au=fT1(θ(f))A1(ϕ(f(u))){{A}_{u}}=\bigcap\nolimits_{f\in{{T}_{1}}}{{{(\theta(f))}_{A}^{-1}}(\phi(f(u)))}. Suppose Au{{A}_{u}}\neq\emptyset. Then by Equation ((7.12)), θ(f)(a)=ϕ(f(u))\theta(f)(a)=\phi(f(u)), aAu\forall a\in{{A}_{u}} and fT1f\in{{T}_{1}}. To define a map α:UVVT2\alpha:U\to V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}} such that ϕ\phi is constructible by α\alpha, let α(u)\alpha(u) be any aAu(A=VVT2)a\in{{A}_{u}}(\subseteq A=V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}}). Then fT1\forall f\in{{T}_{1}}, ϕ(f(u))=θ(f)(a)=θ(f)(α(u))\phi(f(u))=\theta(f)(a)=\theta(f)(\alpha(u)), as desired for ϕ\phi to be constructible by α\alpha.

Therefore, it suffices to show Au{{A}_{u}}\neq\emptyset.

To show Au{{A}_{u}}\neq\emptyset, we shall first show (θ(g))A1(ϕ(g(u)))Au{{(\theta(g))}_{A}^{-1}}(\phi(g(u)))\subseteq A_{u}. Then we shall prove (θ(g))A1(ϕ(g(u))){{(\theta(g))}_{A}^{-1}}(\phi(g(u)))\neq\emptyset.

Since T1=g{{T}_{1}}=\left\langle g\right\rangle, Au=fT1(θ(f))A1(ϕ(f(u)))=k+(θ(gk))A1(ϕ(gk(u))){{A}_{u}}=\bigcap\nolimits_{f\in{{T}_{1}}}{{{(\theta(f))}_{A}^{-1}}(\phi(f(u)))}=\bigcap\nolimits_{k\in{{\mathbb{Z}}^{+}}}{{{(\theta({{g}^{k}}))}_{A}^{-1}}(\phi({{g}^{k}}(u))}), where gk{{g}^{k}} denotes the composite of kk copies of gg.

Let k(+)>1k(\in{{\mathbb{Z}}^{+}})>1. Because g(u)UT1g(u)\in{{\left\langle U\right\rangle}_{{{T}_{1}}}}, by Definition 6.1.8,

ϕ(gk(u))=ϕ(gk1(g(u)))=θ(gk1)(ϕ(g(u)))=(θ(g))k1(ϕ(g(u)))\phi({{g}^{k}}(u))=\phi({{g}^{k-1}}(g(u)))=\theta({{g}^{k-1}})(\phi(g(u)))={{(\theta(g))}^{k-1}}(\phi(g(u)))

because θ\theta is assumed to be distributive over VVT2(Imϕ)V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}}(\supseteq\operatorname{Im}\phi), and so

ϕ(gk(u))=(θ(g))k1(ϕ(g(u))).\phi({{g}^{k}}(u))={{(\theta(g))}^{k-1}}(\phi(g(u))).

Applying (θ(gk))A1{{(\theta({{g}^{k}}))}_{A}^{-1}} to both sides of the above equation, we have

(7.13) (θ(gk))A1(ϕ(gk(u)))=(θ(gk))A1((θ(g))k1(ϕ(g(u)))).{{(\theta({{g}^{k}}))}_{A}^{-1}}(\phi({{g}^{k}}(u)))={{(\theta({{g}^{k}}))}_{A}^{-1}}({{(\theta(g))}^{k-1}}(\phi(g(u)))).

To show (θ(g))A1(ϕ(g(u)))Au{{(\theta(g))}_{A}^{-1}}(\phi(g(u)))\subseteq A_{u}, suppose a(θ(g))A1(ϕ(g(u)))\exists a\in{{(\theta(g))}_{A}^{-1}}(\phi(g(u))). Then by ((7.12)) (with f:=gf:=g and x:=ϕ(g(u))x:=\phi(g(u))), θ(g)(a)=ϕ(g(u))\theta(g)(a)=\phi(g(u)). Hence

(θ(g))k1(θ(g)(a))=(θ(g))k1(ϕ(g(u))).{{(\theta(g))}^{k-1}}(\theta(g)(a))={{(\theta(g))}^{k-1}}(\phi(g(u))).

Since θ\theta is distributive over AA,

(θ(g))k1(θ(g)(a))=(θ(g))k(a)=θ(gk)(a).{{(\theta(g))}^{k-1}}(\theta(g)(a))={{(\theta(g))}^{k}}(a)=\theta({{g}^{k}})(a).

Combining the above three equations, we have θ(gk)(a)=(θ(g))k1(ϕ(g(u)))\theta({{g}^{k}})(a)={{(\theta(g))}^{k-1}}(\phi(g(u))). Then by ((7.12)) (with f:=gkf:=g^{k} and x:=(θ(g))k1(ϕ(g(u)))x:={{(\theta(g))}^{k-1}}(\phi(g(u)))),

a(θ(gk))A1((θ(g))k1(ϕ(g(u)))).a\in{{(\theta({{g}^{k}}))}_{A}^{-1}}({{(\theta(g))}^{k-1}}(\phi(g(u)))).

Since aa is assumed to be any element of (θ(g))A1(ϕ(g(u))){{(\theta(g))}_{A}^{-1}}(\phi(g(u))),

(7.14) (θ(g))A1(ϕ(g(u)))(θ(gk))A1((θ(g))k1(ϕ(g(u)))).{{(\theta(g))}_{A}^{-1}}(\phi(g(u)))\subseteq{{(\theta({{g}^{k}}))}_{A}^{-1}}({{(\theta(g))}^{k-1}}(\phi(g(u)))).

Combining ((7.13)) with ((7.14)), we have

(θ(gk))A1(ϕ(gk(u)))(θ(g))A1(ϕ(g(u))).{{(\theta({{g}^{k}}))}_{A}^{-1}}(\phi({{g}^{k}}(u)))\supseteq{{(\theta(g))}_{A}^{-1}}(\phi(g(u))).

Hence k1,(θ(gk))A1(ϕ(gk(u)))(θ(g))A1(ϕ(g(u)))\forall k\geq 1,{{(\theta({{g}^{k}}))}_{A}^{-1}}(\phi({{g}^{k}}(u)))\supseteq{{(\theta(g))}_{A}^{-1}}(\phi(g(u))), and so

Au(=k+(θ(gk))A1(ϕ(gk(u))))(θ(g))A1(ϕ(g(u)).{{A}_{u}}(=\bigcap\nolimits_{k\in{{\mathbb{Z}}^{+}}}{{{(\theta({{g}^{k}}))}_{A}^{-1}}(\phi({{g}^{k}}(u))}))\supseteq(\theta(g))_{A}^{-1}(\phi(g(u)).

Therefore, to show Au{{A}_{u}}\neq\emptyset, it suffices to prove (θ(g))A1(ϕ(g(u))){{(\theta(g))}_{A}^{-1}}(\phi(g(u)))\neq\emptyset.

Since ϕ(g(u))VT2\phi(g(u))\in{{\left\langle V\right\rangle}_{{{T}_{2}}}} and Imθ=T2\operatorname{Im}\theta={{T}_{2}}, vV\exists v\in V and h+h\in{{\mathbb{Z}}^{+}} such that ϕ(g(u))=θ(gh)(v)\phi(g(u))=\theta({{g}^{h}})(v). If h=1h=1, then by ((7.12)) (with f:=gf:=g and x:=ϕ(g(u))x:=\phi(g(u))), v(θ(g))A1(ϕ(g(u)))v\in{{(\theta(g))}_{A}^{-1}}(\phi(g(u))). And if h>1h>1, then because θ\theta is distributive over VVT2V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}}, ϕ(g(u))=θ(gh)(v)=θ(g)(θ(gh1)(v))\phi(g(u))=\theta({{g}^{h}})(v)=\theta(g)(\theta({{g}^{h-1}})(v)), and hence by ((7.12)) (with f:=gf:=g, x:=ϕ(g(u))x:=\phi(g(u)) and the fact that θ(gh1)(v)VT2A\theta({{g}^{h-1}})(v)\in{{\left\langle V\right\rangle}_{{{T}_{2}}}}\subseteq A), θ(gh1)(v)(θ(g))A1(ϕ(g(u)))\theta({{g}^{h-1}})(v)\in{{(\theta(g))}_{A}^{-1}}(\phi(g(u))). In both cases of hh, (θ(g))A1(ϕ(g(u))){{(\theta(g))}_{A}^{-1}}(\phi(g(u)))\neq\emptyset, as desired. Therefore, Au{{A}_{u}}\neq\emptyset, as desired.

In summary, to define a map α:UVVT2\alpha:U\to V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}}, for each uUu\in U, let α(u)\alpha(u) be any element of Au{A}_{u}. Then α\alpha satisfies ϕ(f(u))=θ(f)(α(u))\phi(f(u))=\theta(f)(\alpha(u)), uU\forall u\in U and fT1f\in{{T}_{1}}. Hence by Definition 7.3.7, ϕ\phi is constructible by α\alpha. ∎

7.4. Another construction of TT-morphisms

Let σ\sigma be a TT-morphism from a TT-space SS. If σ(a)=b\sigma(a)=b, then fT\forall f\in T, σ(f(a))=f(σ(a))=f(b)\sigma(f(a))=f(\sigma(a))=f(b). In other words, if an ordered pair (a,b)σ(a,b)\in\sigma, where σ\sigma is regarded as a binary relation, then fT\forall f\in T, (f(a),f(b))σ(f(a),f(b))\in\sigma. This observation implies that there exists some sort of basic parts of TT-morphisms.

To facilitate our descriptions, we give the following notations only for this subsection.

Notation 7.4.1.

Let T=T{T^{*}=T\bigcup\{Id on DD} and let RT(x,y)={(f(x),f(y))|fT}{{R}_{T}}(x,y)=\{(f(x),f(y))\,|\,f\in T^{*}\}, where (x,y)D×D(x,y)\in D\times D. Besides, both Sa{{S}_{a}} and Sb{{S}_{b}} denote TT-spaces.

Remark.

Clearly, RT(x,y)D×D{{R}_{T}}(x,y)\subseteq D\times D is a binary relation in DD.

Proposition 7.4.2.

Let RSa×SbR\subseteq{{S}_{a}}\times{{S}_{b}} such that DomR\operatorname{Dom}R is a TT-subspace of SaS_{a}. Then the following statements are equivalent:

  1. (i)

    RR is a TT-morphism from DomR\operatorname{Dom}R to Sb{{S}_{b}}.

  2. (ii)

    There exists a binary relation i{(ai,bi)}Sa×Sb\bigcup\nolimits_{i}{\{({{a}_{i}},{{b}_{i}})\}}\subseteq{{S}_{a}}\times{{S}_{b}} such that R=iRT(ai,bi)R=\bigcup\nolimits_{i}{{{R}_{T}}({{a}_{i}},{{b}_{i}})} and RR is a function.

Proof.

(i)\Rightarrow(ii): Suppose R=i{(ai,bi)}R=\bigcup\nolimits_{i}{\{({{a}_{i}},{{b}_{i}})}\}. By Definition 2.3.1, fT\forall f\in T^{*} and (ai,bi)R({{a}_{i}},{{b}_{i}})\in R, R(f(ai))=f(R(ai))=f(bi)R(f({{a}_{i}}))=f(R({{a}_{i}}))=f({{b}_{i}}), and hence (f(ai),f(bi))R(f({{a}_{i}}),f({{b}_{i}}))\in R.

Therefore, R=i{(f(ai),f(bi))|fT}=iRT(ai,bi)R=\bigcup\nolimits_{i}{\{(f({{a}_{i}}),f({{b}_{i}}))\,|\,f\in T^{*}\}}=\bigcup\nolimits_{i}{{{R}_{T}}({{a}_{i}},{{b}_{i}})}. Hence (ii) is true.

(ii)\Rightarrow(i): Let gTg\in T^{*}. Suppose (u,v)R(u,v)\in R.

Since R=(iRT(ai,bi)=)i{(f(ai),f(bi))|fT}R=(\bigcup\nolimits_{i}{{{R}_{T}}({{a}_{i}},{{b}_{i}})}=)\bigcup\nolimits_{i}{\{(f({{a}_{i}}),f({{b}_{i}}))}\,|\,f\in T^{*}\} and TT is a semigroup, (g(u),g(v))R(g(u),g(v))\in R. Because RR is a function, R(g(u))=g(v)=g(R(u))R(g(u))=g(v)=g(R(u)). Hence by Definition 2.3.1, RR is a TT-morphism from DomR\operatorname{Dom}R to Sb{{S}_{b}}. ∎

Proposition 7.4.2 implies that RT(x,y){{R}_{T}}(x,y), where (x,y)Sa×Sb(x,y)\in{{S}_{a}}\times{{S}_{b}}, may be regarded as some sort of basic part of a TT-morphism from a TT-subspace of Sa{{S}_{a}} to Sb{{S}_{b}}. This observation inspires us to define a topology on Sa×Sb{{S}_{a}}\times{{S}_{b}} as follows.

Definition 7.4.3.

Let BT(Sa×Sb)={RT(x,y)|(x,y)Sa×Sb}{B_{T}}({S_{a}}\times{S_{b}})=\{{R_{T}}(x,y)\,|\,(x,y)\in{S_{a}}\times{S_{b}}\}. Let 𝒯T(Sa×Sb){{\mathcal{T}}_{T}}({{S}_{a}}\times{{S}_{b}}) be the topology on Sa×Sb{{S}_{a}}\times{{S}_{b}} generated by BT(Sa×Sb){{B}_{T}}({{S}_{a}}\times{{S}_{b}}) as a subbasis; that is, let 𝒯T(Sa×Sb){{\mathcal{T}}_{T}}({{S}_{a}}\times{{S}_{b}}) be the collection of all unions of finite intersections of elements of BT(Sa×Sb){{B}_{T}}({{S}_{a}}\times{{S}_{b}}).

Proposition 7.4.4.

Let iBi\bigcap\nolimits_{i}{B_{i}} be an arbitrary intersection of elements BiB_{i} of BT(Sa×Sb){{B}_{T}}({{S}_{a}}\times{{S}_{b}}). Then iBi\bigcap\nolimits_{i}{B_{i}} is also a union of elements of BT(Sa×Sb){{B}_{T}}({{S}_{a}}\times{{S}_{b}}).

Proof.

If iBi=\bigcap\nolimits_{i}{B_{i}}=\emptyset, then by convention, iBi\bigcap\nolimits_{i}{B_{i}} may be regarded as the union of an empty family of elements of BT(Sa×Sb){{B}_{T}}({{S}_{a}}\times{{S}_{b}}). So we only need to consider the case where iBi\bigcap\nolimits_{i}{B_{i}}\neq\emptyset.

Any element BiB_{i} of BT(Sa×Sb){{B}_{T}}({{S}_{a}}\times{{S}_{b}}) is a set RT(x,y)={(f(x),f(y))|fT } {R_{T}}(x,y)=\{(f(x),f(y))\,|\,f\in T^{*}\text{ }\!\!\}\!\!\text{ }, where (x,y)Sa×Sb(x,y)\in{{S}_{a}}\times{{S}_{b}}. So for any (u,v)Bi(u,v)\in B_{i} and gTg\in T^{*}, since TT is a semigroup, we can tell (g(u),g(v))Bi(g(u),g(v))\in B_{i}. Then gT\forall g\in T^{*} and (u,v)iBi(u,v)\in\bigcap\nolimits_{i}{B_{i}}, (g(u),g(v))iBi(g(u),g(v))\in\bigcap\nolimits_{i}{B_{i}}.

Therefore,

iBi\bigcap\nolimits_{i}{B_{i}}

=(u,v)iBi{(g(u),g(v))|gT } =\bigcup\nolimits_{(u,v)\in\bigcap\nolimits_{i}{B_{i}}}{\{(g(u),g(v))\,|\,g\in T^{*}\text{ }\!\!\}\!\!\text{ }}

=(u,v)iBiRT(u,v).=\bigcup\nolimits_{(u,v)\in\bigcap\nolimits_{i}{B_{i}}}{{{R}_{T}}(u,v)}.

Thus immediately we have

Corollary 7.4.5.

BT(Sa×Sb){{B}_{T}}({{S}_{a}}\times{{S}_{b}}) is a basis of 𝒯T(Sa×Sb){{\mathcal{T}}_{T}}({{S}_{a}}\times{{S}_{b}}). Thus, 𝒯T(Sa×Sb){{\mathcal{T}}_{T}}({{S}_{a}}\times{{S}_{b}}) is the collection of all unions of elements of BT(Sa×Sb){{B}_{T}}({{S}_{a}}\times{{S}_{b}}).

Combining Proposition 7.4.2 with the above corollary, we immediately obtain the main theorem of this subsection as follows.

Theorem 7.4.6.

Let RSa×SbR\subseteq{{S}_{a}}\times{{S}_{b}} such that DomR\operatorname{Dom}R is a TT-subspace of SaS_{a}. Then RR is a TT-morphism from DomR\operatorname{Dom}R to Sb{{S}_{b}} if and only if R𝒯T(Sa×Sb)R\in{{\mathcal{T}}_{T}}({{S}_{a}}\times{{S}_{b}}) and RR is a function.

Part II. THEORY FOR THE CASE OF MULTIVARIABLE TOTAL OR PARTIAL FUNCTIONS

To make our theory more general, in Section 8, we shall generalize the notion of operator semigroup to incorporate functions of more than one variable. Then we shall find that some familiar concepts such as ring homomorphisms, module homomorphisms, and group homomorphisms can be characterized by TT-morphisms or θ\theta-morphisms. Moreover, in Section 9, we shall allow elements in TT to be partial functions. Then we will find that some notions such as covariant functors between categories can be characterized in terms of TT-morphisms or θ\theta-morphisms.

8. Basic notions for the case of multivariable (total) functions

Subsection 8.1 generalizes the notion of operator semigroup to incorporate multivariable functions. Accordingly, Subsections 8.2, 8.3, 8.4, 8.5 and 8.6 generalize the notions of TT-spaces, TT-morphisms, TT-isomorphisms, θ\theta-morphisms and θ\theta-isomorphisms, respectively.

8.1. Operator generalized-semigroup TT

Definition 8.1.1.

Let DD be a set, let n+n\in{{\mathbb{Z}}^{+}}, and let ff be a function from the cartesian product Dn{{D}^{n}} to DD. i=1,,n\forall i=1,\cdots,n, let ni+n_{i}\in{{\mathbb{Z}}^{+}} and let gi{{g}_{i}} be a function from the cartesian product Dni{{D}^{{{n}_{i}}}} to DD. Let m=inim=\sum\nolimits_{i}{{{n}_{i}}}. Then the composite f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) is the function from Dm{{D}^{m}} to DD given by

(x1,,xm)f(g1(v1),,gn(vn)),({{x}_{1}},\cdots,{{x}_{m}})\mapsto f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})),

where vi:=(xmi+1,,xmi+ni)Dni,i=1,,n{{\text{v}}_{i}}:=({{x}_{{{m}_{i}}+1}},\cdots,{{x}_{{{m}_{i}}+{{n}_{i}}}})\in{{D}^{{{n}_{i}}}},\forall i=1,\cdots,n, m1:=0{{m}_{1}}:=0 and mi:=k=1i1nk,i=2,,n{{m}_{i}}:=\sum\nolimits_{k=1}^{i-1}{{{n}_{k}}},\forall i=2,\cdots,n.

Remark.

If n=1n=1, we can tell that the fg1f\circ g_{1} defined above is the usual composite of functions ff and g1g_{1}.

Now Definition 2.1.1 is generalized to

Definition 8.1.2.

Let DD be a set and let TT be a subset of

{all functions from the cartesian product Dn{{D}^{n}} to D|n+D\,|\,n\in{{\mathbb{Z}}^{+}}}.

If n+\forall n\in{{\mathbb{Z}}^{+}}, fTf\in T of nn variables, and (g1,,gn)Tn({{g}_{1}},\cdots,{{g}_{n}})\in{{T}^{n}} (cartesian product), the composite f(g1,,gn)Tf\circ({{g}_{1}},\cdots,{{g}_{n}})\in T, then we call TT an operator generalized-semigroup, abbreviated as operator gen-semigroup, on DD and DD is called the domain of TT.

Remark.
  1. (1)

    The terminology “generalized-semigroup” is informal and just for convenience because we have not defined it.

  2. (2)

    Obviously, any operator semigroup is an operator gen-semigroup.

The following generalizes Example 2.1.2.

Example 8.1.3.

Let BB be a subfield of a field FF. Let TT be the set

{f:FnF\{f^{*}:{{F}^{n}}\to F given by (a1,,an)f(a1,,an)|fB[x1,,xn],n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in B[{{x}_{1}},\cdots,{{x}_{n}}],n\in{{\mathbb{Z}}^{+}}\},

where B[x1,,xn]B[{{x}_{1}},\cdots,{{x}_{n}}] is the ring of polynomials over BB in nn variables.

Then by Definition 8.1.1, f,g1,,gnT\forall f^{*},{{g}_{1}}^{*},\cdots,{{g}_{n}}^{*}\in T such that fB[x1,,xn]f\in B[{{x}_{1}},\cdots,{{x}_{n}}] and giB[x1,,xni]{{g}_{i}}\in B[{{x}_{1}},\cdots,{{x}_{{{n}_{i}}}}], i=1,,n\forall i=1,\cdots,n, the composite f(g1,,gn)f^{*}\circ({{g}_{1}}^{*},\cdots,{{g}_{n}}^{*}) can be given by

(x1,,xm)f(g1(v1),,gn(vn))({{x}_{1}},\cdots,{{x}_{m}})\mapsto f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})),

where m:=inim:=\sum\nolimits_{i}{{{n}_{i}}}, (x1,,xm)Fm({{x}_{1}},\cdots,{{x}_{m}})\in{{F}^{m}}, vi:=(xmi+1,,xmi+ni)Fni,i=1,,n{{\text{v}}_{i}}:=({{x}_{{{m}_{i}}+1}},\cdots,{{x}_{{{m}_{i}}+{{n}_{i}}}})\in{{F}^{{{n}_{i}}}},\forall i=1,\cdots,n, m1:=0{{m}_{1}}:=0 and mi:=k=1i1nk,i=2,,n{{m}_{i}}:=\sum\nolimits_{k=1}^{i-1}{{{n}_{k}}},\forall i=2,\cdots,n.

For example, let f=g1=g2=x1x2f={{g}_{1}}={{g}_{2}}={{x}_{1}}{{x}_{2}}. Then f(g1,g2)f^{*}\circ({{g}_{1}}^{*},{{g}_{2}}^{*}) can be induced by (with a slight abuse of notation) f(g1,g2)=(x1x2)(x3x4)=x1x2x3x4f({{g}_{1}},{{g}_{2}})=({{x}_{1}}{{x}_{2}})({{x}_{3}}{{x}_{4}})={{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}. That is, f(g1,g2):F4Ff^{*}\circ({{g}_{1}}^{*},{{g}_{2}}^{*}):{{F}^{4}}\to F is given by (a1,a2,a3,a4)a1a2a3a4,(a1,a2,a3,a4)F4({{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}})\mapsto{{a}_{1}}{{a}_{2}}{{a}_{3}}{{a}_{4}},\forall({{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}})\in{{F}^{4}}.

Then from Definition 8.1.2, we can tell that TT is an operator gen-semigroup on FF.

Note that the map which induces TT, i.e.

τ:n+B[x1,,xn]T\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{B[{{x}_{1}},\cdots,{{x}_{n}}]}\to T

given by fff\mapsto f^{*} , is surjective, but it is not necessarily injective.

The following several examples will also be employed.

Example 8.1.4.

Let RR be a ring with identity and let 𝔽2x1,,xn{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle be the ring of polynomials over the prime field of order 2 in nn noncommuting variables. Let

T={f:RnRT=\{f^{*}:{{R}^{n}}\to R given by (a1,,an)f(a1,,an)|f𝔽2x1,,xn,n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle,n\in{{\mathbb{Z}}^{+}}\}.

Then we can tell that TT is an operator gen-semigroup on RR.

The map which induces TT, i.e.

τ:n+𝔽2x1,,xnT\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle}\to T

is given by fff\mapsto f^{*}.

Example 8.1.5.

Let MM be a module over a commutative ring RR with identity. Let

T={f:MnMT=\{f^{*}:{{M}^{n}}\to M given by (a1,,an)f(a1,,an)|f{i=1nrixi|riR},n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\},n\in{{\mathbb{Z}}^{+}}\},

(where i=1nrixiR[x1,,xn]\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\in R[{{x}_{1}},\cdots,{{x}_{n}}]). Then TT is an operator gen-semigroup on MM.

Let τ:n+{i=1nrixi|riR}T\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\}}\to T be the map inducing TT given by fff\mapsto f^{*}.

Example 8.1.6.

Let GG be an abelian group. Let

T=n+{f:GnGT=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{f^{*}:{{G}^{n}}\to G} given by (a1,,an)f(a1,,an)|f{i=1nxipi|pi,i=1,,n}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z},\forall i=1,\cdots,n\}},

(where i=1nxipiFrac(𝔽2[x1,,xn])\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\in\operatorname{Frac}({{\mathbb{F}}_{2}}[{{x}_{1}},\cdots,{{x}_{n}}])). Then TT is an operator gen-semigroup on GG.

Let τ:n+{i=1nxipi|pi}T\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\}}\to T be the map inducing TT given by fff\mapsto f^{*}.

For not necessarily abelian groups, we have

Example 8.1.7.

Let GG be a group. Recall that a word is a finite string of symbols, where repetition is allowed. Let x1,,xn,x11,,xn1\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle denote the set of all words on {x1,,xn,x11,,xn1}\{{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\}. Let

T=n+{f:GnGT=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{f^{*}:{{G}^{n}}\to G} given by (a1,,an)f(a1,,an)|fx1,,xn,x11,,xn1}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle\}.

Then we can tell that TT is an operator gen-semigroup on GG.

Let τ:n+x1,,xn,x11,,xn1T\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle}\to T be the map inducing TT given by fff\mapsto f^{*}.

8.2. TT-spaces

Definition 8.2.1.

Let TT be an operator gen-semigroup on DD and let UDU\subseteq D. Then we call

S:={f(u1,,un)|n+S:=\{f(u_{1},\cdots,u_{n})\,|\,n\in{{\mathbb{Z}}^{+}}, fTf\in T has nn variables, and (u1,,un)Un}(u_{1},\cdots,u_{n})\in U^{n}\}

the TT-space generated by UU and denote SS by UT{{\left\langle U\right\rangle}_{T}}.

Example 8.2.2.

Let BB be a subfield of a field FF. Let TT be the operator gen-semigroup on FF defined in Example 8.1.3. Then ()UF\forall(\emptyset\neq)U\subseteq F, UT{{\left\langle U\right\rangle}_{T}} is the ring B[U]B[U].

Example 8.2.3.

Let RR be a ring with identity. Let TT be the operator gen-semigroup defined in Example 8.1.4. Then ()UR\forall(\emptyset\neq)U\subseteq R, UT{{\left\langle U\right\rangle}_{T}} is a subring of RR. Specifically, RT=R{{\left\langle R\right\rangle}_{T}}=R because IdT\operatorname{Id}\in T.

Example 8.2.4.

Let MM be a module over a commutative ring RR with identity. Let TT be the operator gen-semigroup defined in Example 8.1.5. If XX is a nonempty subset of MM, then XT{{\left\langle X\right\rangle}_{T}} is the submodule of MM generated by XX.

Example 8.2.5.

Let GG be an abelian group. Let TT be the operator gen-semigroup defined in Example 8.1.6. If XX is a nonempty subset of GG, then XT{{\left\langle X\right\rangle}_{T}} is the subgroup of GG generated by XX.

Example 8.2.6.

Let GG be a group. Let TT be the operator gen-semigroup defined in Example 8.1.7. If XX is a nonempty subset of GG, then XT{{\left\langle X\right\rangle}_{T}} is the subgroup of GG generated by XX.

8.3. TT-morphisms

Definition 8.3.1.

Let TT be an operator gen-semigroup and let σ\sigma be a map from a TT-space SS to a TT-space. If n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and fTf\in T of nn variables,

σ(f(a1,,an))=f(σ(a1),,σ(an)),\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})),

then we call σ\sigma a TT-morphism.

Remark.

From Definitions 8.1.2 and 8.2.1 (or from Proposition 10.1.1, which generalizes Proposition 2.2.7), we can tell f(a1,,an)Sf(a_{1},\cdots,a_{n})\in S. Hence σ(f(a1,,an))\sigma(f(a_{1},\cdots,a_{n})) is well-defined.

Note that in Examples 8.1.3 to 8.1.7, all operator gen-semigroups are induced from polynomials, rational functions, or words on a set. Therefore, we introduce an (informal) notion as follows, which will facilitate our study on TT-morphisms and θ\theta-morphisms (defined later).

Definition 8.3.2.

Let ff be an expression associated with a rule RR. Let n+n\in\mathbb{Z}^{+}.

If all of x1,,xnx_{1},\cdots,x_{n} arise as symbols in ff, and there exist sets CC and DD such that, by the rule RR, ff yields a single element, which we denote by f(a1,,an)f(a_{1},\cdots,a_{n}), in CC whenever every x1,,xnx_{1},\cdots,x_{n} in ff is replaced by any a1,,anDa_{1},\cdots,a_{n}\in D, respectively, then we call the couple (f,R)(f,R) a (C-valued) formal function of nn variables x1,,xnx_{1},\cdots,x_{n} and DD is called a domain of (f,R)(f,R).

Moreover, we may call ff a formal function if its associated rule RR is clear from the context.

Remark.
  1. (1)

    For example, 1+0x1+0x2[x1,x2]1+0{{x}_{1}}+0{{x}_{2}}\in\mathbb{Q}[{{x}_{1}},{{x}_{2}}] is a formal function of two variables with the associated rule being the addition and multiplication of field elements. In Examples 8.1.3 to 8.1.7, all operator gen-semigroups are induced from formal functions.

  2. (2)

    By the definition, a formal function (f,R)(f,R) may have more than one domain. For example, any field which contains \mathbb{Q} is a domain of any formal function f[x1,,xn]f\in\mathbb{Q}[{{x}_{1}},\cdots,{{x}_{n}}].

  3. (3)

    Any function ff of nn variables x1,,xnx_{1},\cdots,x_{n} may be regarded as the formal function f(x1,,xn)f(x_{1},\cdots,x_{n}), which is an expression with its associated rule being the mapping defined by the function ff. The converse is not true, however, because a formal function may have more than one domain, and different domains of a formal function may induce different functions. Therefore, the notion of formal function generalizes the notion of function.

When operator gen-semigroups are induced from formal functions, as is the case in Examples 8.1.3 to 8.1.7, the following lemma will be useful.

Lemma 8.3.3.

Let \mathcal{F} be a set of D-valued formal functions on a domain DD and let TT be an operator gen-semigroup on DD induced by \mathcal{F}; that is, there is a surjective map τ:T\tau:\mathcal{F}\to T given by fff\mapsto f^{*}, where f:DnDf^{*}:{{D}^{n}}\to D is given by (a1,,an)f(a1,,an)(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n}) for ff of nn variables, n+\forall n\in{{\mathbb{Z}}^{+}}. Let σ\sigma be a map from a TT-space SS to a TT-space. Then the following statements are equivalent:

  1. (i)

    σ\sigma is a TT-morphism.

  2. (ii)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and ff\in\mathcal{F} of nn variables,

    (8.1) σ(f(a1,,an))=f(σ(a1),,σ(an)).\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).
Proof.

(i)\Rightarrow(ii): By Definition 8.3.1, n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and ff\in\mathcal{F} of nn variables,

σ(τ(f)(a1,,an))=τ(f)(σ(a1),,σ(an)).\sigma(\tau(f)(a_{1},\cdots,a_{n}))=\tau(f)(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

Then by the definition of τ\tau, (ii) must be true.

(ii)\Rightarrow(i): Since τ\tau is surjective, fT\forall f^{*}\in T, f\exists f\in\mathcal{F} such that f=τ(f)f^{*}=\tau(f). Then from statement (ii) and the definition of τ\tau, we can tell that n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and fTf^{*}\in T of nn variables, σ(f(a1,,an))=f(σ(a1),,σ(an))\sigma(f^{*}(a_{1},\cdots,a_{n}))=f^{*}(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})). Hence by Definition 8.3.1, σ\sigma must be a TT-morphism. ∎

Lemma 8.3.3 characterizes a TT-morphism by the formal functions which induce TT. When an operator gen-semigroup is induced from formal functions, the criterion for a TT-morphism by Lemma 8.3.3 is normally more convenient than that by Definition 8.3.1. Thus Lemma 8.3.3 simplifies our proofs of some of our results as follows.

The following several results serve as examples of TT-morphisms.

Firstly, Proposition 2.3.2 is generalized to

Proposition 8.3.4.

Let BB be a subfield of a field FF, let TT be the operator gen-semigroup on FF defined in Example 8.1.3, and let nonempty U,VFU,V\subseteq F. Then a map σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} is a ring homomorphism with BB fixed pointwisely if and only if it is a TT-morphism.

Proof.

We may give a proof which is analogous to that of Proposition 2.3.2. However, now we employ Lemma 8.3.3 instead. By Example 8.1.3,

T={f:FnFT=\{f^{*}:{{F}^{n}}\to F given by (a1,,an)f(a1,,an)|fB[x1,,xn],n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in B[{{x}_{1}},\cdots,{{x}_{n}}],n\in{{\mathbb{Z}}^{+}}\}.

Thus UT{{\left\langle U\right\rangle}_{T}} and VT{{\left\langle V\right\rangle}_{T}} are the rings B[U]B[U] and B[V]B[V], respectively.

Let ={fB[x1,,xn]|n+}\mathcal{F}=\{f\in B[{{x}_{1}},\cdots,{{x}_{n}}]\,|\,n\in{{\mathbb{Z}}^{+}}\}. Then Lemma 8.3.3 applies. Hence we only need to show that σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} is a ring homomorphism with BB fixed pointwisely if and only if statement (ii) in Lemma 8.3.3 is true in this case. Specifically, it suffices to show that the following statements are equivalent:

  1. (1)

    cB\forall c\in B, σ(c)=c\sigma(c)=c, and a,bUT(=B[U])\forall a,b\in{{\left\langle U\right\rangle}_{T}}(=B[U]), σ(a+b)=σ(a)+σ(b)\sigma(a+b)=\sigma(a)+\sigma(b) and σ(ab)=σ(a)σ(b)\sigma(ab)=\sigma(a)\sigma(b).

  2. (2)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)UTn(a_{1},\cdots,a_{n})\in\left\langle U\right\rangle_{T}^{n} and fB[x1,,xn]f\in B[{{x}_{1}},\cdots,{{x}_{n}}],

    (8.2) σ(f(a1,,an))=f(σ(a1),,σ(an)).\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

(1)\Rightarrow(2): Since fB[x1,,xn]f\in B[{{x}_{1}},\cdots,{{x}_{n}}] is a polynomial, it is not hard to see that statement (2) is true.

(2)\Rightarrow(1): Considering the case where fB[x1,,xn]f\in B[{{x}_{1}},\cdots,{{x}_{n}}] is a constant polynomial, we can tell from statement (2) that cB\forall c\in B, σ(c)=c\sigma(c)=c.

Let a,bUTa,b\in{{\left\langle U\right\rangle}_{T}}. Let g=x1+x2g={{x}_{1}}+{{x}_{2}} and h=x1x2h={{x}_{1}}{{x}_{2}}. Then

σ(a+b)\sigma(a+b)

=σ(g(a,b))=\sigma(g(a,b))

=g(σ(a),σ(b))=g(\sigma(a),\sigma(b)) (by Equation ((8.2)))

=σ(a)+σ(b)=\sigma(a)+\sigma(b).

And

σ(ab)\sigma(ab)

=σ(h(a,b))=\sigma(h(a,b))

=h(σ(a),σ(b))=h(\sigma(a),\sigma(b)) (by Equation ((8.2)))

=σ(a)σ(b)=\sigma(a)\sigma(b). ∎

However, ring homomorphism in Proposition 8.3.4 is between extensions over the same field. We will deal with ring homomorphisms between extensions over different fields in Subsection 8.5.

Proposition 8.3.5.

Let MM be a module over a commutative ring RR with identity. Let TT be the operator gen-semigroup defined in Example 8.1.5. If XX and YY are nonempty subsets of MM, then a map σ:XTYT\sigma:{{\left\langle X\right\rangle}_{T}}\to{{\left\langle Y\right\rangle}_{T}} is an RR-homomorphism of RR-modules if and only if it is a TT-morphism.

Proof.

By Example 8.1.5,

T={f:MnMT=\{f^{*}:{{M}^{n}}\to M given by (a1,,an)f(a1,,an)|f{i=1nrixi|riR},n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\},n\in{{\mathbb{Z}}^{+}}\}.

So XT{{\left\langle X\right\rangle}_{T}} and YT{{\left\langle Y\right\rangle}_{T}} are the submodules of MM generated by XX and YY, respectively.

Let ={f{i=1nrixi|riR}|n+}\mathcal{F}=\{f\in\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\}\ |\ n\in{{\mathbb{Z}}^{+}}\}. Then Lemma 8.3.3 applies. Thus we only need to show that σ:XTYT\sigma:{{\left\langle X\right\rangle}_{T}}\to{{\left\langle Y\right\rangle}_{T}} is an RR-homomorphism of RR-modules if and only if statement (ii) in Lemma 8.3.3 is true in this case. Specifically, it suffices to show that the following statements are equivalent:

  1. (1)

    a,bXT\forall a,b\in{{\left\langle X\right\rangle}_{T}} and rRr\in R, σ(a+b)=σ(a)+σ(b)\sigma(a+b)=\sigma(a)+\sigma(b) and σ(ra)=rσ(a)\sigma(ra)=r\sigma(a).

  2. (2)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)XTn(a_{1},\cdots,a_{n})\in\left\langle X\right\rangle_{T}^{n} and f{i=1nrixi|riR}f\in\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\},

    (8.3) σ(f(a1,,an))=f(σ(a1),,σ(an)).\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

(1)\Rightarrow(2): Because f{i=1nrixi|riR}f\in\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\} is a linear polynomial whose constant term is zero, it is not hard to see that statement (2) is true.

(2)\Rightarrow(1): Let a,bXTa,b\in{{\left\langle X\right\rangle}_{T}} and rRr\in R. Let g=x1+x2g={{x}_{1}}+{{x}_{2}} and h=rx1h=r{{x}_{1}}. Then

σ(a+b)\sigma(a+b)

=σ(g(a,b))=\sigma(g(a,b))

=g(σ(a),σ(b))=g(\sigma(a),\sigma(b)) (by Equation ((8.3)))

=σ(a)+σ(b)=\sigma(a)+\sigma(b).

And

σ(ra)\sigma(ra)

=σ(h(a))=\sigma(h(a))

=h(σ(a))=h(\sigma(a)) (by Equation ((8.3)))

=rσ(a)=r\sigma(a). ∎

RR-homomorphism in Proposition 8.3.5 is between submodules of the same RR-module. We will deal with general RR-homomorphisms of RR-modules in Subsection 8.5.

Proposition 8.3.6.

Let GG be an abelian group and let TT be the operator gen-semigroup on GG defined in Example 8.1.6. If XX and YY are nonempty subsets of GG, then a map σ:XTYT\sigma:{{\left\langle X\right\rangle}_{T}}\to{{\left\langle Y\right\rangle}_{T}} is a group homomorphism if and only if it is a TT-morphism.

Proof.

By Example 8.1.6,

T=n+{f:GnGT=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{f^{*}:{{G}^{n}}\to G} given by (a1,,an)f(a1,,an)|f{i=1nxipi|pi}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\}}.

Obviously, XT{{\left\langle X\right\rangle}_{T}} and YT{{\left\langle Y\right\rangle}_{T}} are the subgroups of GG generated by XX and YY, respectively.

We may apply Lemma 8.3.3 as well in this proof. However, our argument as follows seems more straightforward.

Suppose that σ:XTYT\sigma:{{\left\langle X\right\rangle}_{T}}\to{{\left\langle Y\right\rangle}_{T}} is a TT-morphism. Then by Definition 8.3.1, n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)XTn(a_{1},\cdots,a_{n})\in\left\langle X\right\rangle_{T}^{n} and fTf\in T of nn variables,

σ(f(a1,,an))=f(σ(a1),,σ(an)).\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

In words, σ\sigma commutes with the (multiplicative) operation equipped by GG. Hence σ\sigma is a group homomorphism.

Apparently the converse of the above argument is also true. That is, if σ\sigma is a group homomorphism, then σ\sigma commutes with the operation equipped by GG, and hence it is a TT-morphism. ∎

Group homomorphism in Proposition 8.3.6 is between subgroups of the same abelian group. We will deal with group homomorphisms between different abelian groups in Subsection 8.5.

For not necessarily abelian groups, we have

Proposition 8.3.7.

Let GG be a group and let TT be the operator gen-semigroup on GG defined in Example 8.1.7. If XX and YY are nonempty subsets of GG, then a map σ:XTYT\sigma:{{\left\langle X\right\rangle}_{T}}\to{{\left\langle Y\right\rangle}_{T}} is a group homomorphism if and only if it is a TT-morphism.

Proof.

The proof is comparable with that of Proposition 8.3.6.

By Example 8.1.7,

T=n+{f:GnGT=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{f^{*}:{{G}^{n}}\to G} given by (a1,,an)f(a1,,an)|fx1,,xn,x11,,xn1}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle\},

Obviously, XT{{\left\langle X\right\rangle}_{T}} and YT{{\left\langle Y\right\rangle}_{T}} are the subgroups of GG generated by XX and YY, respectively.

Suppose that σ:XTYT\sigma:{{\left\langle X\right\rangle}_{T}}\to{{\left\langle Y\right\rangle}_{T}} is a TT-morphism. Then by Definition 8.3.1, n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)XTn(a_{1},\cdots,a_{n})\in\left\langle X\right\rangle_{T}^{n} and fTf\in T of nn variables,

σ(f(a1,,an))=f(σ(a1),,σ(an)).\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

In words, σ\sigma commutes with the multiplication equipped by GG. Hence σ\sigma is a group homomorphism.

It is not hard to see that the converse of the above argument is also true. Specifically, if σ\sigma is a group homomorphism, then σ\sigma commutes with the multiplication equipped by GG, and hence it is a TT-morphism. ∎

Group homomorphism in Proposition 8.3.7 is between subgroups of the same group. We will deal with general group homomorphisms in Subsection 8.5.

8.4. TT-isomorphisms

Definition 2.4.1 still applies: if a TT-morphism is bijective, then we call it a TT-isomorphism. To justify the definition, we need to show Proposition 2.4.2 with TT being an operator gen-semigroup as follows.

Proposition 8.4.1.

(Proposition 2.4.2) Let σ\sigma be a TT-isomorphism from a TT-space S1S_{1} to a TT-space S2S_{2} and let σ1{{\sigma}^{-1}} be the inverse map of σ\sigma. Then σ1IsoT(S2,S1){{\sigma}^{-1}}\in\operatorname{Iso}_{T}({{S}_{2}},{{S}_{1}}).

Proof.

By Definition 8.3.1, n+\forall n\in{{\mathbb{Z}}^{+}}, fTf\in T of nn variables and (a1,,an)S1n(a_{1},\cdots,a_{n})\in{S_{1}^{n}},

f(a1,,an)=σ1(f(σ(a1),,σ(an))).f(a_{1},\cdots,a_{n})={\sigma}^{-1}(f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}))).

So n+\forall n\in{{\mathbb{Z}}^{+}}, fTf\in T of nn variables and (b1,,bn)S2n(b_{1},\cdots,b_{n})\in{S_{2}^{n}},

f(σ1(b1),,σ1(bn))=σ1(f(b1,,bn)).f({\sigma}^{-1}({b}_{1}),\cdots,{\sigma}^{-1}({b}_{n}))={\sigma}^{-1}(f(b_{1},\cdots,b_{n})).

Then by Definition 8.3.1, σ1{\sigma}^{-1} is a TT-morphism from S2S_{2} to S1S_{1}, and hence σ1IsoT(S2,S1){{\sigma}^{-1}}\in\operatorname{Iso}_{T}({{S}_{2}},{{S}_{1}}) by Definition 2.4.1. ∎

8.5. θ\theta-morphisms

To generalize Definition 6.1.8, we first generalize Notation 6.1.4 to

Notation 8.5.1.

Let D1D_{1} (resp. D2D_{2}) be a set, let M1M_{1} (resp. M2M_{2}) be a subset of {all functions from the cartesian product D1nD_{1}^{n} to D1|n+{D_{1}}\,|\,n\in{{\mathbb{Z}}^{+}}} (resp. a subset of {all functions from D2nD_{2}^{n} to D2|n+{{D}_{2}}\,|\,n\in{{\mathbb{Z}}^{+}}}), let θM1×M2\theta\subseteq{{M}_{1}}\times{{M}_{2}}, and let AD2A\subseteq{{D}_{2}}. We denote by θ|A\theta{{|}_{A}} the binary relation {(f,g|A)|(f,g)θ}\{(f,g{{|}_{A}})\,|\,(f,g)\in\theta\}, where g|Ag{{|}_{A}} denotes the function obtained by restricting every variable of gg to AA.

Definition 6.1.5 is generalized to

Definition 8.5.2.

Let θ\theta and AA be defined as in Notation 8.5.1. Let fDomθf\in\operatorname{Dom}\theta, n+n\in{{\mathbb{Z}}^{+}} and (a1,,an)An(a_{1},\cdots,a_{n})\in A^{n}. If (f,g1),(f,g2)θ\forall(f,g_{1}),(f,g_{2})\in\theta, g1(a1,,an)=g2(a1,,an)g_{1}(a_{1},\cdots,a_{n})=g_{2}(a_{1},\cdots,a_{n}), then we say that θ(f)(a1,,an)\theta(f)(a_{1},\cdots,a_{n}) is well-defined and let

θ(f)(a1,,an)=g1(a1,,an).\theta(f)(a_{1},\cdots,a_{n})=g_{1}(a_{1},\cdots,a_{n}).
Remark.

By Convention 6.1.7,

(f,g1),(f,g2)θ,g1(a1,,an)=g2(a1,,an)\forall(f,g_{1}),(f,g_{2})\in\theta,g_{1}(a_{1},\cdots,a_{n})=g_{2}(a_{1},\cdots,a_{n})

implies that (f,g)θ\forall(f,g)\in\theta, gg has nn variables.

Hence Proposition 6.1.6 is generalized to

Proposition 8.5.3.

Let θ\theta and AA be defined as in Notation 8.5.1. Then the following statements are equivalent:

  1. (i)

    θ|A\theta{{|}_{A}} is a map.

  2. (ii)

    fDomθ\forall f\in\operatorname{Dom}\theta, n+\exists n\in{{\mathbb{Z}}^{+}} such that (a1,,an)An\forall(a_{1},\cdots,a_{n})\in A^{n}, θ(f)(a1,,an)\theta(f)(a_{1},\cdots,a_{n}) is well-defined.

Proof.

θ|A\theta{{|}_{A}} is a map;

\Leftrightarrow (f,g1),(f,g2)θ\forall(f,g_{1}),(f,g_{2})\in\theta, g1|A=g2|Ag_{1}{{|}_{A}}=g_{2}{{|}_{A}};

\Leftrightarrow (f,g1),(f,g2)θ\forall(f,g_{1}),(f,g_{2})\in\theta, g1g_{1} and g2g_{2} have the same number of variables, assumed nn, and (a1,,an)An\forall(a_{1},\cdots,a_{n})\in A^{n}, g1(a1,,an)=g2(a1,,an)g_{1}(a_{1},\cdots,a_{n})=g_{2}(a_{1},\cdots,a_{n});

\Leftrightarrow fDomθ\forall f\in\operatorname{Dom}\theta, n+\exists n\in{{\mathbb{Z}}^{+}} such that (a1,,an)An\forall(a_{1},\cdots,a_{n})\in A^{n}, θ(f)(a1,,an)\theta(f)(a_{1},\cdots,a_{n}) is well-defined. ∎

Then Definition 6.1.8 is generalized to

Definition 8.5.4.

Let T1T_{1} and T2T_{2} be operator gen-semigroups and let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}}. Let ϕ\phi be a map from a T1T_{1}-space SS to a T2T_{2}-space. If n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and fDomθf\in\operatorname{Dom}\theta of nn variables,

ϕ(f(a1,,an))=θ(f)(ϕ(a1),,ϕ(an)),\phi(f(a_{1},\cdots,a_{n}))=\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})),

then ϕ\phi is called a θ\theta-morphism.

Then the first half of Corollary 6.1.9 is generalized to

Corollary 8.5.5.

Let ϕ\phi be a θ\theta-morphism from a T1T_{1}-space SS. Then (f,g)θ\forall(f,g)\in\theta, ff and gg have the same number of variables, and θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map.

Proof.

Suppose nn-variable fDomθf\in\operatorname{Dom}\theta and (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}}. Then by Definition 8.5.4 and Convention 6.1.7, θ(f)(ϕ(a1),,ϕ(an))\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined. Hence by Definition 8.5.2, (f,g)θ\forall(f,g)\in\theta, gg has nn variables. Moreover, by Proposition 8.5.3, θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map. ∎

When operator gen-semigroups are induced from formal functions, the following, which generalizes Lemma 8.3.3, will facilitate study of θ\theta-morphisms.

Lemma 8.5.6.

Let \mathcal{F} ((resp. ){\mathcal{F}}^{\prime}) be a set of D-valued ((resp. D{D}^{\prime}-valued)) formal functions on DD ((resp. D){D}^{\prime}). Let TT ((resp. T){T}^{\prime}) be an operator gen-semigroup on DD ((resp. D{D}^{\prime})) such that there is a map τ:T\tau:\mathcal{F}\to T being given by fff\mapsto f^{*}, where f:DnDf^{*}:{{D}^{n}}\to D is given by (a1,,an)f(a1,,an)(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n}) for ff of nn variables, n+\forall n\in{{\mathbb{Z}}^{+}} ((resp. such that there is a map τ:T{\tau}^{\prime}:{\mathcal{F}}^{\prime}\to{T}^{\prime} being given analogously)). Let ϑ:\vartheta:\mathcal{F}\to{\mathcal{F}}^{\prime} be a map. Let θ={(τ(f),τ(ϑ(f)))|f}\theta=\{(\tau(f),{\tau}^{\prime}(\vartheta(f)))\,|\,f\in\mathcal{F}\} and let ϕ\phi be a map from a TT-space SS to a T{T}^{\prime}-space.

Then the following statements are equivalent:

  1. (i)

    ϕ\phi is a θ\theta-morphism.

  2. (ii)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and ff\in\mathcal{F} of n variables,

    (8.4) ϕ(f(a1,,an))=ϑ(f)(ϕ(a1),,ϕ(an)).\phi(f(a_{1},\cdots,a_{n}))=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).
Proof.

Obviously, θT×T\theta\subseteq T\times T^{\prime}.

(i) \Rightarrow (ii): Assume nn-variable ff\in\mathcal{F} and (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}}. Then

ϕ(f(a1,,an))\phi(f(a_{1},\cdots,a_{n}))

=ϕ(τ(f)(a1,,an))=\phi(\tau(f)(a_{1},\cdots,a_{n}))

=θ(τ(f))(ϕ(a1),,ϕ(an))=\theta(\tau(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) (because τ(f)Domθ\tau(f)\in\operatorname{Dom}\theta and by Definition 8.5.4)

=τ(ϑ(f))(ϕ(a1),,ϕ(an))={\tau}^{\prime}(\vartheta(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) (because (τ(f),τ(ϑ(f)))θ(\tau(f),{\tau}^{\prime}(\vartheta(f)))\in\theta and by Corollary 8.5.5, θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map)

=ϑ(f)(ϕ(a1),,ϕ(an))=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})), as desired.

(ii) \Rightarrow (i): To show that ϕ\phi is a θ\theta-morphism, we first need to show that θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map.

Let (h,g|Imϕ),(h,g|Imϕ)θ|Imϕ(h,g{{|}_{\operatorname{Im}\phi}}),(h,{g}^{\prime}{{|}_{\operatorname{Im}\phi}})\in\theta{{|}_{\operatorname{Im}\phi}} with (h,g),(h,g)θ(h,g),(h,{g}^{\prime})\in\theta. By the definition of θ\theta, l,m\exists l,m\in\mathcal{F} such that (h,g)=(τ(l),τ(ϑ(l)))(h,g)=(\tau(l),{\tau}^{\prime}(\vartheta(l))) and (h,g)=(τ(m),τ(ϑ(m)))(h,{g}^{\prime})=(\tau(m),{\tau}^{\prime}(\vartheta(m))), and hence τ(l)=τ(m)\tau(l)=\tau(m). Thus ll and mm have the same number of variables, which we assume to be nn. Then by Equation (8.4) and Convention 6.1.7, both ϑ(l)\vartheta(l) and ϑ(m)\vartheta(m) have nn variables.

Assume g|Imϕg|Imϕg{{|}_{\operatorname{Im}\phi}}\neq{g}^{\prime}{{|}_{\operatorname{Im}\phi}}; that is, τ(ϑ(l))|Imϕτ(ϑ(m))|Imϕ{\tau}^{\prime}(\vartheta(l)){{|}_{\operatorname{Im}\phi}}\neq{\tau}^{\prime}(\vartheta(m)){{|}_{\operatorname{Im}\phi}}. Then
(z1,,zn)Sn\exists({{z}_{1}},\cdots,{{z}_{n}})\in{{S}^{n}} such that

ϑ(l)(ϕ(z1),,ϕ(zn))ϑ(m)(ϕ(z1),,ϕ(zn)).\vartheta(l)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}}))\neq\vartheta(m)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}})).

However, ϕ(l(z1,,zn))=ϕ(m(z1,,zn))\phi(l({{z}_{1}},\cdots,{{z}_{n}}))=\phi(m({{z}_{1}},\cdots,{{z}_{n}})) because τ(l)=τ(m)\tau(l)=\tau(m), and hence from Equation ((8.4)), we can tell

ϑ(l)(ϕ(z1),,ϕ(zn))=ϑ(m)(ϕ(z1),,ϕ(zn)),\vartheta(l)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}}))=\vartheta(m)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}})),

a contradiction.

Hence θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} must be a map. Then n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and ff\in\mathcal{F} of nn variables,

ϕ(τ(f)(a1,,an))\phi(\tau(f)(a_{1},\cdots,a_{n}))

=ϕ(f(a1,,an))=\phi(f(a_{1},\cdots,a_{n})) (by the definition of τ\tau)

=ϑ(f)(ϕ(a1),,ϕ(an))=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) (by Equation (8.4))

=τ(ϑ(f))(ϕ(a1),,ϕ(an))={\tau}^{\prime}(\vartheta(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) (by the definition of τ{\tau}^{\prime})

=θ(τ(f))(ϕ(a1),,ϕ(an))=\theta(\tau(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) (because (τ(f),τ(ϑ(f)))θ(\tau(f),{\tau}^{\prime}(\vartheta(f)))\in\theta and θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map).

Since Domθ={τ(f)|f}\operatorname{Dom}\theta=\{\tau(f)\,|\,f\in\mathcal{F}\}, by Definition 8.5.4, ϕ\phi is a θ\theta-morphism. ∎

Lemma 8.5.6 characterizes a θ\theta-morphism by formal functions. When operator gen-semigroups are induced from formal functions, the criterion for a θ\theta-morphism by Lemma 8.5.6 is usually more convenient than that by Definition 8.5.4. Thus Lemma 8.5.6 simplifies our proofs of the following several results.

The following generalizes both Propositions 6.1.10 and 8.3.4.

Proposition 8.5.7.

Let F/BF/B ((resp. F/B{F}^{\prime}/{B}^{\prime})) be a field extension, let an operator gen-semigroup TT ((resp. T{T}^{\prime})) be defined as in Example 8.1.3 on FF ((resp. F{F}^{\prime})) over BB ((resp. B{B}^{\prime})), let φ:BB\varphi:B\to{B}^{\prime} be a field isomorphism, and let

ϑ:n+B[x1,,xn]n+B[x1,,xn]\vartheta:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{B[{{x}_{1}},\cdots,{{x}_{n}}]}\to\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{{B}^{\prime}[{{x}_{1}},\cdots,{{x}_{n}}]}

be the map given by fff\mapsto{f}^{\prime} where

f(x1,,xn)=cp1,,pnx1p1xnpnf({{x}_{1}},\cdots,{{x}_{n}})=\sum{{c_{p_{1},\cdots,p_{n}}}x_{1}^{{{p}_{1}}}\cdots x_{n}^{{{p}_{n}}}}

and

f(x1,,xn)=φ(cp1,,pn)x1p1xnpn.{f}^{\prime}({{x}_{1}},\cdots,{{x}_{n}})=\sum{\varphi({c_{p_{1},\cdots,p_{n}}})x_{1}^{{{p}_{1}}}\cdots x_{n}^{{{p}_{n}}}}.

Let θ={(τ(f),τ(ϑ(f)))|fB[x1,,xn],n+}\theta=\{(\tau(f),{\tau}^{\prime}(\vartheta(f)))\,|\,f\in B[{{x}_{1}},\cdots,{{x}_{n}}],n\in{{\mathbb{Z}}^{+}}\}, where τ\tau and τ{\tau}^{\prime} are the maps inducing TT and T{T}^{\prime}, respectively, defined as in Example 8.1.3.

Then ()UF\forall(\emptyset\neq)U\subseteq F and ()VF(\emptyset\neq)V\subseteq{F}^{\prime}, a map ϕ:UTVT\phi:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{{{T}^{\prime}}}} is a ring homomorphism extending φ\varphi if and only if ϕ\phi is a θ\theta-morphism.

Proof.

We may prove Proposition 8.5.7 in a way analogous to the one for Proposition 6.1.10. However, now we employ Lemma 8.5.6 instead.

By Example 8.1.3,

T={f:FnFT=\{f^{*}:{{F}^{n}}\to F given by (a1,,an)f(a1,,an)|fB[x1,,xn],n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in B[{{x}_{1}},\cdots,{{x}_{n}}],n\in{{\mathbb{Z}}^{+}}\}

and

T={f:FnF{T}^{\prime}=\{f^{*}:{{{F}^{\prime}}^{n}}\to{F}^{\prime} given by (a1,,an)f(a1,,an)|fB[x1,,xn],n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in{B}^{\prime}[{{x}_{1}},\cdots,{{x}_{n}}],n\in{{\mathbb{Z}}^{+}}\}.

Thus UT{{\left\langle U\right\rangle}_{T}} and VT{{\left\langle V\right\rangle}_{{{T}^{\prime}}}} are the rings B[U]B[U] and B[V]{B}^{\prime}[V], respectively.

Let =n+B[x1,,xn]\mathcal{F}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{B[{{x}_{1}},\cdots,{{x}_{n}}]} and =n+B[x1,,xn]{\mathcal{F}}^{\prime}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{{B}^{\prime}[{{x}_{1}},\cdots,{{x}_{n}}]}. Then Lemma 8.5.6 applies. Hence it suffices to show that ϕ:UTVT\phi:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{{{T}^{\prime}}}} is a ring homomorphism extending φ\varphi if and only if statement (ii) in Lemma 8.5.6 is true in this case. Specifically, we only need to show that the following statements are equivalent:

  1. (1)

    cB\forall c\in B, ϕ(c)=φ(c)\phi(c)=\varphi(c), and a,bUT\forall a,b\in{{\left\langle U\right\rangle}_{T}}, ϕ(a+b)=ϕ(a)+ϕ(b)\phi(a+b)=\phi(a)+\phi(b) and ϕ(ab)=ϕ(a)ϕ(b)\phi(ab)=\phi(a)\phi(b).

  2. (2)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)UTn(a_{1},\cdots,a_{n})\in\left\langle U\right\rangle_{T}^{n} and fB[x1,,xn]f\in B[{{x}_{1}},\cdots,{{x}_{n}}],

    (8.5) ϕ(f(a1,,an))=ϑ(f)(ϕ(a1),,ϕ(an)).\phi(f(a_{1},\cdots,a_{n}))=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).

(1) \Rightarrow (2): n+\forall n\in{{\mathbb{Z}}^{+}}, f(x1,,xn)=cp1,,pnx1p1xnpnB[x1,,xn]f({{x}_{1}},\cdots,{{x}_{n}})=\sum{{c_{p_{1},\cdots,p_{n}}}x_{1}^{{{p}_{1}}}\cdots x_{n}^{{{p}_{n}}}}\in B[{{x}_{1}},\cdots,{{x}_{n}}] and (a1,,an)UTn(a_{1},\cdots,a_{n})\in\left\langle U\right\rangle_{T}^{n},

ϕ(f(a1,,an))\phi(f(a_{1},\cdots,a_{n}))

=ϕ(cp1,,pna1p1anpn)=\phi(\sum{{c_{p_{1},\cdots,p_{n}}}a_{1}^{{{p}_{1}}}\cdots a_{n}^{{{p}_{n}}}})

=ϕ(cp1,,pn)(ϕ(a1))p1(ϕ(an))pn=\sum{\phi({c_{p_{1},\cdots,p_{n}}})(\phi(a_{1}))^{p_{1}}\cdots(\phi(a_{n}))^{p_{n}}} (because statement (1) is true)

=φ(cp1,,pn)(ϕ(a1))p1(ϕ(an))pn=\sum{\varphi({c_{p_{1},\cdots,p_{n}}})(\phi(a_{1}))^{p_{1}}\cdots(\phi(a_{n}))^{p_{n}}} (because cB\forall c\in B, ϕ(c)=φ(c)\phi(c)=\varphi(c))

=ϑ(f)(ϕ(a1),,ϕ(an))=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) (by the definition of ϑ\vartheta).

(2) \Rightarrow (1): Considering the case where fB[x1,,xn]f\in B[{{x}_{1}},\cdots,{{x}_{n}}] is a constant polynomial, we can tell from Equation ((8.5)) and the definition of ϑ\vartheta that cB\forall c\in B, ϕ(c)=φ(c)\phi(c)=\varphi(c).

Let a,bUTa,b\in{{\left\langle U\right\rangle}_{T}}. Let g=x1+x2g={{x}_{1}}+{{x}_{2}} and let h=x1x2h={{x}_{1}}{{x}_{2}}. Then

ϕ(a+b)\phi(a+b)

=ϕ(g(a,b))=\phi(g(a,b))

=ϑ(g)(ϕ(a),ϕ(b))=\vartheta(g)(\phi(a),\phi(b)) (by Equation ((8.5)))

=g(ϕ(a),ϕ(b))=g(\phi(a),\phi(b)) (ϑ(g)=g\vartheta(g)=g by the definition of ϑ\vartheta)

=ϕ(a)+ϕ(b)=\phi(a)+\phi(b).

And

ϕ(ab)\phi(ab)

=ϕ(h(a,b))=\phi(h(a,b))

=ϑ(h)(ϕ(a),ϕ(b))=\vartheta(h)(\phi(a),\phi(b)) (by Equation ((8.5)))

=h(ϕ(a),ϕ(b))=h(\phi(a),\phi(b)) (ϑ(h)=h\vartheta(h)=h by the definition of ϑ\vartheta)

=ϕ(a)ϕ(b)=\phi(a)\phi(b). ∎

The following is for ring homomorphisms between not necessarily commutative rings.

Proposition 8.5.8.

Let AA and RR be rings with identity, let TA{{T}_{A}} ((resp. TR{{T}_{R}})) be the operator gen-semigroup on AA ((resp. RR)) defined as in Example 8.1.4, and let θ={(τA(f),τR(f))|f𝔽2x1,,xn,n+}\theta=\{({{\tau}_{A}}(f),{{\tau}_{R}}(f))\,|\,f\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle,n\in{{\mathbb{Z}}^{+}}\}, where τA{{\tau}_{A}} and τR{{\tau}_{R}} are the maps inducing TA{{T}_{A}} and TR{{T}_{R}}, respectively, defined as in Example 8.1.4. Then a map ϕ:AR\phi:A\to R is a ring homomorphism if and only if ϕ\phi is a θ\theta-morphism.

Proof.

By Example 8.1.4,

TA={f:AnA{{T}_{A}}=\{f^{*}:{{A}^{n}}\to A given by (a1,,an)f(a1,,an)|f𝔽2x1,,xn,n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle,n\in{{\mathbb{Z}}^{+}}\},

and

TR={f:RnR{{T}_{R}}=\{f^{*}:{{R}^{n}}\to R given by (a1,,an)f(a1,,an)|f𝔽2x1,,xn,n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle,n\in{{\mathbb{Z}}^{+}}\}.

Obviously ATA=A{{\left\langle A\right\rangle}_{{{T}_{A}}}}=A and RTR=R{{\left\langle R\right\rangle}_{{{T}_{R}}}}=R. Hence ϕ:AR\phi:A\to R is ϕ:ATARTR\phi:{{\left\langle A\right\rangle}_{{{T}_{A}}}}\to{{\left\langle R\right\rangle}_{{{T}_{R}}}}.

Let ==n+𝔽2x1,,xn\mathcal{F}={\mathcal{F}}^{\prime}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle} and let ϑ=Id\vartheta=\operatorname{Id}. Then Lemma 8.5.6 applies. By Lemma 8.5.6, we only need to show that ϕ:ATARTR\phi:{{\left\langle A\right\rangle}_{{{T}_{A}}}}\to{{\left\langle R\right\rangle}_{{{T}_{R}}}} is a ring homomorphism if and only if statement (ii) in Lemma 8.5.6 is true in this case. Specifically, it suffices to show that the following statements are equivalent:

  1. (1)

    ϕ(1A)=1R\phi({{1}_{A}})={{1}_{R}} and a,bA\forall a,b\in A, ϕ(a+b)=ϕ(a)+ϕ(b)\phi(a+b)=\phi(a)+\phi(b) and ϕ(ab)=ϕ(a)ϕ(b)\phi(ab)=\phi(a)\phi(b).

  2. (2)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)An(a_{1},\cdots,a_{n})\in{{A}^{n}} and f𝔽2x1,,xnf\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle,

    (8.6) ϕ(f(a1,,an))=f(ϕ(a1),,ϕ(an)).\phi(f(a_{1},\cdots,a_{n}))=f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).

(1) \Rightarrow (2): Since f𝔽2x1,,xnf\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle is a polynomial, it is not hard to see that statement (2) is true.

(2) \Rightarrow (1): Considering the case where f𝔽2x1,,xnf\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle is the constant polynomial 1, we can tell from Equation ((8.6)) that ϕ(1A)=1R\phi({{1}_{A}})={{1}_{R}}.

Let a,bAa,b\in A. Let g=x1+x2g={{x}_{1}}+{{x}_{2}} and h=x1x2h={{x}_{1}}{{x}_{2}}. Then

ϕ(a+b)\phi(a+b)

=ϕ(g(a,b))=\phi(g(a,b))

=g(ϕ(a),ϕ(b))=g(\phi(a),\phi(b)) (by Equation ((8.6)))

=ϕ(a)+ϕ(b)=\phi(a)+\phi(b).

And

ϕ(ab)\phi(ab)

=ϕ(h(a,b))=\phi(h(a,b))

=h(ϕ(a),ϕ(b))=h(\phi(a),\phi(b)) (by Equation ((8.6)))

=ϕ(a)ϕ(b)=\phi(a)\phi(b). ∎

Proposition 8.3.5 is generalized to

Proposition 8.5.9.

Let MM and NN be modules over a commutative ring RR with identity, let TM{{T}_{M}} ((resp. TN{{T}_{N}})) be the operator gen-semigroup on MM ((resp. NN)) defined as in Example 8.1.5, and let

θ={(τM(f),τN(f))|f{i=1nrixi|riR},n+},\theta=\{({{\tau}_{M}}(f),{{\tau}_{N}}(f))\,|\,f\in\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\},n\in{{\mathbb{Z}}^{+}}\},

where τM{{\tau}_{M}} and τN{{\tau}_{N}} are the maps inducing TM{{T}_{M}} and TN{{T}_{N}}, respectively, defined as in Example 8.1.5.

If XX and YY are nonempty subsets of MM and NN, respectively, then a map ϕ:XTMYTN\phi:{{\left\langle X\right\rangle}_{{{T}_{M}}}}\to{{\left\langle Y\right\rangle}_{{{T}_{N}}}} is an R-homomorphism of R-modules if and only if ϕ\phi is a θ\theta-morphism.

Proof.

Our proof is comparable with that of Proposition 8.3.5.

By Example 8.1.5,

TM={f:MnM{{T}_{M}}=\{f^{*}:{{M}^{n}}\to M given by (a1,,an)f(a1,,an)|f{i=1nrixi|riR},n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\},n\in{{\mathbb{Z}}^{+}}\}

and

TN={f:NnN{{T}_{N}}=\{f^{*}:{{N}^{n}}\to N given by (a1,,an)f(a1,,an)|f{i=1nrixi|riR},n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\},n\in{{\mathbb{Z}}^{+}}\}.

Clearly, XTM{{\left\langle X\right\rangle}_{{{T}_{M}}}} and YTN{{\left\langle Y\right\rangle}_{{{T}_{N}}}} are the submodules of MM and NN generated by XX and YY, respectively.

Let ==n+{i=1nrixi|riR}\mathcal{F}={\mathcal{F}}^{\prime}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\}} and let ϑ=Id\vartheta=\operatorname{Id}. Then Lemma 8.5.6 applies. By Lemma 8.5.6, it suffices to show that ϕ:XTMYTN\phi:{{\left\langle X\right\rangle}_{{{T}_{M}}}}\to{{\left\langle Y\right\rangle}_{{{T}_{N}}}} is an RR-homomorphism of RR-modules if and only if statement (ii) in Lemma 8.5.6 is true in this case. Specifically, we only need to show that the following statements are equivalent:

  1. (1)

    a,bXTM\forall a,b\in{{\left\langle X\right\rangle}_{{{T}_{M}}}} and rRr\in R, ϕ(a+b)=ϕ(a)+ϕ(b)\phi(a+b)=\phi(a)+\phi(b) and ϕ(ra)=rϕ(a)\phi(ra)=r\phi(a).

  2. (2)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)XTMn(a_{1},\cdots,a_{n})\in\left\langle X\right\rangle_{{{T}_{M}}}^{n} and f{i=1nrixi|riR}f\in\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\},

    (8.7) ϕ(f(a1,,an))=f(ϕ(a1),,ϕ(an)).\phi(f(a_{1},\cdots,a_{n}))=f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).

(1) \Rightarrow (2): Because f{i=1nrixi|riR}f\in\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\} is a linear polynomial whose constant term is zero, it is not hard to see that statement (2) is true.

(2) \Rightarrow (1): Let a,bXTMa,b\in{{\left\langle X\right\rangle}_{{{T}_{M}}}} and let rRr\in R. Let g=x1+x2g={{x}_{1}}+{{x}_{2}} and let h=rx1h=r{{x}_{1}}. Then

ϕ(a+b)\phi(a+b)

=ϕ(g(a,b))=\phi(g(a,b))

=g(ϕ(a),ϕ(b))=g(\phi(a),\phi(b)) (by Equation ((8.7)))

=ϕ(a)+ϕ(b)=\phi(a)+\phi(b).

And

ϕ(ra)\phi(ra)

=ϕ(h(a))=\phi(h(a))

=h(ϕ(a))=h(\phi(a)) (by Equation ((8.7)))

=rϕ(a)=r\phi(a). ∎

Proposition 8.3.6 is generalized as follows.

Proposition 8.5.10.

Let GG and HH be abelian groups, let TG{{T}_{G}} ((resp. TH{{T}_{H}})) be the operator gen-semigroup on GG ((resp. HH)) defined as in Example 8.1.6, and let θ={(τG(f),τH(f))|f{i=1nxipi|pi},n+}\theta=\{({{\tau}_{G}}(f),{{\tau}_{H}}(f))\,|\,f\in\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\},n\in{{\mathbb{Z}}^{+}}\} , where τG{{\tau}_{G}} and τH{{\tau}_{H}} are the maps inducing TG{{T}_{G}} and TH{{T}_{H}}, respectively, defined as in Example 8.1.6.

If XX and YY are nonempty subsets of GG and HH respectively, then a map ϕ:XTGYTH\phi:{{\left\langle X\right\rangle}_{{{T}_{G}}}}\to{{\left\langle Y\right\rangle}_{{{T}_{H}}}} is a group homomorphism if and only if ϕ\phi is a θ\theta-morphism.

Proof.

By Example 8.1.6,

TG=n+{f:GnG{{T}_{G}}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{f^{*}:{{G}^{n}}\to G} given by (a1,,an)f(a1,,an)|f{i=1nxipi|pi}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\}}

and

TH=n+{f:HnH{{T}_{H}}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{f^{*}:{{H}^{n}}\to H} given by (a1,,an)f(a1,,an)|f{i=1nxipi|pi}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\}}.

Besides, XTG{{\left\langle X\right\rangle}_{{{T}_{G}}}} and YTH{{\left\langle Y\right\rangle}_{{{T}_{H}}}} are the subgroups of GG and HH generated by XX and YY, respectively.

Let ==n+{i=1nxipi|pi}\mathcal{F}={\mathcal{F}}^{\prime}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\}} and let ϑ=Id\vartheta=\operatorname{Id}. Then Lemma 8.5.6 applies. By Lemma 8.5.6, it suffices to show that ϕ:XTGYTH\phi:{{\left\langle X\right\rangle}_{{{T}_{G}}}}\to{{\left\langle Y\right\rangle}_{{{T}_{H}}}} is a group homomorphism if and only if statement (ii) in Lemma 8.5.6 is true in this case. Specifically, we only need to show that the following statements are equivalent:

  1. (1)

    a,bXTG\forall a,b\in{{\left\langle X\right\rangle}_{{{T}_{G}}}}, ϕ(ab)=ϕ(a)ϕ(b)\phi(ab)=\phi(a)\phi(b).

  2. (2)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)XTGn(a_{1},\cdots,a_{n})\in\left\langle X\right\rangle_{{{T}_{G}}}^{n} and f{i=1nxipi|pi}f\in\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\},

    ϕ(f(a1,,an))=f(ϕ(a1),,ϕ(an)).\phi(f(a_{1},\cdots,a_{n}))=f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).

It is not hard to see (1)\Leftrightarrow(2). ∎

Proposition 8.3.7 is generalized to

Proposition 8.5.11.

Let GG and HH be groups, let TG{{T}_{G}} ((resp. TH{{T}_{H}})) be the operator gen-semigroup on GG ((resp. HH)) defined as in Example 8.1.7, and let θ={(τG(f),τH(f))|fx1,,xn,x11,,xn1,n+}\theta=\{({{\tau}_{G}}(f),{{\tau}_{H}}(f))\,|\,f\in\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle,n\in{{\mathbb{Z}}^{+}}\}, where τG{{\tau}_{G}} and τH{{\tau}_{H}} are the maps inducing TG{{T}_{G}} and TH{{T}_{H}}, respectively, defined as in Example 8.1.7.

If XX and YY are nonempty subsets of GG and HH, respectively, then a map ϕ:XTGYTH\phi:{{\left\langle X\right\rangle}_{{{T}_{G}}}}\to{{\left\langle Y\right\rangle}_{{{T}_{H}}}} is a group homomorphism if and only if ϕ\phi is a θ\theta-morphism.

Proof.

Our proof is comparable with that of Proposition 8.5.10.

By Example 8.1.7,

T=n+{f:GnGT=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{f^{*}:{{G}^{n}}\to G} given by (a1,,an)f(a1,,an)|fx1,,xn,x11,,xn1}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle\}.

and

T=n+{f:HnHT=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{f^{*}:{{H}^{n}}\to H} given by (a1,,an)f(a1,,an)|fx1,,xn,x11,,xn1}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle\}.

Clearly, XTG{{\left\langle X\right\rangle}_{{{T}_{G}}}} and YTH{{\left\langle Y\right\rangle}_{{{T}_{H}}}} are the subgroups of GG and HH generated by XX and YY, respectively.

Let ==n+x1,,xn,x11,,xn1\mathcal{F}={\mathcal{F}}^{\prime}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle} and let ϑ=Id\vartheta=\operatorname{Id}. Then Lemma 8.5.6 applies. By Lemma 8.5.6, we only need to show that ϕ:XTGYTH\phi:{{\left\langle X\right\rangle}_{{{T}_{G}}}}\to{{\left\langle Y\right\rangle}_{{{T}_{H}}}} is a group homomorphism if and only if statement (ii) in Lemma 8.5.6 is true in this case. Specifically, it suffices to show that the following statements are equivalent:

  1. (1)

    a,bXTG\forall a,b\in{{\left\langle X\right\rangle}_{{{T}_{G}}}}, ϕ(ab)=ϕ(a)ϕ(b)\phi(ab)=\phi(a)\phi(b).

  2. (2)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)XTGn(a_{1},\cdots,a_{n})\in\left\langle X\right\rangle_{{{T}_{G}}}^{n} and fx1,,xn,x11,,xn1f\in\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle,

    ϕ(f(a1,,an))=f(ϕ(a1),,ϕ(an)).\phi(f(a_{1},\cdots,a_{n}))=f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).

It is not hard to see (1)\Leftrightarrow(2). ∎

8.6. θ\theta-isomorphisms

For T1T_{1} and T2T_{2} being operator gen-semigroups, Definition 6.2.1 still applies: a θ\theta-morphism from a T1T_{1}-space S1S_{1} to a T2T_{2}-space is a θ\theta-isomorphism if it is bijective. To justify the definition, we need to show Lemma 6.2.2 and Proposition 6.2.3 for the case of operator gen-semigroups as follows.

Lemma 8.6.1.

(Lemma 6.2.2) Let ϕ\phi be a θ\theta-isomorphism from a T1T_{1}-space S1S_{1} to a T2T_{2}-space S2S_{2}. Then θ1|S1\theta^{-1}{{|}_{S_{1}}} is a map, where θ1:={(g,f)|(f,g)θ}\theta^{-1}:=\{(g,f)\,|\,(f,g)\in\theta\}.

Proof.

Let (g,f1),(g,f2)θ1(g,f_{1}),(g,f_{2})\in\theta^{-1} with gg of nn variables. Then by Corollary 8.5.5, both f1f_{1} and f2f_{2} have nn variables. Let (a1,,an)S1n(a_{1},\cdots,a_{n})\in S_{1}^{n}. Then to show that θ1|S1\theta^{-1}{{|}_{S_{1}}} is a map, by Proposition 8.5.3 and Definition 8.5.2, it suffices to show that f1(a1,,an)=f2(a1,,an)f_{1}(a_{1},\cdots,a_{n})=f_{2}(a_{1},\cdots,a_{n}). Moreover, since ϕ\phi is injective, we only need to show ϕ(f1(a1,,an))=ϕ(f2(a1,,an))\phi(f_{1}(a_{1},\cdots,a_{n}))=\phi(f_{2}(a_{1},\cdots,a_{n})):
ϕ(f1(a1,,an))\phi(f_{1}(a_{1},\cdots,a_{n}))
=θ(f1)(ϕ(a1),,ϕ(an))=\theta(f_{1})(\phi(a_{1}),\cdots,\phi(a_{n})) (by Definition 8.5.4)
=g(ϕ(a1),,ϕ(an))=g(\phi(a_{1}),\cdots,\phi(a_{n})) (because (f1,g)θ(f_{1},g)\in\theta and θ|Imϕ\theta{|}_{\operatorname{Im}\phi} is a map by Corollary 8.5.5)
=θ(f2)(ϕ(a1),,ϕ(an))=\theta(f_{2})(\phi(a_{1}),\cdots,\phi(a_{n})) (because (f2,g)θ(f_{2},g)\in\theta and θ|Imϕ\theta{|}_{\operatorname{Im}\phi} is a map)
=ϕ(f2(a1,,an))=\phi(f_{2}(a_{1},\cdots,a_{n})) (by Definition 8.5.4). ∎

Proposition 8.6.2.

(Proposition 6.2.3) Let S1S_{1} and S2S_{2} be a T1T_{1}-space and a T2T_{2}-space, respectively, let ϕIsoθ(S1,S2)\phi\in\operatorname{Iso}_{\theta}({{S}_{1}},{{S}_{2}}) and let ϕ1{{\phi}^{-1}} be the inverse map of ϕ\phi. Then ϕ1Isoθ1(S2,S1){{\phi}^{-1}}\in\operatorname{Iso}_{\theta^{-1}}({{S}_{2}},{{S}_{1}}).

Proof.

By Definition 8.5.4, n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)S1n(a_{1},\cdots,a_{n})\in{S_{1}^{n}} and fDomθf\in\operatorname{Dom}\theta of nn variables,

f(a1,,an)=ϕ1(θ(f)(ϕ(a1),,ϕ(an))).f(a_{1},\cdots,a_{n})=\phi^{-1}(\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))).

Hence n+,fDomθ\forall n\in{{\mathbb{Z}}^{+}},f\in\operatorname{Dom}\theta of nn variables and (b1,,bn)S2n(b_{1},\cdots,b_{n})\in{S_{2}^{n}},

(8.8) f(ϕ1(b1),,ϕ1(bn))=ϕ1(θ(f)(b1,,bn)).f({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n}))=\phi^{-1}(\theta(f)(b_{1},\cdots,b_{n})).

By Lemma 8.6.1, θ1|S1\theta^{-1}{{|}_{S_{1}}} is a map. Then by Proposition 8.5.3, gDomθ1(=Imθ)\forall g\in\operatorname{Dom}\theta^{-1}(=\operatorname{Im}\theta), n+\exists n\in{{\mathbb{Z}}^{+}} such that (a1,,an)S1n\forall(a_{1},\cdots,a_{n})\in S_{1}^{n}, θ1(g)(a1,,an)\theta^{-1}(g)(a_{1},\cdots,a_{n}) is well-defined, and from (8.8), we can tell that the number of variables of gg is also nn.

Thus n+\forall n\in{{\mathbb{Z}}^{+}}, gDomθ1g\in\operatorname{Dom}\theta^{-1} of nn variables and (b1,,bn)S2n(b_{1},\cdots,b_{n})\in{S_{2}^{n}}, it follows from (8.8) that

θ1(g)(ϕ1(b1),,ϕ1(bn))=ϕ1(g(b1,,bn)).\theta^{-1}(g)({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n}))={{\phi}^{-1}}(g(b_{1},\cdots,b_{n})).

Hence by Definition 8.5.4, ϕ1\phi^{-1} is a θ1\theta^{-1}-morphism from S2S_{2} to S1S_{1}.

Moreover, since ϕ1\phi^{-1} is bijective, by Definition 6.2.1, ϕ1Isoθ1(S2,S1){{\phi}^{-1}}\in\operatorname{Iso}_{\theta^{-1}}({{S}_{2}},{{S}_{1}}). ∎

For operator gen-semigroups, we will generalize most results obtained in Sections 2 to 7. Before doing this in Sections 10 to 14, however, we shall generalize the notion of operator gen-semigroup further in the next section.

9. Basic notions for the case of (multivariable) partial functions

In this section, we shall allow elements in TT to be partial functions. Then we will find that some more familiar concepts such as ring homomorphisms between fields and covariant functors between categories can be characterized in terms of TT-morphisms or θ\theta-morphisms.

Subsection 9.1 generalizes the notion of operator gen-semigroup. Then correspondingly, the notions of TT-spaces, TT-morphisms, TT-isomorphisms, θ\theta-morphisms and θ\theta-isomorphisms are generalized in Sections 9.2, 9.3, 9.4, 9.5 and 9.6, respectively.

9.1. Partial-operator generalized-semigroup TT

Recall the concept of partial functions as follows.

Definition 9.1.1.

Let XX and YY be sets. A partial function ff of n(+)n(\in{{\mathbb{Z}}^{+}}) variables from the cartesian product Xn{X}^{n} to YY is a (total) function from a subset SS of Xn{X}^{n} to YY, and SS, which may be empty, is called the domain of definition of ff.

Moreover, let (x1,,xm)Xm(x_{1},\cdots,x_{m})\in X^{m}, where m+m\in{{\mathbb{Z}}^{+}}. If (x1,,xm)(x_{1},\cdots,x_{m}) lies in the domain of definition of ff, then f(x1,,xm)f(x_{1},\cdots,x_{m}) is said to be well-defined.

Remark.

By the definition, any function is a partial function.

We are going to extend our study to partial functions. First, we generalize Definition 8.1.1 to

Definition 9.1.2.

Let DD be a set, let n+n\in{{\mathbb{Z}}^{+}}, and let ff be a partial function from the cartesian product Dn{{D}^{n}} to DD. i=1,,n\forall i=1,\cdots,n, let ni+n_{i}\in{{\mathbb{Z}}^{+}} and let gi{{g}_{i}} be a partial function from the cartesian product Dni{{D}^{{{n}_{i}}}} to DD. Let m=inim=\sum\nolimits_{i}{{n_{i}}}. Then the composite f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) is the partial function from Dm{{D}^{m}} to DD given by

(x1,,xm)f(g1(v1),,gn(vn)),({{x}_{1}},\cdots,{{x}_{m}})\mapsto f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})),

where vi:=(xmi+1,,xmi+ni)Dni,i=1,,n{{\text{v}}_{i}}:=({{x}_{{{m}_{i}}+1}},\cdots,{{x}_{{{m}_{i}}+{{n}_{i}}}})\in{{D}^{{{n}_{i}}}},\forall i=1,\cdots,n, m1:=0{{m}_{1}}:=0 and mi:=k=1i1nk,i=2,,n{{m}_{i}}:=\sum\nolimits_{k=1}^{i-1}{{{n}_{k}}},\forall i=2,\cdots,n.

Remark.
  1. (1)

    Note that (x1,,xm)({{x}_{1}},\cdots,{{x}_{m}}) is in the domain of definition of f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) if and only if each gi(vi){{g}_{i}}({{\text{v}}_{i}}) is well-defined and f(g1(v1),,gn(vn))f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})) is well-defined.

  2. (2)

    We may denote the nn-tuple (g,g,,g)(g,g,\cdots,g) by gn{{g}^{n}} (in e.g. Section 14) for brevity.

Before we generalize Definition 8.1.2, we need a terminology as follows.

Terminology 9.1.3.

Let ff and gg be partial functions from AA to BB. If the domain of definition of ff contains the domain of definition of gg, which is denoted by D(g)D(g), and f|D(g)=g|D(g)f{{|}_{D(g)}}=g{{|}_{D(g)}}, then we say that gg is a restriction of ff.

Remark.

Trivially, if D(g)=D(g)=\emptyset, then gg is a restriction of any ff.

Definition 9.1.4.

Let DD be a set and let TT be a subset of

{all partial functions from the cartesian product Dn{{D}^{n}} to D|n+D\,|\,n\in{{\mathbb{Z}}^{+}}}.

If n+\forall n\in{{\mathbb{Z}}^{+}}, fTf\in T of nn variables, and (g1,,gn)Tn({{g}_{1}},\cdots,{{g}_{n}})\in{{T}^{n}} (cartesian product), the composite f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) is a restriction of some element of TT, then we call TT a partial-operator generalized-semigroup, abbreviated as par-operator gen-semigroup, on DD and DD is called the domain of TT.

Remark.
  1. (1)

    Trivially, \emptyset is a par-operator gen-semigroup.

  2. (2)

    Obviously, any operator gen-semigroup is a par-operator gen-semigroup.

Definition 2.1.6 is generalized to

Definition 9.1.5.

Let DD be a set and GG be a subset of

{all partial functions from the cartesian product Dn{{D}^{n}} to D|n+D\,|\,n\in{{\mathbb{Z}}^{+}}}.

Then we call the intersection of all par-operator gen-semigroups on DD containing GG the par-operator gen-semigroup on DD generated by GG, and denote it by G\left\langle G\right\rangle.

Remark.

From Definition 9.1.4, we can tell that G\left\langle G\right\rangle is the smallest par-operator gen-semigroup on DD which contains GG.

Example 9.1.6.

Let BB be a subfield of a field FF. Let

T={T=\{the partial function f:FnFf^{*}:{{F}^{n}}\to F given by (a1,,an)f(a1,,an)|fFrac(B[x1,,xn]),n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\text{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}]),n\in{{\mathbb{Z}}^{+}}\}.

Then it is not hard to see that TT is a par-operator gen-semigroup on FF. The map which induces TT, i.e. τ:n+Frac(B[x1,,xn])T\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\text{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}])}\to T given by fff\mapsto f^{*}, is surjective, but it is not necessarily injective.

Example 9.1.7.

Let BB be a differential subfield of a differential field FF; that is, field FF is endowed with a derivation and the derivation of FF restricts to the derivation of BB. Let T=TFTT=\left\langle{{T}_{F}}\bigcup{{T}_{\partial}}\right\rangle (Definition 9.1.5), where TF{{T}_{F}} is the par-operator gen-semigroup on FF defined as in Example 9.1.6 and T{{T}_{\partial}} is the operator semigroup defined in Example 2.1.5. Then by Definition 9.1.5, TT is a par-operator gen-semigroup on FF.

Example 9.1.8.

For this example, DD in Definitions 9.1.2 and 9.1.4 is generalized to be a class. Let 𝒞\mathcal{C} be a category and let DD be the collection of all morphisms in 𝒞\mathcal{C}. n+\forall n\in{{\mathbb{Z}}^{+}}, let

n={{{\mathcal{F}}_{n}}=\{word ww on {x1,,xn}|\{x_{1},\cdots,x_{n}\}\,| each of x1,,xnx_{1},\cdots,x_{n} arises in ww}.

Let T={T=\{the partial function f:DnDf^{*}:{{D}^{n}}\to D given by

(a1,,an)f(a1,,an)|fn,n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in{{\mathcal{F}}_{n}},n\in{{\mathbb{Z}}^{+}}\},

where f(a1,,an)f(a_{1},\cdots,a_{n}) is defined to be the composite ai1aik{{a}_{{{i}_{1}}}}\circ\cdots\circ{{a}_{{{i}_{k}}}} for f=xi1xikf={{x}_{{{i}_{1}}}}\cdots{{x}_{{{i}_{k}}}} (where i1,,ik{1,,n}{{i}_{1}},\cdots,{{i}_{k}}\in\{1,\cdots,n\}).

In particular, for f=x11f=x_{1}\in{{\mathcal{F}}_{1}}, ff^{*} is the identity function on DD.

Thus, for fnf\in{{\mathcal{F}}_{n}}, (a1,,an)Dn(a_{1},\cdots,a_{n})\in D^{n} and i1,,ik{1,,n}{{i}_{1}},\cdots,{{i}_{k}}\in\{1,\cdots,n\}, the following four statements are equivalent:

  1. (i)

    (a1,,an)(a_{1},\cdots,a_{n}) is in the domain of definition of the ff^{*} induced by f=xi1xikf={{x}_{{{i}_{1}}}}\cdots{{x}_{{{i}_{k}}}}.

  2. (ii)

    for f=xi1xikf={{x}_{{{i}_{1}}}}\cdots{{x}_{{{i}_{k}}}}, f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined.

  3. (iii)

    ai1aik{{a}_{{{i}_{1}}}}\circ\cdots\circ{{a}_{{{i}_{k}}}} is well-defined.

  4. (iv)

    j=1,,(k1)\forall j=1,\cdots,(k-1), the domain (object) of aij{{a}_{{{i}_{j}}}} is the target (object) of aij+1{{a}_{{{i}_{j+1}}}}.

Then it is not hard to see that TT is a par-operator gen-semigroup on DD. Let τ:n+nT\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{{{\mathcal{F}}_{n}}}\to T be the map inducing TT given by fff\mapsto f^{*}.

9.2. TT-spaces

Definition 9.2.1.

Let TT be a par-operator gen-semigroup on DD and let UDU\subseteq D.

Then we call

S:={S:=\{well-defined f(u1,,un)|fTf({{u}_{1}},\cdots,{{u}_{n}})\,|\,f\in T, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}}, n+}n\in{{\mathbb{Z}}^{+}}\}

the TT-space generated by UU and denote SS by UT{{\left\langle U\right\rangle}_{T}}.

Moreover, if BSB\subseteq S is also a TT-space, then we say that BB is a TT-subspace of SS and write BSB\leq S or SBS\geq B.

Example 9.2.2.

Let BB be a subfield of a field FF. Let TT be the par-operator gen-semigroup on FF defined in Example 9.1.6. Then ()UF\forall(\emptyset\neq)U\subseteq F, we can tell that the TT-space UT{{\left\langle U\right\rangle}_{T}} is the field B(U)B(U).

Example 9.2.3.

Let BB be a differential subfield of a differential field FF. Let T(=TFT)T(=\left\langle{{T}_{F}}\bigcup{{T}_{\partial}}\right\rangle) be the par-operator gen-semigroup on FF defined in Example 9.1.7. Then we claim that ()UF\forall(\emptyset\neq)U\subseteq F, the TT-space UT{{\left\langle U\right\rangle}_{T}} is the differential subfield of FF generated by UU over BB.

To show the claim, let K=UTK={{\left\langle U\right\rangle}_{T}}. Then by Proposition 10.1.1 (to be shown in Section 10), which generalizes Proposition 2.2.7, KTK{{\left\langle K\right\rangle}_{T}}\subseteq K. And KTFKT{{\left\langle K\right\rangle}_{{{T}_{F}}}}\subseteq{{\left\langle K\right\rangle}_{T}} because TFT{{T}_{F}}\subseteq T. Hence KTFK{{\left\langle K\right\rangle}_{{{T}_{F}}}}\subseteq K. On the other hand, KTFK{{\left\langle K\right\rangle}_{{{T}_{F}}}}\supseteq K because IdTF\operatorname{Id}\in{{T}_{F}}. Therefore KTF=K{{\left\langle K\right\rangle}_{{{T}_{F}}}}=K, and thus by Example 9.2.2, KK is a field. Analogously, we can show KT=K{{\left\langle K\right\rangle}_{{{T}_{\partial}}}}=K, and hence KK is a differential field.

Example 9.2.4.

Let 𝒞\mathcal{C} be a category. Let DD and TT be defined as in Example 9.1.8. Then the TT-space DT=D{{\left\langle D\right\rangle}_{T}}=D since IdT\operatorname{Id}\in T. Because there is a bijection between the objects AA in 𝒞\mathcal{C} and their identity morphisms 1A1_{A}, we may view DT{{\left\langle D\right\rangle}_{T}} as 𝒞\mathcal{C}.

9.3. TT-morphisms

Now we generalize Definition 8.3.1 to

Definition 9.3.1.

Let TT be a par-operator gen-semigroup. Let σ\sigma be a map from a TT-space SS to a TT-space. If n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and fTf\in T, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor f(σ(a1),,f(\sigma({{a}_{1}}),\cdots, σ(an))\sigma({{a}_{n}})) is well-defined or

σ(f(a1,,an))=f(σ(a1),,σ(an)),\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})),

then we call σ\sigma a TT-morphism.

Remark.

From Definitions 9.1.4 and 9.2.1 (or from Proposition 10.1.1), we can tell that if f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined, then f(a1,,an)Sf(a_{1},\cdots,a_{n})\in S, and so σ(f(a1,,an))\sigma(f(a_{1},\cdots,a_{n})) is well-defined.

Note that in Examples 9.1.6 and 9.1.8, par-operator gen-semigroups are induced from rational functions or words on a set. Hence we generalize Definition 8.3.2 as follows.

Definition 9.3.2.

Let ff be an expression associated with a rule RR. Let n+n\in\mathbb{Z}^{+}.

If all of x1,,xnx_{1},\cdots,x_{n} arise as symbols in ff, and there are sets CC and DD such that, by the rule RR, ff yields nothing or a single element in CC, which we denote by f(a1,,an)(C)f(a_{1},\cdots,a_{n})(\in C), whenever every x1,,xnx_{1},\cdots,x_{n} in ff is replaced by any a1,,anDa_{1},\cdots,a_{n}\in D, respectively, then we call the couple (f,R)(f,R) a (CC-valued) formal partial function of nn variables x1,,xnx_{1},\cdots,x_{n} and DD is called a domain of (f,R)(f,R).

Moreover, in the case where a1,,anDa_{1},\cdots,a_{n}\in D and ff yields f(a1,,an)f(a_{1},\cdots,a_{n}), we say that f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined or (a1,,an)(a_{1},\cdots,a_{n}) is in the domain of definition of f(x1,,xn)f(x_{1},\cdots,x_{n}).

Besides, we may call ff a formal partial function if its associated rule RR is clear from the context.

Remark.

Analogous to a formal function, a formal partial function may have more than one domain. And the notion of formal partial function generalizes the notion of partial function.

When par-operator gen-semigroups are induced from formal partial functions, the following lemma, which generalizes Lemma 8.3.3, will be useful.

Lemma 9.3.3.

Let \mathcal{F} be a set of DD-valued formal partial functions on a domain DD and let TT be a par-operator gen-semigroup on DD induced by \mathcal{F}; that is, there is a surjective map τ:T\tau:\mathcal{F}\to T being given by fff\mapsto f^{*}, where f:DnDf^{*}:{{D}^{n}}\to D is given by (a1,,an)f(a1,,an)(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n}) for ff of nn variables, n+\forall n\in{{\mathbb{Z}}^{+}}. Let σ\sigma be a map from a TT-space SS to a TT-space. Then the following statements are equivalent:

  1. (i)

    σ\sigma is a TT-morphism.

  2. (ii)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and ff\in\mathcal{F} of nn variables, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor f(σ(a1),,σ(an))f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) is well-defined or

    (9.1) σ(f(a1,,an))=f(σ(a1),,σ(an)).\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).
Proof.

(i) \Rightarrow (ii): By Definition 9.3.1, n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and ff\in\mathcal{F} of nn variables, neither τ(f)(a1,,an)\tau(f)(a_{1},\cdots,a_{n}) nor τ(f)(σ(a1),,σ(an))\tau(f)(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) is well-defined or σ(τ(f)(a1,,an))=τ(f)(σ(a1),,σ(an))\sigma(\tau(f)(a_{1},\cdots,a_{n}))=\tau(f)(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

Then by the definition of τ\tau, (ii) must be true.

(ii) \Rightarrow (i): Since τ\tau is surjective, fT\forall f^{*}\in T, f\exists f\in\mathcal{F} such that f=τ(f)f^{*}=\tau(f). Then from statement (ii) and the definition of τ\tau, we can tell that n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and fTf^{*}\in T of nn variables, neither f(a1,,an)f^{*}(a_{1},\cdots,a_{n}) nor f(σ(a1),,σ(an))f^{*}(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) is well-defined or

σ(f(a1,,an))=f(σ(a1),,σ(an))\sigma(f^{*}(a_{1},\cdots,a_{n}))=f^{*}(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

So by Definition 9.3.1, σ\sigma must be a TT-morphism. ∎

Lemma 9.3.3 characterizes a TT-morphism by formal partial functions which induce TT. When a par-operator gen-semigroup is induced from formal partial functions, the criterion for a TT-morphism by Lemma 9.3.3 is normally more convenient than that by Definition 9.3.1.

The following is comparable with Proposition 8.3.4, except that now TT is defined by Example 9.1.6, and hence UT{{\left\langle U\right\rangle}_{T}} and VT{{\left\langle V\right\rangle}_{T}} must be fields.

Proposition 9.3.4.

Let BB be a subfield of a field FF, let TT be the par-operator gen-semigroup on FF defined in Example 9.1.6, and let nonempty U,VFU,V\subseteq F. Then a map σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} is a ring homomorphism with BB fixed pointwisely if and only if it is a TT-morphism.

Proof.

By the definition of TT in Example 9.1.6,

T={T=\{the partial function f:FnFf^{*}:{{F}^{n}}\to F given by (a1,,an)f(a1,,an)|fFrac(B[x1,,xn]),n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\text{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}]),n\in{{\mathbb{Z}}^{+}}\}.

Then UT{{\left\langle U\right\rangle}_{T}} and VT{{\left\langle V\right\rangle}_{T}} are the fields B(U)B(U) and B(V)B(V), respectively. Let ={fFrac(B[x1,,xn])|n+}\mathcal{F}=\{f\in\text{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}])\ |\ n\in{{\mathbb{Z}}^{+}}\}. Then Lemma 9.3.3 applies. Hence it suffices to show that σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} is a ring homomorphism with BB fixed pointwisely if and only if statement (ii) in Lemma 9.3.3 is true in this case. Specifically, we only need to show that the following statements are equivalent:

  1. (1)

    cB\forall c\in B, σ(c)=c\sigma(c)=c, and a,bUT(=B(U))\forall a,b\in{{\left\langle U\right\rangle}_{T}}(=B(U)), σ(a+b)=σ(a)+σ(b)\sigma(a+b)=\sigma(a)+\sigma(b) and σ(ab)=σ(a)σ(b)\sigma(ab)=\sigma(a)\sigma(b).

  2. (2)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)UTn(a_{1},\cdots,a_{n})\in\left\langle U\right\rangle_{T}^{n} and fFrac(B[x1,,xn])f\in\text{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}]), neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor f(σ(a1),,σ(an))f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) is well-defined or

    (9.2) σ(f(a1,,an))=f(σ(a1),,σ(an)).\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

(1) \Rightarrow (2): Let f(x1,,xn)=g(x1,,xn)/h(x1,,xn)f({{x}_{1}},\cdots,{{x}_{n}})=g({{x}_{1}},\cdots,{{x}_{n}})/h({{x}_{1}},\cdots,{{x}_{n}}), where n+n\in{{\mathbb{Z}}^{+}} and h,gB[x1,,xn]h,g\in B[{{x}_{1}},\cdots,{{x}_{n}}]. Let (a1,,an)UTn(a_{1},\cdots,a_{n})\in\left\langle U\right\rangle_{T}^{n}. Then

f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined;

h(a1,,an)0\Leftrightarrow h(a_{1},\cdots,a_{n})\neq 0;

σ(h(a1,,an))0\Leftrightarrow\sigma(h(a_{1},\cdots,a_{n}))\neq 0 (by statement (1));

f(σ(a1),,σ(an))\Leftrightarrow f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}))

=g(σ(a1),,σ(an))/h(σ(a1),,σ(an))=g(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}))/h(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}))

=σ(g(a1,,an))/σ(h(a1,,an))=\sigma(g(a_{1},\cdots,a_{n}))/\sigma(h(a_{1},\cdots,a_{n})) (because h,gB[x1,,xn]h,g\in B[{{x}_{1}},\cdots,{{x}_{n}}] and statement (1) is true)

is well-defined.

Moreover, in the case where any of the above equivalent conditions is true,

σ(f(a1,,an))\sigma(f(a_{1},\cdots,a_{n}))

=σ(g(a1,,an)/h(a1,,an))=\sigma(g(a_{1},\cdots,a_{n})/h(a_{1},\cdots,a_{n}))

=σ(g(a1,,an))/σ(h(a1,,an))=\sigma(g(a_{1},\cdots,a_{n}))/\sigma(h(a_{1},\cdots,a_{n})).

Combining the above four equations, we have

σ(f(a1,,an))=f(σ(a1),,σ(an)).\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

(2) \Rightarrow (1): Considering the case where fFrac(B[x1,,xn])f\in\text{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}]) is a constant polynomial, we can tell from statement (2) that cB\forall c\in B, σ(c)=c\sigma(c)=c.

Let a,bUTa,b\in{{\left\langle U\right\rangle}_{T}}. Let g=x1+x2g={{x}_{1}}+{{x}_{2}} and h=x1x2h={{x}_{1}}{{x}_{2}}.

Then

σ(a+b)\sigma(a+b)

=σ(g(a,b))=\sigma(g(a,b))

=g(σ(a),σ(b))=g(\sigma(a),\sigma(b)) (by Equation ((9.2)))

=σ(a)+σ(b)=\sigma(a)+\sigma(b).

And

σ(ab)\sigma(ab)

=σ(h(a,b))=\sigma(h(a,b))

=h(σ(a),σ(b))=h(\sigma(a),\sigma(b)) (by Equation ((9.2)))

=σ(a)σ(b)=\sigma(a)\sigma(b). ∎

However, σ\sigma in Proposition 9.3.4 is between field extensions over the same field BB. We will deal with ring homomorphisms between field extensions over different fields in Subsection 9.5.

To characterize differential ring homomorphisms between differential fields by TT-morphisms, we need a lemma as follows.

Lemma 9.3.5.

Let TT be a par-operator gen-semigroup, let σ\sigma be a map from a TT-space SS to a TT-space, let fTf\in T have nn variables, and i=1,,n\forall i=1,\cdots,n, let giT{{g}_{i}}\in T have ni{{n}_{i}} variables. Suppose that both of the following two conditions are satisfied,

  1. (i)

    (a1,,an)Sn\forall(a_{1},\cdots,a_{n})\in{{S}^{n}}, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor f(σ(a1),,σ(an))f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) is well-defined or σ(f(a1,,an))=f(σ(a1),,σ(an))\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}));

  2. (ii)

    i=1,,n\forall i=1,\cdots,n and (a1,,ani)Sni({{a}_{1}},\cdots,{{a}_{{{n}_{i}}}})\in{{S}^{{{n}_{i}}}},
    neither gi(a1,,ani){{g}_{i}}({{a}_{1}},\cdots,{{a}_{{{n}_{i}}}}) nor gi(σ(a1),,σ(ani)){{g}_{i}}(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{{{n}_{i}}}})) is well-defined or

    σ(gi(a1,,ani))=gi(σ(a1),,σ(ani)).\sigma({{g}_{i}}({{a}_{1}},\cdots,{{a}_{{{n}_{i}}}}))={{g}_{i}}(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{{{n}_{i}}}})).

Then (a1,,am)Sm\forall({{a}_{1}},\cdots,{{a}_{m}})\in{{S}^{m}}, where m=inim=\sum\nolimits_{i}{{{n}_{i}}},
neither (f(g1,,gn))(a1,,am)(f\circ({{g}_{1}},\cdots,{{g}_{n}}))({{a}_{1}},\cdots,{{a}_{m}}) nor (f(g1,,gn))(σ(a1),,σ(am))(f\circ({{g}_{1}},\cdots,{{g}_{n}}))(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{m}})) is well-defined or

σ((f(g1,,gn))(a1,,am))=(f(g1,,gn))(σ(a1),,σ(am)).\sigma((f\circ({{g}_{1}},\cdots,{{g}_{n}}))({{a}_{1}},\cdots,{{a}_{m}}))=(f\circ({{g}_{1}},\cdots,{{g}_{n}}))(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{m}})).
Proof.

Let (a1,,am)Sm({{a}_{1}},\cdots,{{a}_{m}})\in{{S}^{m}}, where m=inim=\sum\nolimits_{i}{{{n}_{i}}}. Then

(f(g1,,gn))(a1,,am)(f\circ({{g}_{1}},\cdots,{{g}_{n}}))({{a}_{1}},\cdots,{{a}_{m}}) is well-defined;

f(g1(u1),,gn(un))\Leftrightarrow f({{g}_{1}}({{\text{u}}_{1}}),\cdots,{{g}_{n}}({{\text{u}}_{n}})) is well-defined, where ui=(ami+1,,ami+ni){{\text{u}}_{i}}=({{a}_{{{m}_{i}}+1}},\cdots,{{a}_{{{m}_{i}}+{{n}_{i}}}}), i=1,,n\forall i=1,\cdots,n, m1=0{{m}_{1}}=0, and mi=k=1i1nk,i=2,,n{{m}_{i}}=\sum\nolimits_{k=1}^{i-1}{{{n}_{k}}},\forall i=2,\cdots,n; (by Definition 9.1.2)

f(σ(g1(u1)),,σ(gn(un)))\Leftrightarrow f(\sigma({{g}_{1}}({{\text{u}}_{1}})),\cdots,\sigma({{g}_{n}}({{\text{u}}_{n}}))) is well-defined; (by condition (i) and the fact that each gi(ui)S{{g}_{i}}({{\text{u}}_{i}})\in S (cf. Proposition 10.1.1 in Section 10))

f(g1(σ(u1)),,gn(σ(un)))\Leftrightarrow f({{g}_{1}}(\sigma({{\text{u}}_{1}})),\cdots,{{g}_{n}}(\sigma({{\text{u}}_{n}}))) is well-defined, where i=1,,n\forall i=1,\cdots,n, σ(ui):=(σ(ami+1),,σ(ami+ni))\sigma\text{(}{{\text{u}}_{i}}):=(\sigma\text{(}{{a}_{{{m}_{i}}+1}}),\cdots,\sigma\text{(}{{a}_{{{m}_{i}}+{{n}_{i}}}})); (by condition (ii))

(f(g1,,gn))(σ(a1),,σ(am))\Leftrightarrow(f\circ({{g}_{1}},\cdots,{{g}_{n}}))(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{m}})) is well-defined. (by Definition 9.1.2)

And in the case where any of the above five equivalent conditions is true,

(f(g1,,gn))(a1,,am)S(f\circ({{g}_{1}},\cdots,{{g}_{n}}))({{a}_{1}},\cdots,{{a}_{m}})\in S (by Proposition 10.1.1)

and

σ((f(g1,,gn))(a1,,am))\sigma((f\circ({{g}_{1}},\cdots,{{g}_{n}}))({{a}_{1}},\cdots,{{a}_{m}}))

=σ(f(g1(u1),,gn(un)))=\sigma(f({{g}_{1}}({{\text{u}}_{1}}),\cdots,{{g}_{n}}({{\text{u}}_{n}})))(by Definition 9.1.2)

=f(σ(g1(u1)),,σ(gn(un)))=f(\sigma({{g}_{1}}({{\text{u}}_{1}})),\cdots,\sigma({{g}_{n}}({{\text{u}}_{n}}))) (by (i) and the fact that each gi(ui)S{{g}_{i}}({{\text{u}}_{i}})\in S (by Proposition 10.1.1))

=f(g1(σ(u1)),,gn(σ(un)))=f({{g}_{1}}(\sigma({{\text{u}}_{1}})),\cdots,{{g}_{n}}(\sigma({{\text{u}}_{n}}))) (by condition (ii))

=(f(g1,,gn))(σ(a1),,σ(am))=(f\circ({{g}_{1}},\cdots,{{g}_{n}}))(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{m}})) (by Definition 9.1.2) ∎

Proposition 9.3.6.

Let BB be a differential subfield of a differential field FF, let TT be the par-operator gen-semigroup on FF defined in Example 9.1.7, and let nonempty U,VFU,V\subseteq F. Then a map σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} is a differential ring homomorphism with BB fixed pointwisely if and only if it is a TT-morphism.

Proof.

Let A=UTA={{\left\langle U\right\rangle}_{T}} and R=VTR={{\left\langle V\right\rangle}_{T}}. By Example 9.2.3, AA and RR are the differential subfields of FF generated by UU and VV over BB, respectively.

Recall that a ring homomorphism σ:AR\sigma:A\to R is a differential homomorphism if and only if σ\sigma commutes with \partial. Then by Definition 9.3.1, we only need to show that the following statements are equivalent:

  1. (1)

    σ\sigma is a ring homomorphism with BB fixed pointwisely and aA\forall a\in A, σ((a))=(σ(a))\sigma(\partial(a))=\partial(\sigma(a)).

  2. (2)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)An(a_{1},\cdots,a_{n})\in{{A}^{n}} and fTf\in T of nn variables, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor f(σ(a1),,σ(an))f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) is well-defined or

    σ(f(a1,,an))=f(σ(a1),,σ(an)).\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

(1) \Rightarrow (2): Let TF{{T}_{F}} be the par-operator gen-semigroup on FF defined as in Example 9.1.6 and let T{{T}_{\partial}} be the operator semigroup defined in Example 2.1.5.

Since σ\sigma is a ring homomorphism with BB fixed pointwisely, by Proposition 9.3.4, σ\sigma is a TF{{T}_{F}}-morphism. By Definition 9.3.1, n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)An(a_{1},\cdots,a_{n})\in{{A}^{n}}, and gTFg\in{{T}_{F}} of nn variables, neither g(a1,,an)g(a_{1},\cdots,a_{n}) nor g(σ(a1),,σ(an))g(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) is well-defined or σ(g(a1,,an))=g(σ(a1),,σ(an))\sigma(g(a_{1},\cdots,a_{n}))=g(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})). Moreover, since σ((a))=(σ(a)),aA\sigma(\partial(a))=\partial(\sigma(a)),\forall a\in A, we can tell that gT={n|n0}\forall g\in{{T}_{\partial}}=\{{{\partial}^{n}}\,|\,n\in{{\mathbb{N}}_{0}}\} and aAa\in A, σ(g(a))=g(σ(a))\sigma(g(a))=g(\sigma(a)).

Therefore, n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)An(a_{1},\cdots,a_{n})\in{{A}^{n}}, and gTFTg\in{{T}_{F}}\bigcup{{T}_{\partial}} of nn variables, neither g(a1,,an)g(a_{1},\cdots,a_{n}) nor g(σ(a1),,σ(an))g(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) is well-defined or σ(g(a1,,an))=g(σ(a1),,σ(an))\sigma(g(a_{1},\cdots,a_{n}))=g(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

Let fTFTf\in{{T}_{F}}\bigcup{{T}_{\partial}} have nn variables, let giTFT{{g}_{i}}\in{{T}_{F}}\bigcup{{T}_{\partial}} have ni{{n}_{i}} variables, i=1,,n\forall i=1,\cdots,n, and let (a1,,am)Am({{a}_{1}},\cdots,{{a}_{m}})\in{{A}^{m}}, where m=inim=\sum\nolimits_{i}{{{n}_{i}}}. By the statement in the preceding paragraph, Lemma 9.3.5 applies (with T:=TFTT:=\left\langle{{T}_{F}}\bigcup{{T}_{\partial}}\right\rangle). It follows that neither (f(g1,,gn))(a1,,am)(f\circ({{g}_{1}},\cdots,{{g}_{n}}))({{a}_{1}},\cdots,{{a}_{m}}) nor (f(g1,,gn))(σ(a1),,σ(am))(f\circ({{g}_{1}},\cdots,{{g}_{n}}))(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{m}})) is well-defined or

σ((f(g1,,gn))(a1,,am))=(f(g1,,gn))(σ(a1),,σ(am)).\sigma((f\circ({{g}_{1}},\cdots,{{g}_{n}}))({{a}_{1}},\cdots,{{a}_{m}}))=(f\circ({{g}_{1}},\cdots,{{g}_{n}}))(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{m}})).

Moreover, since T=TFTT=\left\langle{{T}_{F}}\bigcup{{T}_{\partial}}\right\rangle, from Definitions 9.1.4 and 9.1.5, we can tell that the above f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) is a restriction of some element of TT, and hT\forall h\in T, hh can be obtained by a finite number of compositions (defined by Definition 9.1.2) of elements of TFT{{T}_{F}}\bigcup{{T}_{\partial}} (or elements which “originally come from” TFT{{T}_{F}}\bigcup{{T}_{\partial}}). Thus, by induction and application of Lemma 9.3.5 as in the preceding paragraph, we can show that n+\forall n\in{{\mathbb{Z}}^{+}}, hTh\in T of nn variables and (a1,,an)An(a_{1},\cdots,a_{n})\in{{A}^{n}}, neither h(a1,,an)h(a_{1},\cdots,a_{n}) nor h(σ(a1),,σ(an))h(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) is well-defined or

σ(h(a1,,an))=h(σ(a1),,σ(an)).\sigma(h(a_{1},\cdots,a_{n}))=h(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).

(2) \Rightarrow (1): As shown in Example 9.2.3, ATF=A{{\left\langle A\right\rangle}_{{{T}_{F}}}}=A. Thus AA is also a TF{{T}_{F}}-space. Then because TFT{{T}_{F}}\subseteq T, by Definition 9.3.1, σ\sigma is a TF{{T}_{F}}-morphism. Hence by Proposition 9.3.4, σ\sigma is a ring homomorphism with BB fixed pointwisely. Moreover, if let f=(T)f=\partial(\in T), then by statement (2), σ((a))=(σ(a)),aA\sigma(\partial(a))=\partial(\sigma(a)),\forall a\in A. ∎

9.4. TT-isomorphisms

To par-operator gen-semigroups, Definition 2.4.1 still applies: if a TT-morphism is bijective, then we call it a TT-isomorphism. To justify the definition, we need to show Proposition 2.4.2 with a different proof as follows.

Proposition 9.4.1.

(Proposition 2.4.2) Let σ\sigma be a TT-isomorphism from a TT-space S1{{S}_{1}} to a TT-space S2{{S}_{2}} and let σ1{{\sigma}^{-1}} be the inverse map of σ\sigma. Then σ1IsoT(S2,S1){{\sigma}^{-1}}\in\operatorname{Iso}_{T}({{S}_{2}},{{S}_{1}}).

Proof.

By Definition 9.3.1, n+\forall n\in{{\mathbb{Z}}^{+}}, fTf\in T of nn variables and (a1,,an)S1n(a_{1},\cdots,a_{n})\in{S_{1}^{n}}, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor f(σ(a1),,f(\sigma({{a}_{1}}),\cdots, σ(an))\sigma({{a}_{n}})) is well-defined or

f(a1,,an)=σ1(f(σ(a1),,σ(an))).f(a_{1},\cdots,a_{n})={\sigma}^{-1}(f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}))).

Hence n+\forall n\in{{\mathbb{Z}}^{+}}, fTf\in T of nn variables and (b1,,bn)S2n(b_{1},\cdots,b_{n})\in{S_{2}^{n}}, neither f(b1,,bn)f({b}_{1},\cdots,{b}_{n}) nor f(σ1(b1),,σ1(bn))f({\sigma}^{-1}({b}_{1}),\cdots,{\sigma}^{-1}({b}_{n})) is well-defined or

f(σ1(b1),,σ1(bn))=σ1(f(b1,,bn)).f({\sigma}^{-1}({b}_{1}),\cdots,{\sigma}^{-1}({b}_{n}))={\sigma}^{-1}(f(b_{1},\cdots,b_{n})).

Then by Definition 9.3.1, σ1{\sigma}^{-1} is a TT-morphism from S2S_{2} to S1S_{1}, and hence σ1IsoT(S2,S1){{\sigma}^{-1}}\in\operatorname{Iso}_{T}({{S}_{2}},{{S}_{1}}) by Definition 2.4.1. ∎

9.5. θ\theta-morphisms

We first generalize Notation 8.5.1 to

Notation 9.5.1.

Let D1D_{1} (resp. D2D_{2}) be a set, let M1M_{1} (resp. M2M_{2}) be a subset of {all partial functions from the cartesian product D1nD_{1}^{n} to D1|n+{D_{1}}\,|\,n\in{{\mathbb{Z}}^{+}}} (resp. a subset of {all partial functions from D2nD_{2}^{n} to D2|n+{{D}_{2}}\,|\,n\in{{\mathbb{Z}}^{+}}}), let θM1×M2\theta\subseteq{{M}_{1}}\times{{M}_{2}}, and let AD2A\subseteq{{D}_{2}}. We denote by θ|A\theta{{|}_{A}} the set {(f,g|A)|(f,g)θ}\{(f,g{{|}_{A}})\,|\,(f,g)\in\theta\}, where g|Ag{{|}_{A}} denotes the partial function obtained by restricting every variable of gg to AA.

Then for the case of partial functions, Proposition 8.5.3 needs to be modified as follows.

Proposition 9.5.2.

Let θ\theta and AA be defined as in Notation 9.5.1. Then the following statements are equivalent:

  1. (i)

    θ|A\theta{{|}_{A}} is a map.

  2. (ii)

    (f,g1),(f,g2)θ\forall(f,g_{1}),(f,g_{2})\in\theta, g1|A=g2|Ag_{1}{{|}_{A}}=g_{2}{{|}_{A}}.

  3. (iii)

    (f,g1),(f,g2)θ\forall(f,g_{1}),(f,g_{2})\in\theta, n+n\in{{\mathbb{Z}}^{+}} and (a1,,an)An(a_{1},\cdots,a_{n})\in A^{n}, g1(a1,,an)=g2(a1,,an)g_{1}(a_{1},\cdots,a_{n})=g_{2}(a_{1},\cdots,a_{n}) or neither g1(a1,,an)g_{1}(a_{1},\cdots,a_{n}) nor g2(a1,,an)g_{2}(a_{1},\cdots,a_{n}) is well-defined.

Proof.

Obvious. ∎

Definitions 6.1.5 and 8.5.2 are generalized to

Definition 9.5.3.

Let θ\theta and AA be defined as in Notation 9.5.1. Let fDomθf\in\operatorname{Dom}\theta, n+n\in{{\mathbb{Z}}^{+}} and (a1,,an)An(a_{1},\cdots,a_{n})\in A^{n}. If g1(a1,,an)=g2(a1,,an)g_{1}(a_{1},\cdots,a_{n})=g_{2}(a_{1},\cdots,a_{n}), (f,g1),(f,g2)θ\forall(f,g_{1}),(f,g_{2})\in\theta, then we say that θ(f)(a1,,an)\theta(f)(a_{1},\cdots,a_{n}) is well-defined and let θ(f)(a1,,an)=g1(a1,,an)\theta(f)(a_{1},\cdots,a_{n})=g_{1}(a_{1},\cdots,a_{n}).

Remark.

By Convention 6.1.7,

g1(a1,,an)=g2(a1,,an),(f,g1),(f,g2)θg_{1}(a_{1},\cdots,a_{n})=g_{2}(a_{1},\cdots,a_{n}),\forall(f,g_{1}),(f,g_{2})\in\theta

implies that (f,g)θ\forall(f,g)\in\theta, gg has nn variables and g(a1,,an)g(a_{1},\cdots,a_{n}) is well-defined.

Then by Proposition 9.5.2, we immediately get

Corollary 9.5.4.

Let θ\theta and AA be defined as in Notation 9.5.1. Suppose that θ|A\theta{{|}_{A}} is a map. Let (f,g)θ(f,g)\in\theta. If n+\exists n\in{{\mathbb{Z}}^{+}} and (a1,,an)An(a_{1},\cdots,a_{n})\in A^{n} such that g(a1,,an)g(a_{1},\cdots,a_{n}) is well-defined, then θ(f)(a1,,an)\theta(f)(a_{1},\cdots,a_{n}) is well-defined and

θ(f)(a1,,an)=g(a1,,an).\theta(f)(a_{1},\cdots,a_{n})=g(a_{1},\cdots,a_{n}).

Definition 8.5.4 is generalized to

Definition 9.5.5.

Let T1T_{1} and T2T_{2} be par-operator gen-semigroups and let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}}. Let ϕ\phi be a map from a T1T_{1}-space SS to a T2T_{2}-space. If

  1. (i)

    θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map and

  2. (ii)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and fDomθf\in\operatorname{Dom}\theta, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor θ(f)(ϕ(a1),,ϕ(an))\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined or

    ϕ(f(a1,,an))=θ(f)(ϕ(a1),,ϕ(an)),\phi(f(a_{1},\cdots,a_{n}))=\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})),

then ϕ\phi is called a θ\theta-morphism.

Remark.

Suppose that f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined, where (a1,,an)Sn(a_{1},\cdots,a_{n})\in S^{n} and fDomθf\in\operatorname{Dom}\theta. Then by condition (ii) in the definition and Convention 6.1.7, θ(f)(ϕ(a1),,ϕ(an))\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is also well-defined. Hence (f,g)θ\forall(f,g)\in\theta, by Definition 9.5.3, gg has nn variables.

The following explains why, unlike Definition 8.5.4, we put condition (i) in Definition 9.5.5 explicitly.

Proposition 9.5.6.

Let θ\theta, SS and ϕ\phi be given as in Definition 9.5.5. Suppose that condition (ii) in Definition 9.5.5 is satisfied. Then θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is possibly not a map.

Proof.

By Proposition 9.5.2, θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map if and only if (f,g1),(f,g2)θ\forall(f,g_{1}),(f,g_{2})\in\theta, n+n\in{{\mathbb{Z}}^{+}} and (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}}, g1(ϕ(a1),,ϕ(an))=g2(ϕ(a1),,ϕ(an))g_{1}(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))=g_{2}(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) or neither g1(ϕ(a1),,ϕ(an))g_{1}(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) nor g2(ϕ(a1),,ϕ(an))g_{2}(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined. We can tell that this condition for θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} being a map is not guaranteed by condition (ii) in Definition 9.5.5. For example, condition (ii) in Definition 9.5.5 cannot rule out the possibility where there exist (f,g1),(f,g2)θ(f,g_{1}),(f,g_{2})\in\theta, n+n\in{{\mathbb{Z}}^{+}} and (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} such that only one of g1(ϕ(a1),,ϕ(an))g_{1}(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) and g2(ϕ(a1),,ϕ(an))g_{2}(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined (because in this case, it is possible that neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor θ(f)(ϕ(a1),,ϕ(an))\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined). ∎

When par-operator gen-semigroups are induced from formal partial functions, the following lemma, which generalizes both Lemmas 8.5.6 and 9.3.3, will be useful.

Lemma 9.5.7.

Let \mathcal{F} ((resp. {\mathcal{F}}^{\prime})) be a set of D-valued ((resp. D{D}^{\prime}-valued)) formal partial functions on DD ((resp. D{D}^{\prime})). Let TT ((resp. T{T}^{\prime})) be a par-operator gen-semigroup on DD ((resp. D{D}^{\prime})) such that there is a map τ:T\tau:\mathcal{F}\to T being given by fff\mapsto f^{*}, where f:DnDf^{*}:{{D}^{n}}\to D is given by (a1,,an)f(a1,,an)(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n}) for ff of nn variables, n+\forall n\in{{\mathbb{Z}}^{+}} ((resp. such that there is a map τ:T{\tau}^{\prime}:{\mathcal{F}}^{\prime}\to{T}^{\prime} being given analogously)). Let ϑ:\vartheta:\mathcal{F}\to{\mathcal{F}}^{\prime} be a map. Let θ={(τ(f),τ(ϑ(f)))|f}\theta=\{(\tau(f),{\tau}^{\prime}(\vartheta(f)))\,|\,f\in\mathcal{F}\} and let ϕ\phi be a map from a TT-space SS to a T{T}^{\prime}-space. Then the following statements are equivalent:

  1. (i)

    ϕ\phi is a θ\theta-morphism.

  2. (ii)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and ff\in\mathcal{F} of nn variables, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor ϑ(f)(ϕ(a1),,ϕ(an))\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined or

    (9.3) ϕ(f(a1,,an))=ϑ(f)(ϕ(a1),,ϕ(an)).\phi(f(a_{1},\cdots,a_{n}))=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).
Proof.

Clearly, θT×T\theta\subseteq T\times T^{\prime}.

(i) \Rightarrow (ii): By Definition 9.5.5, θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map and n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and ff\in\mathcal{F} of nn variables, neither τ(f)(a1,,an)\tau(f)(a_{1},\cdots,a_{n}) nor θ(τ(f))(ϕ(a1),,ϕ(an))\theta(\tau(f))(\phi({{a}_{\text{1}}}),\cdots,\phi({{a}_{n}})) is well-defined or

ϕ(τ(f)(a1,,an))=θ(τ(f))(ϕ(a1),,ϕ(an)).\phi(\tau(f)(a_{1},\cdots,a_{n}))=\theta(\tau(f))(\phi({{a}_{\text{1}}}),\cdots,\phi({{a}_{n}})).

Let n+n\in{{\mathbb{Z}}^{+}}, let (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and let ff\in\mathcal{F} have nn variables. Then

f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined;

τ(f)(a1,,an)\Leftrightarrow\tau(f)(a_{1},\cdots,a_{n}) is well-defined;

θ(τ(f))(ϕ(a1),,ϕ(an))\Leftrightarrow\theta(\tau(f))(\phi({{a}_{\text{1}}}),\cdots,\phi({{a}_{n}})) is well-defined;

τ(ϑ(f))(ϕ(a1),,ϕ(an))\Leftrightarrow{\tau}^{\prime}(\vartheta(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined (because (τ(f),τ(ϑ(f)))θ(\tau(f),{\tau}^{\prime}(\vartheta(f)))\in\theta and θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map);

ϑ(f)(ϕ(a1),,ϕ(an))\Leftrightarrow\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined.

And in the case where the above equivalent conditions are true,

ϕ(f(a1,,an))\phi(f(a_{1},\cdots,a_{n}))

=ϕ(τ(f)(a1,,an))=\phi(\tau(f)(a_{1},\cdots,a_{n}))

=θ(τ(f))(ϕ(a1),,ϕ(an))=\theta(\tau(f))(\phi({{a}_{\text{1}}}),\cdots,\phi({{a}_{n}}))

=τ(ϑ(f))(ϕ(a1),,ϕ(an))={\tau}^{\prime}(\vartheta(f))(\phi({{a}_{\text{1}}}),\cdots,\phi({{a}_{n}})) (because (τ(f),τ(ϑ(f)))θ(\tau(f),{\tau}^{\prime}(\vartheta(f)))\in\theta and θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map)

=ϑ(f)(ϕ(a1),,ϕ(an))=\vartheta(f)(\phi({{a}_{\text{1}}}),\cdots,\phi({{a}_{n}})), as desired.

(ii) \Rightarrow (i): By Definition 9.5.5, to show that ϕ\phi is a θ\theta-morphism, we first need to show that θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map.

Let (h,g|Imϕ),(h,g|Imϕ)θ|Imϕ(h,g{{|}_{\operatorname{Im}\phi}}),(h,{g}^{\prime}{{|}_{\operatorname{Im}\phi}})\in\theta{{|}_{\operatorname{Im}\phi}} with (h,g),(h,g)θ(h,g),(h,{g}^{\prime})\in\theta. By the definition of θ\theta, l,m\exists l,m\in\mathcal{F} such that (h,g)=(τ(l),τ(ϑ(l)))(h,g)=(\tau(l),{\tau}^{\prime}(\vartheta(l))) and (h,g)=(τ(m),τ(ϑ(m)))(h,{g}^{\prime})=(\tau(m),{\tau}^{\prime}(\vartheta(m))), and thus τ(l)=τ(m)\tau(l)=\tau(m). Hence ll and mm have the same number of variables, assumed nn. Then by Equation (9.3) and Convention 6.1.7, ϑ(l)\vartheta(l) also has nn variables if ll is well-defined somewhere in SnS^{n}. Analogously, ϑ(m)\vartheta(m) also has nn variables if mm is well-defined somewhere in SnS^{n}.

Assume that g|Imϕg|Imϕg{{|}_{\operatorname{Im}\phi}}\neq{g}^{\prime}{{|}_{\operatorname{Im}\phi}}; that is, τ(ϑ(l))|Imϕτ(ϑ(m))|Imϕ{\tau}^{\prime}(\vartheta(l)){{|}_{\operatorname{Im}\phi}}\neq{\tau}^{\prime}(\vartheta(m)){{|}_{\operatorname{Im}\phi}}. Then k+\exists k\in{{\mathbb{Z}}^{+}} and (z1,,zk)Sk({{z}_{1}},\cdots,{{z}_{k}})\in{{S}^{k}} such that only one of ϑ(l)(ϕ(z1),,ϕ(zk))\vartheta(l)(\phi({{z}_{1}}),\cdots,\phi({{z}_{k}})) and ϑ(m)(ϕ(z1),,ϕ(zk))\vartheta(m)(\phi({{z}_{1}}),\cdots,\phi({{z}_{k}})) is well-defined or both of them are well-defined but

ϑ(l)(ϕ(z1),,ϕ(zk))ϑ(m)(ϕ(z1),,ϕ(zk)).\vartheta(l)(\phi({{z}_{1}}),\cdots,\phi({{z}_{k}}))\neq\vartheta(m)(\phi({{z}_{1}}),\cdots,\phi({{z}_{k}})).

By statement (ii), if ϑ(l)(ϕ(z1),,ϕ(zk))\vartheta(l)(\phi({{z}_{1}}),\cdots,\phi({{z}_{k}})) (resp. ϑ(m)(ϕ(z1),,ϕ(zk))\vartheta(m)(\phi({{z}_{1}}),\cdots,\phi({{z}_{k}}))) is well-defined, then l(z1,,zk)l(z_{1},\cdots,z_{k}) (resp. m(z1,,zk))m(z_{1},\cdots,z_{k}))) is also well-defined. Hence k=nk=n.

However, τ(l)=τ(m)\tau(l)=\tau(m), and hence neither l(z1,,zn)l({{z}_{1}},\cdots,{{z}_{n}}) nor m(z1,,zn)m({{z}_{1}},\cdots,{{z}_{n}}) is well-defined or both of them are well-defined and l(z1,,zn)=m(z1,,zn)l({{z}_{1}},\cdots,{{z}_{n}})=m({{z}_{1}},\cdots,{{z}_{n}}). It follows from (ii) that neither ϑ(l)(ϕ(z1),,ϕ(zn))\vartheta(l)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}})) nor ϑ(m)(ϕ(z1),,ϕ(zn))\vartheta(m)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}})) is well-defined or both of them are well-defined and

ϑ(l)(ϕ(z1),,ϕ(zn))=ϑ(m)(ϕ(z1),,ϕ(zn)),\vartheta(l)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}}))=\vartheta(m)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}})),

a contradiction.

Hence θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} must be a map.

Moreover, by statement (ii), n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and ff\in\mathcal{F} of nn variables, there are only two cases as follows.

Case (1): Neither f(a1,,an))f(a_{1},\cdots,a_{n})) nor ϑ(f)(ϕ(a1),,ϕ(an))\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined, and hence neither τ(f)(a1,,an)\tau(f)(a_{1},\cdots,a_{n}) nor τ(ϑ(f))(ϕ(a1),,ϕ(an))\tau^{\prime}(\vartheta(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined. Because (τ(f),τ(ϑ(f)))θ(\tau(f),{\tau}^{\prime}(\vartheta(f)))\in\theta, by Definition 9.5.3, neither τ(f)(a1,,an)\tau(f)(a_{1},\cdots,a_{n}) nor θ(τ(f))(ϕ(a1),,ϕ(an))\theta(\tau(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined.

Case (2):

ϕ(τ(f)(a1,,an))\phi(\tau(f)(a_{1},\cdots,a_{n}))

=ϕ(f(a1,,an))=\phi(f(a_{1},\cdots,a_{n}))

=ϑ(f)(ϕ(a1),,ϕ(an))=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) (by Equation (9.3))

=τ(ϑ(f))(ϕ(a1),,ϕ(an))={\tau}^{\prime}(\vartheta(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))

=θ(τ(f))(ϕ(a1),,ϕ(an))=\theta(\tau(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) (because (τ(f),τ(ϑ(f)))θ(\tau(f),{\tau}^{\prime}(\vartheta(f)))\in\theta and θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map).

Therefore, n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and ff\in\mathcal{F} of nn variables, neither τ(f)(a1,,an)\tau(f)(a_{1},\cdots,a_{n}) nor θ(τ(f))(ϕ(a1),,ϕ(an))\theta(\tau(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined or

ϕ(τ(f)(a1,,an))=θ(τ(f))(ϕ(a1),,ϕ(an)).\phi(\tau(f)(a_{1},\cdots,a_{n}))=\theta(\tau(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).

Since Domθ={τ(f)|f}\operatorname{Dom}\theta=\{\tau(f)\,|\,f\in\mathcal{F}\}, by Definition 9.5.5, ϕ\phi is a θ\theta-morphism. ∎

Lemma 9.5.7 characterizes a θ\theta-morphism by formal partial functions. When par-operator gen-semigroups are induced from formal partial functions, the criterion for a θ\theta-morphism by Lemma 9.5.7 is more convenient than that by Definition 9.5.5.

The following, which generalizes Proposition 9.3.4, is analogous to Proposition 8.5.7, except that now TT is defined by Example 9.1.6, and hence UT{{\left\langle U\right\rangle}_{T}} and VT{{\left\langle V\right\rangle}_{T}} must be fields.

Proposition 9.5.8.

Let F/BF/B (resp. F/B{F}^{\prime}/{B}^{\prime}) be a field extension, let a par-operator gen-semigroup TT ((resp. T{T}^{\prime})) be defined as in Example 9.1.6 on FF ((resp. F{F}^{\prime})) over BB ((resp. B{B}^{\prime})), let φ:BB\varphi:B\to{B}^{\prime} be a field isomorphism, and let

ϑ:n+Frac(B[x1,,xn])n+Frac(B[x1,,xn])\vartheta:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\emph{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}])}\to\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\emph{Frac}({B}^{\prime}[{{x}_{1}},\cdots,{{x}_{n}}])}

be the map given by fff\mapsto{f}^{\prime} where

f(x1,,xn)=(cp1,,pnx1p1xnpn)/(cp1,,pnx1p1xnpn)f({{x}_{1}},\cdots,{{x}_{n}})=(\sum{{c_{p_{1},\cdots,p_{n}}}x_{1}^{{{p}_{1}}}\cdots x_{n}^{{{p}_{n}}}})/(\sum{{c^{\prime}_{p_{1},\cdots,p_{n}}}x_{1}^{{{p}_{1}}}\cdots x_{n}^{{{p}_{n}}}})

and

f(x1,,xn)=(φ(cp1,,pn)x1p1xnpn)/(φ(cp1,,pn)x1p1xnpn){f}^{\prime}({{x}_{1}},\cdots,{{x}_{n}})=(\sum{\varphi({c_{p_{1},\cdots,p_{n}}})x_{1}^{{{p}_{1}}}\cdots x_{n}^{{{p}_{n}}}})/(\sum{\varphi({c^{\prime}_{p_{1},\cdots,p_{n}}})x_{1}^{{{p}_{1}}}\cdots x_{n}^{{{p}_{n}}}}).

Let θ={(τ(f),τ(ϑ(f))|fFrac(B[x1,,xn]),n+}\theta=\{(\tau(f),{\tau}^{\prime}(\vartheta(f))\,|\,f\in\emph{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}]),n\in{{\mathbb{Z}}^{+}}\}, where τ\tau and τ{\tau}^{\prime} are the maps inducing TT and T{T}^{\prime}, respectively, defined as in Example 9.1.6.

Then ()UF\forall(\emptyset\neq)U\subseteq F and ()VF(\emptyset\neq)V\subseteq{F}^{\prime}, a map ϕ:UTVT\phi:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{{{T}^{\prime}}}} is a ring homomorphism extending φ\varphi if and only if ϕ\phi is a θ\theta-morphism.

Proof.

By the definition of TT in Example 9.1.6,

T={T=\{the partial function f:FnFf^{*}:{{F}^{n}}\to F given by (a1,,an)f(a1,,an)|fFrac(B[x1,,xn]),n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\text{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}]),n\in{{\mathbb{Z}}^{+}}\}.

and

T={{T}^{\prime}=\{the partial function f:FnFf^{*}:{{{F}^{\prime}}^{n}}\to{F}^{\prime} given by (a1,,an)f(a1,,an)|fFrac(B[x1,,xn]),n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in\text{Frac}({B}^{\prime}[{{x}_{1}},\cdots,{{x}_{n}}]),n\in{{\mathbb{Z}}^{+}}\}.

Thus UT{{\left\langle U\right\rangle}_{T}} and VT{{\left\langle V\right\rangle}_{{{T}^{\prime}}}} are the fields B(U)B(U) and B(V){B}^{\prime}(V), respectively.

Let =n+Frac(B[x1,,xn])\mathcal{F}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\text{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}])} and =n+Frac(B[x1,,xn]){\mathcal{F}}^{\prime}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\text{Frac}({B}^{\prime}[{{x}_{1}},\cdots,{{x}_{n}}])}. Then Lemma 9.5.7 applies. Hence it suffices to show that ϕ:UTVT\phi:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{{{T}^{\prime}}}} is a ring homomorphism extending φ\varphi if and only if statement (ii) in Lemma 9.5.7 is true in this case. Specifically, it is sufficient to show that the following statements are equivalent:

  1. (1)

    cB\forall c\in B, ϕ(c)=φ(c)\phi(c)=\varphi(c), and a,bUT\forall a,b\in{{\left\langle U\right\rangle}_{T}}, ϕ(a+b)=ϕ(a)+ϕ(b)\phi(a+b)=\phi(a)+\phi(b) and ϕ(ab)=ϕ(a)ϕ(b)\phi(ab)=\phi(a)\phi(b).

  2. (2)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)UTn(a_{1},\cdots,a_{n})\in\left\langle U\right\rangle_{T}^{n} and fFrac(B[x1,,xn])f\in\text{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}]), neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor ϑ(f)(ϕ(a1),,ϕ(an))\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined or

    (9.4) ϕ(f(a1,,an))=ϑ(f)(ϕ(a1),,ϕ(an)).\phi(f(a_{1},\cdots,a_{n}))=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).

(1) \Rightarrow (2): Let f(x1,,xn)=g(x1,,xn)/h(x1,,xn)f({{x}_{1}},\cdots,{{x}_{n}})=g({{x}_{1}},\cdots,{{x}_{n}})/h({{x}_{1}},\cdots,{{x}_{n}}), where n+n\in{\mathbb{Z}}^{+} and h,gB[x1,,xn]h,g\in B[{{x}_{1}},\cdots,{{x}_{n}}]. Let (a1,,an)UTn(a_{1},\cdots,a_{n})\in\left\langle U\right\rangle_{T}^{n}. Then

f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined;

h(a1,,an)0\Leftrightarrow h(a_{1},\cdots,a_{n})\neq 0;

ϑ(h)(a1,,an)0\Leftrightarrow\vartheta(h)(a_{1},\cdots,a_{n})\neq 0 (by the definition of ϑ\vartheta);

ϑ(f)(ϕ(a1),,ϕ(an))\Leftrightarrow\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))

=ϑ(g)(ϕ(a1),,ϕ(an))/ϑ(h)(ϕ(a1),,ϕ(an))=\vartheta(g)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))/\vartheta(h)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) (by the definition of ϑ\vartheta)

is well-defined.

Moreover, in the case where the above equivalent conditions are true,

ϕ(f(a1,,an))\phi(f(a_{1},\cdots,a_{n}))

=ϕ(cp1,,pna1p1anpn/cp1,,pna1p1anpn)=\phi(\sum{{c_{p_{1},\cdots,p_{n}}}a_{1}^{{{p}_{1}}}\cdots a_{n}^{{{p}_{n}}}}/\sum{{c^{\prime}_{p_{1},\cdots,p_{n}}}a_{1}^{{{p}_{1}}}\cdots a_{n}^{{{p}_{n}}}})

=ϕ(cp1,,pn)(ϕ(a1))p1(ϕ(an))pn/ϕ(cp1,,pn)(ϕ(a1))p1(ϕ(an))pn=\sum{\phi({c_{p_{1},\cdots,p_{n}}})(\phi(a_{1}))^{p_{1}}\cdots(\phi(a_{n}))^{p_{n}}}/\sum{\phi({c^{\prime}_{p_{1},\cdots,p_{n}}})(\phi(a_{1}))^{p_{1}}\cdots(\phi(a_{n}))^{p_{n}}} (because statement (1) is true)

=φ(cp1,,pn)(ϕ(a1))p1(ϕ(an))pn/φ(cp1,,pn)(ϕ(a1))p1(ϕ(an))pn=\sum{\varphi({c_{p_{1},\cdots,p_{n}}})(\phi(a_{1}))^{p_{1}}\cdots(\phi(a_{n}))^{p_{n}}}/\sum{\varphi({c^{\prime}_{p_{1},\cdots,p_{n}}})(\phi(a_{1}))^{p_{1}}\cdots(\phi(a_{n}))^{p_{n}}} (because cB\forall c\in B, ϕ(c)=φ(c)\phi(c)=\varphi(c))

=ϑ(f)(ϕ(a1),,ϕ(an))=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) (by the definition of ϑ\vartheta),

as desired.

(2) \Rightarrow (1): Considering the case where fFrac(B[x1,,xn])f\in\text{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}]) is a constant polynomial, we can tell from Equation ((9.4)) and the definition of ϑ\vartheta that cB\forall c\in B, ϕ(c)=φ(c)\phi(c)=\varphi(c).

Let a,bUTa,b\in{{\left\langle U\right\rangle}_{T}}. Let g=x1+x2g={{x}_{1}}+{{x}_{2}} and h=x1x2h={{x}_{1}}{{x}_{2}}.

Then

ϕ(a+b)\phi(a+b)

=ϕ(g(a,b))=\phi(g(a,b))

=ϑ(g)(ϕ(a),ϕ(b))=\vartheta(g)(\phi(a),\phi(b)) (by Equation ((9.4)))

=g(ϕ(a),ϕ(b))=g(\phi(a),\phi(b)) (ϑ(g)=g\vartheta(g)=g by the definition of ϑ\vartheta)

=ϕ(a)+ϕ(b)=\phi(a)+\phi(b).

And

ϕ(ab)\phi(ab)

=ϕ(h(a,b))=\phi(h(a,b))

=ϑ(h)(ϕ(a),ϕ(b))=\vartheta(h)(\phi(a),\phi(b)) (by Equation ((9.4)))

=h(ϕ(a),ϕ(b))=h(\phi(a),\phi(b)) (ϑ(h)=h\vartheta(h)=h by the definition of ϑ\vartheta)

=ϕ(a)ϕ(b)=\phi(a)\phi(b). ∎

We are going to show that a covariant functor can be characterized by a θ\theta-morphism. Because in every category, there is a bijection between objects AA and their identity morphisms 1A1_{A}, we view DD as 𝒞\mathcal{C} and D{D}^{\prime} as 𝒞{\mathcal{C}}^{\prime} in the following.

Proposition 9.5.9.

Let 𝒞\mathcal{C} (resp. 𝒞{\mathcal{C}}^{\prime}) be a category, let DD ((resp. D{D}^{\prime})) be the collection of all morphisms in 𝒞\mathcal{C} ((resp. 𝒞{\mathcal{C}}^{\prime})), and let n{{\mathcal{F}}_{n}}, n+\forall n\in{{\mathbb{Z}}^{+}}, and the par-operator gen-semigroup TT ((resp. T{T}^{\prime})) on DD ((resp. D{D}^{\prime})) be defined as in Example 9.1.8. Let θ={(τ(f),τ(f)|fn+n}\theta=\{(\tau(f),{\tau}^{\prime}(f)\,|\,f\in\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{{{\mathcal{F}}_{n}}}\}, where τ\tau and τ{\tau}^{\prime} are the maps inducing TT and T{T}^{\prime}, respectively, defined as in Example 9.1.8.

Suppose that a function ϕ:DD\phi:D\to{D}^{\prime} maps each identity morphism in DD to an identity morphism in D{D}^{\prime}. Then ϕ\phi is a covariant functor if and only if it is a θ\theta-morphism.

Proof.

By Example 9.1.8,

T={f:DnDT=\{f^{*}:{{D}^{n}}\to D given by (a1,,an)f(a1,,an)|fn,n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in{{\mathcal{F}}_{n}},n\in{{\mathbb{Z}}^{+}}\},

and

T={f:DnD{T}^{\prime}=\{f^{*}:{{{D}^{\prime}}^{n}}\to{D}^{\prime} given by (a1,,an)f(a1,,an)|fn,n+}(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})\,|\,f\in{{\mathcal{F}}_{n}},n\in{{\mathbb{Z}}^{+}}\}.

By Example 9.2.4, DD and DD^{\prime} are TT-space and TT^{\prime}-space, respectively. Let ==n+n\mathcal{F}={\mathcal{F}}^{\prime}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{{{\mathcal{F}}_{n}}} and let ϑ:\vartheta:\mathcal{F}\to{\mathcal{F}}^{\prime} be the identity function. Then Lemma 9.5.7 applies. Hence it suffices to show that ϕ\phi is a covariant functor if and only if statement (ii) in Lemma 9.5.7 is true in this case. Specifically, we only need to show that the following statements are equivalent:

  1. (1)

    Xobj(𝒞)\forall X\in\text{obj}(\mathcal{C}), ϕ(1X)=1ϕ(X)\phi({{1}_{X}})={{1}_{\phi(X)}}; if m:ABm:A\to B in 𝒞\mathcal{C}, then ϕ(m):ϕ(A)ϕ(B)\phi(m):\phi(A)\to\phi(B) in 𝒞{\mathcal{C}}^{\prime}; and if AmBhCA\xrightarrow{m}B\xrightarrow{h}C in 𝒞\mathcal{C}, then ϕ(A)ϕ(m)ϕ(B)ϕ(h)ϕ(C)\phi(A)\xrightarrow{\phi(m)}\phi(B)\xrightarrow{\phi(h)}\phi(C) in 𝒞{\mathcal{C}}^{\prime} and ϕ(hm)=ϕ(h)ϕ(m)\phi(h\circ m)=\phi(h)\circ\phi(m).

  2. (2)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Dn(a_{1},\cdots,a_{n})\in{{D}^{n}} and ff\in\mathcal{F} of nn variables, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor f(ϕ(a1),,ϕ(an))f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined or

    (9.5) ϕ(f(a1,,an))=f(ϕ(a1),,ϕ(an)).\phi(f(a_{1},\cdots,a_{n}))=f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).

(1) \Rightarrow (2): Let (a1,,an)Dn(a_{1},\cdots,a_{n})\in{{D}^{n}} and let f=xi1xiknf={{x}_{{{i}_{1}}}}\cdots{{x}_{{{i}_{k}}}}\in{{\mathcal{F}}_{n}} (cf. Example 9.1.8). Then

f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined;

ai1aik\Leftrightarrow{{a}_{{{i}_{1}}}}\circ\cdots\circ{{a}_{{{i}_{k}}}} is well-defined (by the definition of f(a1,,an)f(a_{1},\cdots,a_{n}) in Example 9.1.8);

ϕ(ai1)ϕ(aik)\Leftrightarrow\phi({{a}_{{{i}_{1}}}})\circ\cdots\circ\phi({{a}_{{{i}_{k}}}}) is well-defined (explained below);

f(ϕ(a1),,ϕ(an))\Leftrightarrow f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined (by the definition of f(a1,,an)f(a_{1},\cdots,a_{n}) in Example 9.1.8).

The second equivalence is explained as follows. DD (resp. D{D}^{\prime}) is the collection of all morphisms in category 𝒞\mathcal{C} (resp. 𝒞{\mathcal{C}}^{\prime}), and hence i=1,,n\forall i=1,\cdots,n, ai{{a}_{i}} (resp. ϕ(ai)\phi({{a}_{i}})) is a morphism in 𝒞\mathcal{C} (resp. 𝒞{\mathcal{C}}^{\prime}). Then ai1aik{{a}_{{{i}_{1}}}}\circ\cdots\circ{{a}_{{{i}_{k}}}} is well-defined if and only if j=1,,(k1)\forall j=1,\cdots,(k-1), the domain of aij{{a}_{{{i}_{j}}}} is the target of aij+1{{a}_{{{i}_{j+1}}}}. And this occurs if and only if ϕ(ai1)ϕ(aik)\phi({{a}_{{{i}_{1}}}})\circ\cdots\circ\phi({{a}_{{{i}_{k}}}}) is well-defined because ϕ\phi is a covariant functor.

Therefore, f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined if and only if f(ϕ(a1),,ϕ(an))f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined.

Moreover, in the case where the above equivalent conditions are satisfied, since ϕ\phi is a covariant functor,

ϕ(ai1aik)=ϕ(ai1)ϕ(aik);\phi({{a}_{{{i}_{1}}}}\circ\cdots\circ{{a}_{{{i}_{k}}}})=\phi({{a}_{{{i}_{1}}}})\circ\cdots\circ\phi({{a}_{{{i}_{k}}}});

that is (by the definition of f(a1,,an)f(a_{1},\cdots,a_{n}) in Example 9.1.8),

ϕ(f(a1,,an))=f(ϕ(a1),,ϕ(an)).\phi(f(a_{1},\cdots,a_{n}))=f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).

(2) \Rightarrow (1): Firstly, we view DD (resp. D{D}^{\prime}) as 𝒞\mathcal{C} (resp. 𝒞{\mathcal{C}}^{\prime}). That is, Xobj(𝒞)\forall X\in\text{obj}(\mathcal{C}), 1XD{{1}_{X}}\in D may be regarded as XX, and vice versa, and so do 𝒞{\mathcal{C}}^{\prime} and D{D}^{\prime}. It is supposed that ϕ\phi maps each identity morphism in DD to an identity morphism in D{D}^{\prime}, and hence with a slight abuse of notation, Xobj(𝒞)\forall X\in\text{obj}(\mathcal{C}), ϕ(X)obj(𝒞)\phi(X)\in\text{obj}({\mathcal{C}}^{\prime}), more precisely, ϕ(1X)=1ϕ(X)\phi({{1}_{X}})={{1}_{\phi(X)}}.

Secondly, let m:ABm:A\to B be a morphism in 𝒞\mathcal{C}. Let gg\in\mathcal{F} be x1x2x3{{x}_{1}}{{x}_{2}}{{x}_{3}}. Then

ϕ(m)\phi(m)

=ϕ(1Bm1A)=\phi({{1}_{B}}\circ m\circ{{1}_{A}})

=ϕ(g(1B,m,1A))=\phi(g({{1}_{B}},m,{{1}_{A}})) (by the definition of f(a1,,an)f(a_{1},\cdots,a_{n}) in Example 9.1.8)

=g(ϕ(1B),ϕ(m),ϕ(1A))=g(\phi({{1}_{B}}),\phi(m),\phi({{1}_{A}})) (by Equation ((9.5)))

=ϕ(1B)ϕ(m)ϕ(1A)=\phi({{1}_{B}})\circ\phi(m)\circ\phi({{1}_{A}}) (by the definition of f(a1,,an)f(a_{1},\cdots,a_{n}) in Example 9.1.8).

Hence ϕ(m):ϕ(A)ϕ(B)\phi(m):\phi(A)\to\phi(B) in 𝒞{\mathcal{C}}^{\prime} since ϕ(1A)=1ϕ(A)\phi({{1}_{A}})={{1}_{\phi(A)}} and ϕ(1B)=1ϕ(B)\phi({{1}_{B}})={{1}_{\phi(B)}}.

Thirdly, if AmBhCA\xrightarrow{m}B\xrightarrow{h}C in 𝒞\mathcal{C}, then again by the above argument, ϕ(A)ϕ(m)ϕ(B)ϕ(h)ϕ(C)\phi(A)\xrightarrow{\phi(m)}\phi(B)\xrightarrow{\phi(h)}\phi(C) in 𝒞{\mathcal{C}}^{\prime}. Moreover, let ff\in\mathcal{F} be x1x2{{x}_{1}}{{x}_{2}}. Then by the definition of f(a1,,an)f(a_{1},\cdots,a_{n}) in Example 9.1.8 and Equation ((9.5)),

ϕ(hm)=ϕ(f(h,m))=f(ϕ(h),ϕ(m))=ϕ(h)ϕ(m).\phi(h\circ m)=\phi(f(h,m))=f(\phi(h),\phi(m))=\phi(h)\circ\phi(m).

9.6. θ\theta-isomorphisms

To T1T_{1} and T2T_{2} as par-operator gen-semigroups, Definition 6.2.1 still applies: a θ\theta-morphism from a T1T_{1}-space S1S_{1} to a T2T_{2}-space is a θ\theta-isomorphism if it is bijective. To justify the definition, we need to show Lemma 6.2.2 and Proposition 6.2.3 for the case of par-operator gen-semigroups as follows.

Lemma 9.6.1.

(Lemma 6.2.2) Let ϕ\phi be a θ\theta-isomorphism from a T1T_{1}-space S1S_{1} to a T2T_{2}-space S2S_{2}. Then θ1|S1\theta^{-1}{{|}_{S_{1}}} is a map, where θ1:={(g,f)|(f,g)θ}\theta^{-1}:=\{(g,f)\,|\,(f,g)\in\theta\}.

Proof.

Let (g,f1),(g,f2)θ1(g,f_{1}),(g,f_{2})\in\theta^{-1}, n+n\in\mathbb{Z}^{+} and (a1,,an)S1n(a_{1},\cdots,a_{n})\in S_{1}^{n}. Then to show that θ1|S1\theta^{-1}{{|}_{S_{1}}} is a map, by Proposition 9.5.2, it suffices to show that neither f1(a1,,an)f_{1}(a_{1},\cdots,a_{n}) nor f2(a1,,an)f_{2}(a_{1},\cdots,a_{n}) is well-defined or f1(a1,,an)=f2(a1,,an)f_{1}(a_{1},\cdots,a_{n})=f_{2}(a_{1},\cdots,a_{n}).

f1(a1,,an)f_{1}(a_{1},\cdots,a_{n}) is well-defined;
\Leftrightarrow θ(f1)(ϕ(a1),,ϕ(an))\theta(f_{1})(\phi(a_{1}),\cdots,\phi(a_{n})) is well-defined (by Definition 9.5.5) ;
\Leftrightarrow g(ϕ(a1),,ϕ(an))g(\phi(a_{1}),\cdots,\phi(a_{n})) is well-defined (because (f1,g)θ(f_{1},g)\in\theta and θ|Imϕ\theta{|}_{\operatorname{Im}\phi} is a map by Definition 9.5.5);
\Leftrightarrow θ(f2)(ϕ(a1),,ϕ(an))\theta(f_{2})(\phi(a_{1}),\cdots,\phi(a_{n})) is well-defined (because (f2,g)θ(f_{2},g)\in\theta and θ|Imϕ\theta{|}_{\operatorname{Im}\phi} is a map);
\Leftrightarrow f2(a1,,an)f_{2}(a_{1},\cdots,a_{n}) is well-defined (by Definition 9.5.5).

And in the case where the above equivalent conditions are satisfied,
ϕ(f1(a1,,an))\phi(f_{1}(a_{1},\cdots,a_{n}))
=θ(f1)(ϕ(a1),,ϕ(an))=\theta(f_{1})(\phi(a_{1}),\cdots,\phi(a_{n})) (by Definition 9.5.5)
=g(ϕ(a1),,ϕ(an))=g(\phi(a_{1}),\cdots,\phi(a_{n}))
=θ(f2)(ϕ(a1),,ϕ(an))=\theta(f_{2})(\phi(a_{1}),\cdots,\phi(a_{n}))
=ϕ(f2(a1,,an))=\phi(f_{2}(a_{1},\cdots,a_{n})) (by Definition 9.5.5),
and hence f1(a1,,an)=f2(a1,,an)f_{1}(a_{1},\cdots,a_{n})=f_{2}(a_{1},\cdots,a_{n}) (because ϕ\phi is injective), as desired. ∎

Proposition 9.6.2.

(Proposition 6.2.3) Let S1S_{1} and S2S_{2} be a T1T_{1}-space and a T2T_{2}-space, respectively, let ϕIsoθ(S1,S2)\phi\in\operatorname{Iso}_{\theta}({{S}_{1}},{{S}_{2}}) and let ϕ1{{\phi}^{-1}} be the inverse map of ϕ\phi. Then ϕ1Isoθ1(S2,S1){{\phi}^{-1}}\in\operatorname{Iso}_{\theta^{-1}}({{S}_{2}},{{S}_{1}}).

Proof.

By Definition 9.5.5, θ|S2\theta{{|}_{S_{2}}} is a map and n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)S1n(a_{1},\cdots,a_{n})\in{S_{1}^{n}} and fDomθf\in\operatorname{Dom}\theta, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor θ(f)(ϕ(a1),,ϕ(an))\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined or

f(a1,,an)=ϕ1(θ(f)(ϕ(a1),,ϕ(an))).f(a_{1},\cdots,a_{n})=\phi^{-1}(\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))).

Hence we have

Fact (A): n+,fDomθ\forall n\in{{\mathbb{Z}}^{+}},f\in\operatorname{Dom}\theta and (b1,,bn)S2n(b_{1},\cdots,b_{n})\in{S_{2}^{n}},
neither f(ϕ1(b1),,ϕ1(bn))f({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n})) nor θ(f)(b1,,bn)\theta(f)(b_{1},\cdots,b_{n}) is well-defined or

(9.6) f(ϕ1(b1),,ϕ1(bn))=ϕ1(θ(f)(b1,,bn)).f({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n}))=\phi^{-1}(\theta(f)(b_{1},\cdots,b_{n})).

Let gImθg\in\operatorname{Im}\theta, n+n\in{{\mathbb{Z}}^{+}} and (b1,,bn)S2n(b_{1},\cdots,b_{n})\in{S_{2}^{n}}. Then there are two cases as follows.

Case (1): g(b1,,bn)g(b_{1},\cdots,b_{n}) is well-defined.

We are showing that θ1(g)(ϕ1(b1),,ϕ1(bn))\theta^{-1}(g)({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n})) is also well-defined and

θ1(g)(ϕ1(b1),,ϕ1(bn))=ϕ1(g(b1,,bn)).\theta^{-1}(g)({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n}))=\phi^{-1}(g(b_{1},\cdots,b_{n})).

Since θ|S2\theta{{|}_{S_{2}}} is a map, by Corollary 9.5.4, (f,g)θ\forall(f,g)\in\theta, θ(f)(b1,,bn)\theta(f)(b_{1},\cdots,b_{n}) is well-defined and

θ(f)(b1,,bn)=g(b1,,bn).\theta(f)(b_{1},\cdots,b_{n})=g(b_{1},\cdots,b_{n}).

Then by Fact (A) and Equation (9.6), (f,g)θ\forall(f,g)\in\theta,

f(ϕ1(b1),,ϕ1(bn))=ϕ1(g(b1,,bn)).f({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n}))=\phi^{-1}(g(b_{1},\cdots,b_{n})).

Thus by (the θ1\theta^{-1} version of) Definition 9.5.3, θ1(g)(ϕ1(b1),,ϕ1(bn))\theta^{-1}(g)({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n})) is well-defined and

θ1(g)(ϕ1(b1),,ϕ1(bn))=ϕ1(g(b1,,bn)).\theta^{-1}(g)({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n}))=\phi^{-1}(g(b_{1},\cdots,b_{n})).

Case (2): g(b1,,bn)g(b_{1},\cdots,b_{n}) is not well-defined.

We are showing that θ1(g)(ϕ1(b1),,ϕ1(bn))\theta^{-1}(g)({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n})) is not well-defined, either.

Let (f,g)θ(f,g)\in\theta. Assume that θ(f)(b1,,bn)\theta(f)(b_{1},\cdots,b_{n}) is well-defined. Then by Definition 9.5.3, g(b1,,bn)g(b_{1},\cdots,b_{n}) is also well-defined, a contradiction. Hence θ(f)(b1,,bn)\theta(f)(b_{1},\cdots,b_{n}) is not well-defined. Thus by Fact (A), f(ϕ1(b1),,ϕ1(bn))f({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n})) is not well-defined, either, and hence θ1(g)(ϕ1(b1),,ϕ1(bn))\theta^{-1}(g)({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n})) is not well-defined by (the θ1\theta^{-1} version of) Definition 9.5.3.

Combining Cases (1) and (2), we can tell that n+\forall n\in{{\mathbb{Z}}^{+}}, gDomθ1(=Imθ)g\in\operatorname{Dom}\theta^{-1}(=\operatorname{Im}\theta) and (b1,,bn)S2n(b_{1},\cdots,b_{n})\in{S_{2}^{n}},

θ1(g)(ϕ1(b1),,ϕ1(bn))=ϕ1(g(b1,,bn)),\theta^{-1}(g)({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n}))={{\phi}^{-1}}(g(b_{1},\cdots,b_{n})),

or neither θ1(g)(ϕ1(b1),,ϕ1(bn))\theta^{-1}(g)({\phi}^{-1}({b}_{1}),\cdots,{\phi}^{-1}({b}_{n})) nor g(b1,,bn)g(b_{1},\cdots,b_{n}) is well-defined.

Moreover, by Lemma 9.6.1, θ1|S1\theta^{-1}|_{S_{1}} is a map. Then by Definition 9.5.5, ϕ1\phi^{-1} is a θ1\theta^{-1}-morphism from S2S_{2} to S1S_{1}.

Furthermore, since ϕ1\phi^{-1} is bijective, by Definition 6.2.1, ϕ1Isoθ1(S2,S1){{\phi}^{-1}}\in\operatorname{Iso}_{\theta^{-1}}({{S}_{2}},{{S}_{1}}). ∎

10. Basic properties and more notions

In Sections 10 to 14 as follows, our study will focus on par-operator gen-semigroups. The following statement is obvious by Definitions 2.1.1, 8.1.2, and 9.1.4: Each operator semigroup is an operator gen-semigroup, and each operator gen-semigroup is a par-operator gen-semigroup.

Thus our research on par-operator gen-semigroups also applies to operator gen-semigroups and operator semigroups. From now on, unless otherwise specified, DD always denotes a set and TT always denotes a par-operator gen-semigroup on DD defined by Definition 9.1.4. In Sections 10 to 14, unless otherwise specified, TT-morphisms and θ\theta-morphisms are always defined by Definitions 9.3.1 and 9.5.5, respectively.

For par-operator gen-semigroups, Sections 10, 11, 12, 13 and 14 generalize results obtained in Sections 2, 3, 4, 5 and 7, respectively. Throughout Sections 10 to 14, by statements such as “Proposition (or Definition, etc.) x.x.x (which is the index) still holds (or still applies)” we mean that the proposition (or definition, etc.) still holds for (or still applies to) the case of par-operator gen-semigroups given specifically in the preceding paragraph.

Roughly speaking, a result for an operator semigroup TT still holds for a par-operator gen-semigroup if the correctness of the proof does not depend on the number of the variables or the domain of definition of any element of TT. However, if the correctness of the proof of a result does depend on the number of the variables or the domain of definition of an element of TT, the result may still hold but need a different proof.

10.1. Basic properties of TT-spaces and quasi-TT-spaces

In this subsection, we shall generalize properties of TT-spaces or quasi-TT-spaces obtained in Subsection 2.2.

Proposition 2.2.7 is generalized to

Proposition 10.1.1.

Let SS be a TT-space. Then n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and fTf\in T of nn variables, f(a1,,an)Sf(a_{1},\cdots,a_{n})\in S if f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined. Hence AS\forall A\subseteq S, ATS{{\left\langle A\right\rangle}_{T}}\leq S. In particular, STS{{\left\langle S\right\rangle}_{T}}\leq S.

Proof.

Let n+n\in{{\mathbb{Z}}^{+}}, let (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and let fTf\in T have nn variables. By Definition 9.2.1, UD\exists U\subseteq D such that UT=S{{\left\langle U\right\rangle}_{T}}=S. And i=1,,n\forall i=1,\cdots,n, giT\exists{{g}_{i}}\in T of ni{{n}_{i}} variables and uiUni{{\text{u}}_{i}}\in{{U}^{{{n}_{i}}}} (cartesian product) such that gi(ui)=ai.{{g}_{i}}({{\text{u}}_{i}})={{a}_{i}}. By Definition 9.1.4, the partial function f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) defined by Definition 9.1.2 is a restriction of some element of TT. So f(a1,,an)=f(g1(u1),,gn(un))UT=Sf(a_{1},\cdots,a_{n})=f({{g}_{1}}({{\text{u}}_{1}}),\cdots,{{g}_{n}}({{\text{u}}_{n}}))\in{{\left\langle U\right\rangle}_{T}}=S if f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined.

Therefore, AS\forall A\subseteq S, n+n\in{{\mathbb{Z}}^{+}}, (a1,,an)An(a_{1},\cdots,a_{n})\in{{A}^{n}} and fTf\in T of nn variables, f(a1,,an)Sf(a_{1},\cdots,a_{n})\in S if f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined. Hence ATS{{\left\langle A\right\rangle}_{T}}\subseteq S, and thus ATS{{\left\langle A\right\rangle}_{T}}\leq S (by Definition 9.2.1). ∎

Definition 2.2.8 still applies. Then by Proposition 10.1.1, Proposition 2.2.9 still holds. Propositions 2.2.10 and 2.2.12 still hold because their proofs still apply. However, we could not generalize Propositions 2.2.13 and 2.2.14 because now the elements of TT may have more than one variable. Proposition 2.2.16 and its proof still apply.

10.2. Basic properties of TT-morphisms and TT-isomorphisms

In this subsection, we shall generalize properties of TT-morphisms and TT-isomorphisms obtained in Subsections 2.3 and 2.4.

Proposition 2.3.6 still holds, but now it requires a different proof as follows.

Proposition 10.2.1.

(Proposition 2.3.6) Let σ\sigma be a TT-morphism from S1{{S}_{1}} to S2{{S}_{2}}. Then ImσqS2\operatorname{Im}\sigma\leq_{q}{S_{2}}. Moreover, if IdT\operatorname{Id}\in T or more generally, ImσImσT\operatorname{Im}\sigma\subseteq{{\left\langle\operatorname{Im}\sigma\right\rangle}_{T}}, then ImσS2\operatorname{Im}\sigma\leq{{S}_{2}}.

Remark.

We keep the original index of the proposition (i.e. 2.3.6) because we shall use it later (explicitly or implicitly).

Proof.

By Definition 9.3.1, n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)S1n(a_{1},\cdots,a_{n})\in S_{1}^{n} and fTf\in T of nn variables, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor f(σ(a1),,σ(an))f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) is well-defined or

f(σ(a1),,σ(an))=σ(f(a1,,an))Imσ.f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}))=\sigma(f(a_{1},\cdots,a_{n}))\in\operatorname{Im}\sigma.

Hence ImσTImσ{{\left\langle\operatorname{Im}\sigma\right\rangle}_{T}}\subseteq\operatorname{Im}\sigma, and thus by Definition 2.2.8, Imσ\operatorname{Im}\sigma is a quasi-TT-subspace of S2S_{2}.

Moreover, if IdT\operatorname{Id}\in T or ImσImσT\operatorname{Im}\sigma\subseteq{{\left\langle\operatorname{Im}\sigma\right\rangle}_{T}}, then by Proposition 2.2.10, Imσ\operatorname{Im}\sigma is a TT-space, and hence ImσS2\operatorname{Im}\sigma\leq{{S}_{2}}. ∎

By the way, we generalize Proposition 10.2.1 to θ\theta-morphisms as follows, which generalizes Proposition 6.1.11 to the case of par-operator gen-semigroups.

Proposition 10.2.2.

Let ϕ\phi be a θ\theta-morphism from a T1T_{1}-space S1S_{1} to a T2T_{2}-space S2S_{2}. Suppose Imθ=T2\operatorname{Im}\theta=T_{2}. Then ImϕqS2\operatorname{Im}\phi\leq_{q}{S_{2}}. Moreover, if IdT2\operatorname{Id}\in T_{2} or more generally, ImϕImϕT2\operatorname{Im}\phi\subseteq{{\left\langle\operatorname{Im}\phi\right\rangle}_{T_{2}}}, then ImϕS2\operatorname{Im}\phi\leq{{S}_{2}}.

Proof.

By Definition 9.5.5, n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)S1n(a_{1},\cdots,a_{n})\in{S_{1}^{n}} and fDomθf\in\operatorname{Dom}\theta, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor θ(f)(ϕ(a1),,ϕ(an))\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined or

θ(f)(ϕ(a1),,ϕ(an))=ϕ(f(a1,,an))Imϕ.\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))=\phi(f(a_{1},\cdots,a_{n}))\in\operatorname{Im}\phi.

Since Imθ=T2\operatorname{Im}\theta=T_{2}, we can tell ImϕT2Imϕ{{\left\langle\operatorname{Im}\phi\right\rangle}_{T_{2}}}\subseteq\operatorname{Im}\phi, and so by Definition 2.2.8, Imϕ\operatorname{Im}\phi is a quasi-T2T_{2}-subspace of S2S_{2}.

Moreover, if IdT2\operatorname{Id}\in T_{2} or ImϕImϕT2\operatorname{Im}\phi\subseteq{{\left\langle\operatorname{Im}\phi\right\rangle}_{T_{2}}}, then by Proposition 2.2.10, Imϕ\operatorname{Im}\phi is a T2T_{2}-space and hence ImϕS2\operatorname{Im}\phi\leq{{S}_{2}}. ∎

Definition 2.3.7 still applies. And Proposition 2.3.8 still holds as shown below.

Proposition 10.2.3.

(Proposition 2.3.8) Let SS be a TT-space. Then EndT(S)\operatorname{End}_{T}(S) defined by Definition 2.3.7 constitutes a monoid, which we still denote by EndT(S)\operatorname{End}_{T}(S), with composition of functions as the binary operation.

Proof.

The identity map on SS lies in EndT(S)\operatorname{End}_{T}(S), and hence we only need to show that EndT(S)\operatorname{End}_{T}(S) is a semigroup with composition of functions as the binary operation.

Let σ1,σ2EndT(S){{\sigma}_{1}},{{\sigma}_{2}}\in\operatorname{End}_{T}(S), fTf\in T, n+n\in{{\mathbb{Z}}^{+}} and (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}}. Then by Definition 9.3.1,

f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined;

f(σ2(a1),,σ2(an))\Leftrightarrow f({{\sigma}_{2}}({{a}_{1}}),\cdots,{{\sigma}_{2}}({{a}_{n}})) is well-defined;

f(σ1(σ2(a1)),,σ1(σ2(an)))\Leftrightarrow f({{\sigma}_{1}}({{\sigma}_{2}}({{a}_{1}})),\cdots,{{\sigma}_{1}}({{\sigma}_{2}}({{a}_{n}}))) is well-defined.

Moreover, in the case where the above equivalent conditions are satisfied,

(σ1σ2)(f(a1,,an))({{\sigma}_{1}}\circ{{\sigma}_{2}})(f(a_{1},\cdots,a_{n}))

=σ1(σ2(f(a1,,an)))={{\sigma}_{1}}({{\sigma}_{2}}(f(a_{1},\cdots,a_{n})))

=σ1(f(σ2(a1),,σ2(an)))={{\sigma}_{1}}(f({{\sigma}_{2}}({{a}_{1}}),\cdots,{{\sigma}_{2}}({{a}_{n}})))

=f(σ1(σ2(a1)),,σ1(σ2(an)))=f({{\sigma}_{1}}({{\sigma}_{2}}({{a}_{1}})),\cdots,{{\sigma}_{1}}({{\sigma}_{2}}({{a}_{n}}))).

Then again by Definition 9.3.1, σ1σ2EndT(S){{\sigma}_{1}}\circ{{\sigma}_{2}}\in\operatorname{End}_{T}(S).

Composition of functions is associative. Therefore, EndT(S)\operatorname{End}_{T}(S) is a monoid with composition of functions as the binary operation. ∎

Proposition 2.3.9 is still obvious.

As already said before, Definition 2.4.1 still applies to the case of par-operator gen-semigroups. Proposition 2.4.2 was generalized to TT being a par-operator gen-semigroup by Proposition 9.4.1. Proposition 2.4.3 still holds because its proof still applies. Proposition 2.4.4 is still obvious.

10.3. Galois TT-extensions

Definition 2.5.1 and the remark right after it still apply. Proposition 2.5.2 still holds, but now it requires a different proof as follows.

Proposition 10.3.1.

(Proposition 2.5.2) Let SS be a TT-space and let HH be a subset of EndT(S)\operatorname{End}_{T}(S). Then SHqS{S^{H}}\leq_{q}S. Moreover, if IdT\operatorname{Id}\in T or SHSHT{{S}^{H}}\subseteq{{\left\langle{{S}^{H}}\right\rangle}_{T}}, then SHS{{S}^{H}}\leq S.

Proof.

Let n+n\in{{\mathbb{Z}}^{+}}, (a1,,an)(SH)n(a_{1},\cdots,a_{n})\in{{({{S}^{H}})}^{n}} and fTf\in T of nn variables. By Definition 9.3.1, σH\forall\sigma\in H, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor f(σ(a1),,σ(an))f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) is well-defined, or

σ(f(a1,,an))=f(σ(a1),,σ(an))=f(a1,,an).\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}))=f(a_{1},\cdots,a_{n}).

Therefore, f(a1,,an)SHf(a_{1},\cdots,a_{n})\in{{S}^{H}} if f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined. Thus SHTSH{{\left\langle{{S}^{H}}\right\rangle}_{T}}\subseteq{{S}^{H}}. Hence SH{{S}^{H}} is a quasi-TT-space, and so SHqS{S^{H}}\leq_{q}S.

Moreover, if IdT\operatorname{Id}\in T or SHSHT{{S}^{H}}\subseteq{{\left\langle{{S}^{H}}\right\rangle}_{T}}, then by Proposition 2.2.10, SH{S^{H}} is a TT-space, and hence SHS{{S}^{H}}\leq S. ∎

Proposition 2.5.3 is still obvious.

10.4. Galois TT-monoids and Galois TT-groups

Definition 2.6.1 and the remark right after it still apply. Propositions 2.6.2 and 2.6.3 are still true because their proofs still apply. Propositions 2.6.4, 2.6.5 and 2.6.6 are still obvious.

10.5. Generated submonoids of EndT(S)\operatorname{End}_{T}(S) and subgroups of AutT(S)\operatorname{Aut}_{T}(S)

Definition 2.7.1 and the remark right after it still apply. Proposition 2.7.2 still holds because its proof still applies.

11. Galois correspondences

In this section, we shall find that all results obtained in Subsection 3.2 still apply to the case where TT is a par-operator gen-semigroup.

Notations 3.0.1 (and the two remarks for it) and 3.0.2 still apply.

Definition 3.2.1 and the remark right after it still apply. Lemma 3.2.2 still holds because its proof still applies.

Definition 3.2.3 and the remark right after it still apply. Lemma 3.2.4 still holds because its proof still applies.

Corollaries 3.2.5 and 3.2.6 still hold because their proofs still apply.

Then by Propositions 8.3.5, 8.3.6, 8.3.7, 8.5.8, 9.3.4, 9.3.6 and 9.5.9, we may apply Corollaries 3.2.5 and 3.2.6 to modules, abelian groups, non-abelian groups, rings, field, differential field, and categories, respectively. We can tell that Galois correspondences exist not only for Galois TT-groups (by Corollary 3.2.6), but also for Galois TT-monoids (by Corollary 3.2.5).

Moreover, if we want to build up the Galois correspondences in terms of topology, then we may employ results in Section 13, which generalize results in Section 5.

12. Lattice structures of objects arising in Galois correspondences on a TT-space SS

In this section, results obtained in Section 4 are generalized for par-operator gen-semigroups.

12.1. SubT(S)\operatorname{Sub}_{T}(S), SMn(EndT(S))\operatorname{SMn}(\operatorname{End}_{T}(S)) and SGr(AutT(S))\operatorname{SGr}(\operatorname{Aut}_{T}(S))

As said in Subsection 10.1, we could not generalize Proposition 2.2.13 because now the elements of TT may have more than one variable. So to generalize Proposition 4.1.1, we need a notion as follows.

Definition 12.1.1.

Let SS be a TT-space. If XSX\subseteq S, then the intersection of all the quasi-TT-subspaces of SS containing XX, denoted by XS{{\left\langle X\right\rangle}_{S}}, is called the quasi-TT-subspace of SS generated by XX.

Remark.

By Proposition 2.2.12, XS{{\left\langle X\right\rangle}_{S}} is the smallest quasi-TT-subspace of SS which contains XX.

Then Proposition 4.1.1 is generalized to

Proposition 12.1.2.

Let SS be a TT-space. Then SubT(S)\operatorname{Sub}_{T}(S) is a lattice with inclusion as the binary relation if S1,S2SubT(S)\forall{{S}_{1}},{{S}_{2}}\in\operatorname{Sub}_{T}(S), S1S2:=S1S2{{S}_{1}}\wedge{{S}_{2}}:={{S}_{1}}\bigcap{{S}_{2}} and S1S2:=S1S2S{{S}_{1}}\vee{{S}_{2}}:={{\left\langle{{S}_{1}}\bigcup{{S}_{2}}\right\rangle}_{S}}. Moreover, SubT(S)\operatorname{Sub}_{T}(S) is a complete lattice if let A=Si\wedge A=\bigcap{{{S}_{i}}} and let A=SiS\vee A={{\left\langle\bigcup{{{S}_{i}}}\right\rangle}_{S}}, A={Si}SubT(S)\forall A=\{{{S}_{i}}\}\subseteq\operatorname{Sub}_{T}(S).

Proof.

Straightforward consequences of Proposition 2.2.12 and Definition 12.1.1. ∎

Propositions 4.1.2 and 4.1.3 still hold because Propositions 2.3.9 and 2.4.4 still hold and Definition 2.7.1 and the remark right after it still apply. Proposition 4.1.4 still holds because Proposition 2.7.2 still holds.

12.2. IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) and IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)

Lemma 4.2.1 still holds because its proof still applies. Proposition 4.2.2 would still hold if Proposition 4.1.1 in it were replaced by Proposition 12.1.2 as follows.

Proposition 12.2.1.

IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) is a complete \wedge-sublattice of the complete lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 12.1.2.

Proof.

Almost the same as the proof of Proposition 4.2.2 except that Proposition 4.1.1 is replaced by Proposition 12.1.2. ∎

Analogously, Lemma 4.2.4 still holds, and Proposition 4.2.5 would still hold if Proposition 4.1.1 in it were replaced by Proposition 12.1.2 as follows.

Proposition 12.2.2.

IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) is a complete \wedge-sublattice of the complete lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 12.1.2.

Proof.

Almost the same as the proof of Proposition 4.2.5 except that Proposition 4.1.1 is replaced by Proposition 12.1.2. ∎

12.3. GSMnT(S/B)\operatorname{GSMn}_{T}(S/B) and GSGrT(S/B)\operatorname{GSGr}_{T}(S/B)

Lemma 4.3.1, Proposition 4.3.2, Lemma 4.3.4 and Proposition 4.3.5 still hold because their proofs still apply.

13. Topologies employed to construct Galois correspondences

In this section, results obtained in Section 5 are generalized for par-operator gen-semigroups.

13.1. Topologies on a TT-space SS for IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B)

Lemma 5.1.1 and its proof still apply.

However, we could not generalize a part of Theorem 5.1.2 straightforwardly because we could not generalize Proposition 2.2.13. Instead, we have the following, which in a sense is analogous to Theorem 5.3.2.

Theorem 13.1.1.

Let SS be a TT-space and let BSB\subseteq S. Let

P1={all topologies on S such that Equation (5.1) is satisfied}{{P}_{1}}=\{\text{all topologies on $S$ such that Equation $($\ref{5.1}$)$ is satisfied}\}

and let

Q1={all topologies on S which are finer than 𝒯1},{{Q}_{1}}=\{\text{all topologies on $S$ which are finer than ${{\mathcal{T}}_{1}}$}\},

where 𝒯1{{\mathcal{T}}_{1}} is defined in Lemma 5.1.1. Then P1Q1{{P}_{1}}\subseteq{{Q}_{1}} and the following statements are equivalent:

  1. (i)

    P1{{P}_{1}}\neq\emptyset.

  2. (ii)

    For any intersection KK of finite unions of elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B), KSubT(S)K\in\operatorname{Sub}_{T}(S) implies KIntTEnd(S/B){}K\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\bigcup\{\emptyset\}.

  3. (iii)

    𝒯1P1(Q1){{\mathcal{T}}_{1}}\in{{P}_{1}}(\subseteq{{Q}_{1}}); that is, 𝒯1{{\mathcal{T}}_{1}} is the coarsest topology on SS such that Equation ((5.1)) is satisfied.

Moreover, if IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) is a complete \vee-sublattice of the complete lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 12.1.2, then the above three statements are true.

Proof.

By Lemma 5.1.1, 𝒯1{{\mathcal{T}}_{1}} is the smallest topology on SS such that the collection of all closed sets contains IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B). Suppose 𝒯P1\mathcal{T}\in{{P}_{1}}. Then all elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) are closed in 𝒯\mathcal{T}, and thus 𝒯1𝒯\mathcal{T}_{1}\subseteq\mathcal{T}. Hence 𝒯Q1\mathcal{T}\in{{Q}_{1}}. Therefore P1Q1{{P}_{1}}\subseteq{{Q}_{1}}.

(i) \Rightarrow (ii): Let 𝒯P1\mathcal{T}\in{{P}_{1}}. Let KK be any intersection of finite unions of elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B). By the definition of P1P_{1}, each element of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) is closed in 𝒯\mathcal{T}, and so is KK. Suppose KSubT(S)K\in\operatorname{Sub}_{T}(S). Then K{S|BSqS}K\in\{{S}^{\prime}\,|\,B\subseteq{S}^{\prime}\leq_{q}S\}. Again by the definition of P1P_{1}, KIntTEnd(S/B){}K\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\bigcup\{\emptyset\}. Hence (ii) is true.

(ii) \Rightarrow (iii): By the definition of 𝒯1{{\mathcal{T}}_{1}} in Lemma 5.1.1, all elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) are closed in 𝒯1{{\mathcal{T}}_{1}}. So by Definition 3.2.1 and Proposition 10.3.1,

IntTEnd(S/B)\{}\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\}\subseteq {BSqS|SB\subseteq{S}^{\prime}\leq_{q}S\,|\,{S}^{\prime} is nonempty and closed in 𝒯1{{\mathcal{T}}_{1}}}.

On the other hand, we prove

IntTEnd(S/B)\{}\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\}\supseteq {BSqS|SB\subseteq{S}^{\prime}\leq_{q}S\,|\,{S}^{\prime} is nonempty and closed in 𝒯1{{\mathcal{T}}_{1}}}

as follows.

Let K{BSqS|SK\in\{B\subseteq{S}^{\prime}\leq_{q}S\,|\,{S}^{\prime} is nonempty and closed in 𝒯1}{{\mathcal{T}}_{1}}\}. We need to show KIntTEnd(S/B)\{}K\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\}.

Because 𝒯1{{\mathcal{T}}_{1}} is generated by subbasis β1{{\beta}_{1}}, 𝒯1{{\mathcal{T}}_{1}} is the collection of all unions of finite intersections of elements of β1={S\A|AIntTEnd(S/B)}{S}{{\beta}_{1}}=\{S\backslash A\,|\,A\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\}\bigcup\{S\}. Thus S\KS\backslash K is a union of finite intersections of elements of β1{{\beta}_{1}}. It follows from DeMorgan’s Laws that KK is an intersection of finite unions of elements of IntTEnd(S/B){}\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\bigcup\{\emptyset\}. Because KK\neq\emptyset and KSubT(S)K\in\operatorname{Sub}_{T}(S), by (ii), KIntTEnd(S/B)\{}K\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\}, as desired. Hence

IntTEnd(S/B)\{}\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\{\emptyset\}\supseteq {BSqS|SB\subseteq{S}^{\prime}\leq_{q}S\,|\,{S}^{\prime} is nonempty and closed in 𝒯1{{\mathcal{T}}_{1}}}

Therefore, Equation ((5.1)) is satisfied for 𝒯1{{\mathcal{T}}_{1}}.

Thus 𝒯1P1{{\mathcal{T}}_{1}}\in{{P}_{1}}. We showed P1Q1{{P}_{1}}\subseteq{{Q}_{1}} at the beginning of our proof, and thus by the definition of Q1{{Q}_{1}}, 𝒯1{\mathcal{T}}_{1} is the coarsest topology in P1P_{1}. Hence by the definition of P1{{P}_{1}}, 𝒯1{\mathcal{T}}_{1} is the coarsest topology on SS such that Equation ((5.1)) is satisfied.

(iii) \Rightarrow (i): Obvious.

Let KK be any intersection of finite unions of elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B). By Proposition 12.2.1, any intersection of elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) lies in IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B). Then as a result of the first distributive law of set operations, KK is a union of elements of IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B). Suppose that IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B) is a complete \vee-sublattice of the complete lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 12.1.2. Then KSubT(S)K\in\operatorname{Sub}_{T}(S) implies KIntTEnd(S/B)K\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B), and hence (ii) is true (in this case). ∎

If TT is an operator semigroup, then by Propositions 2.2.13 and 2.5.2, S1,S2IntTEnd(S/B)\forall{{S}_{1}},{{S}_{2}}\in\operatorname{Int}_{T}^{\operatorname{End}}(S/B), S1S2SubT(S){{S}_{1}}\bigcup{{S}_{2}}\in\operatorname{Sub}_{T}(S), which implies that, in this case, statement (ii) in Theorem 5.1.2 coincides with statement (ii) in Theorem 13.1.1. So Theorem 13.1.1 generalizes Theorem 5.1.2.

13.2. Topologies on a TT-space SS for IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)

Lemma 5.2.1 and its proof still apply.

We could not generalize a part of Theorem 5.2.2 straightforwardly because we could not generalize Proposition 2.2.13. Instead, we have the following, which is analogous to Theorem 13.1.1.

Theorem 13.2.1.

Let SS be a TT-space and let BSB\subseteq S. Let

P2={all topologies on S such that Equation (5.3) is satisfied}{{P}_{2}}=\{\text{all topologies on $S$ such that Equation $($\ref{5.3}$)$ is satisfied}\}

and let

Q2={all topologies on S which are finer (larger) than 𝒯2},{{Q}_{2}}=\{\text{all topologies on $S$ which are finer $($larger$)$ than ${{\mathcal{T}}_{2}}$}\},

where 𝒯2{{\mathcal{T}}_{2}} is defined in Lemma 5.2.1. Then P2Q2{{P}_{2}}\subseteq{{Q}_{2}} and the following statements are equivalent:

  1. (i)

    P2{{P}_{2}}\neq\emptyset.

  2. (ii)

    For any intersection KK of finite unions of elements of IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B), KSubT(S)K\in\operatorname{Sub}_{T}(S) implies KIntTAut(S/B){}K\in\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)\bigcup\{\emptyset\}.

  3. (iii)

    𝒯2P2(Q2){{\mathcal{T}}_{2}}\in{{P}_{2}}(\subseteq{{Q}_{2}}); that is, 𝒯2{{\mathcal{T}}_{2}} is the coarsest topology on SS such that Equation ((5.3)) is satisfied.

Moreover, if IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B) is a complete \vee-sublattice of the complete lattice SubT(S)\operatorname{Sub}_{T}(S) defined in Proposition 12.1.2, then the above three statements are true.

Proof.

Almost the same as the proof of Theorem 13.1.1 except that IntTEnd(S/B)\operatorname{Int}_{T}^{\operatorname{End}}(S/B), 𝒯1{{\mathcal{T}}_{1}}, β1{{\beta}_{1}}, P1{{P}_{1}} and Q1{{Q}_{1}} are replaced by IntTAut(S/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B), 𝒯2{{\mathcal{T}}_{2}}, β2{{\beta}_{2}}, P2{{P}_{2}} and Q2{{Q}_{2}}, respectively, and accordingly, Lemma 5.1.1 and Proposition 12.2.1 are replaced by Lemma 5.2.1 and Proposition 12.2.2, respectively. ∎

Just as Theorem 13.1.1 generalizes Theorem 5.1.2, Theorem 13.2.1 generalizes Theorem 5.2.2.

13.3. Topologies on EndT(S)\operatorname{End}_{T}(S)

Lemma 5.3.1 and its proof still apply. Theorem 5.3.2 still holds because its proof still applies.

13.4. Topologies on AutT(S)\operatorname{Aut}_{T}(S)

Lemma 5.4.1 and its proof still apply. Theorem 5.4.2 still holds because its proof still applies.

14. Constructions of the generalized morphisms and isomorphisms

In this section, results obtained in Section 7 are generalized for par-operator gen-semigroups.

14.1. A construction of TT-morphisms from uT(uD){{\left\langle u\right\rangle}_{T}}(u\in D)

Although the contents of this subsection will be further generalized in the next subsection, we still keep this subsection because we will need it in Sections 19 and 20.

Notation 7.1.1 is generalized for partial functions of multiple variables as follows.

Notation 14.1.1.

Let Mn+{M\subseteq\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{}all partial functions from Dn{{D}^{n}} to DD}, where DD is a set. Let u,vDu,v\in D. By uMvu\xrightarrow{M}v we mean that

n,m+\forall n,m\in{{\mathbb{Z}}^{+}}, fMf\in M of nn variables and gMg\in M of mm variables,

f(un)=g(um)f({{u}^{n}})=g({{u}^{m}}) implies f(vn)=g(vm)f({{v}^{n}})=g({{v}^{m}}),

where xk{{x}^{k}} denotes the kk-tuple (x,,x)(x,\cdots,x).

Besides, let [u)M{{[u)}_{M}} denote the set {wD|uMw}\{w\in D\,|\,u\xrightarrow{M}w\}.

Moreover, by uMvu\overset{M}{\longleftrightarrow}v we mean that uMvu\xrightarrow{M}v and vMuv\xrightarrow{M}u.

Besides, let [u]M{{[u]}_{M}} denote the set {wD|uMw}\{w\in D\,|\,u\overset{M}{\longleftrightarrow}w\}.

Remark.
  1. (1)

    Suppose uMvu\xrightarrow{M}v. Then n+\forall n\in{{\mathbb{Z}}^{+}} and fMf\in M of nn variables, f(un)=f(un)f({{u}^{n}})=f({{u}^{n}}) implies f(vn)=f(vn)f({{v}^{n}})=f({{v}^{n}}), and hence by Convention 6.1.7, that f(un)f({{u}^{n}}) is well-defined implies that f(vn)f({{v}^{n}}) is well-defined.

  2. (2)

    Apparently, uMvu\overset{M}{\longleftrightarrow}v defines an equivalence relation on DD.

Then Proposition 7.1.3 still holds but its proof changes as follows.

Proposition 14.1.2.

(Proposition 7.1.3) Let σ\sigma be a TT-morphism from a TT-space SS. Then aS\forall a\in S, aTσ(a)a\xrightarrow{T}\sigma(a).

Proof.

Let fTf\in T have nn variables, let gTg\in T have mm variables and let aSa\in S.

If f(an)=g(am)f({{a}^{n}})=g({{a}^{m}}), where xk{{x}^{k}} denotes the kk-tuple (x,,x)(x,\cdots,x), then by Definition 9.3.1, both f((σ(a))n)f({{(\sigma(a))}^{n}}) and g((σ(a))m)g({{(\sigma(a))}^{m}}) are well-defined and

f((σ(a))n)=σ(f(an))=σ(g(am))=g((σ(a))m).f({{(\sigma(a))}^{n}})=\sigma(f({{a}^{n}}))=\sigma(g({{a}^{m}}))=g({{(\sigma(a))}^{m}}).

Therefore, aTσ(a)a\xrightarrow{T}\sigma(a). ∎

Analogously, it is not hard to tell that Proposition 7.1.4 still holds.

Proposition 7.1.6 can be generalized for par-operator gen-semigroups, but we omit it for brevity.

Proposition 7.1.7 is generalized to

Proposition 14.1.3.

Let u,vDu,v\in D.

  1. (a)

    The following statements are equivalent:

    1. (i)

      uTvu\xrightarrow{T}v.

    2. (ii)

      There exists a map σ:uTvT\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}} given by f(un)f(vn)f({{u}^{n}})\mapsto f({{v}^{n}}), n+\forall n\in{{\mathbb{Z}}^{+}} and fTf\in T of nn variables such that f(un)f({{u}^{n}}) is well-defined.

  2. (b)

    The map σ\sigma given in (ii) is a TT-morphism.

Proof.

For (a):

uTvu\xrightarrow{T}v;
n,m+\Leftrightarrow\forall n,m\in{{\mathbb{Z}}^{+}}, fTf\in T of nn variables and gTg\in T of mm variables, f(un)=g(um)f({{u}^{n}})=g({{u}^{m}}) implies f(vn)=g(vm)f({{v}^{n}})=g({{v}^{m}});
σ:uTvT\Leftrightarrow\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}} given by f(un)f(vn)f({{u}^{n}})\mapsto f({{v}^{n}}), n+\forall n\in{{\mathbb{Z}}^{+}} and fTf\in T of nn variables such that f(un)f({{u}^{n}}) is well-defined, is a well-defined map (cf. Remark (1) right after Notation 14.1.1).

For (b):

Let (a1,,an)uTn(a_{1},\cdots,a_{n})\in\left\langle u\right\rangle_{T}^{n}. Then i=1,,n\forall i=1,\cdots,n, giT\exists{{g}_{i}}\in T of ni{{n}_{i}} variables such that gi(uni)=ai{{g}_{i}}({{u}^{{{n}_{i}}}})={{a}_{i}}.

Let fTf\in T have nn variables. By Definition 9.1.2, f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) is a partial function from Dm{{D}^{m}} to DD where m=inim=\sum\nolimits_{i}{{{n}_{i}}}. By Definition 9.1.4, f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) is a restriction of some element of TT. Hence

f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined;
σ(f(a1,,an))\Leftrightarrow\sigma(f(a_{1},\cdots,a_{n}))

=σ(f(g1(un1),,gn(unn)))=\sigma(f({{g}_{1}}({{u}^{{{n}_{1}}}}),\cdots,{{g}_{n}}({{u}^{{{n}_{n}}}})))

=σ((f(g1,,gn))(um))=\sigma((f\circ({{g}_{1}},\cdots,{{g}_{n}}))({{u}^{m}})) (by Definition 9.1.2)

=(f(g1,,gn))(vm)=(f\circ({{g}_{1}},\cdots,{{g}_{n}}))({{v}^{m}}) (by the definition of σ\sigma in (ii))

=f(g1(vn1),,gn(vnn))=f({{g}_{1}}({{v}^{{{n}_{1}}}}),\cdots,{{g}_{n}}({{v}^{{{n}_{n}}}})) (by Definition 9.1.2)

=f(σ(g1(un1)),,σ(gn(unn)))=f(\sigma({{g}_{1}}({{u}^{{{n}_{1}}}})),\cdots,\sigma({{g}_{n}}({{u}^{{{n}_{n}}}}))) (by the definition of σ\sigma in (ii))

=f(σ(a1),,σ(an))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}))

is well-defined;
as desired for σ\sigma to be a TT-morphism (by Definition 9.3.1). ∎

Corollary 7.1.8 is generalized to the following, which is a straightforward result of Proposition 14.1.3.

Corollary 14.1.4.

Let u,vDu,v\in D.

  1. (a)

    The following statements are equivalent:

    1. (i)

      uTvu\overset{T}{\longleftrightarrow}v.

    2. (ii)

      There exists a bijective map σ:uTvT\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}} given by f(un)f(vn)f({{u}^{n}})\mapsto f({{v}^{n}}), n+\forall n\in{{\mathbb{Z}}^{+}} and fTf\in T of nn variables such that f(un)f({{u}^{n}}) is well-defined.

  2. (b)

    The bijective map σ\sigma given in (ii) is a TT-isomorphism.

14.2. A construction of TT-morphisms from UT(UD){{\left\langle U\right\rangle}_{T}}(U\subseteq D)

For partial functions of multiple variables, we generalize Notation 7.2.1 as follows.

Notation 14.2.1.

Let DD be a set, let α:U(D)V(D)\alpha:U(\subseteq D)\to V(\subseteq D) be a map and let MM be a subset of n+{\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{}all partial functions from Dn{{D}^{n}} to DD}.

By UM,αVU\xrightarrow{M,\alpha}V we mean that n,m+\forall n,m\in{{\mathbb{Z}}^{+}}, f,gMf,g\in M, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}}, and (w1,,wm)Um({{w}_{1}},\cdots,{{w}_{m}})\in{{U}^{m}},
f(u1,,un)=g(w1,,wm)f(α(u1),,α(un))=g(α(w1),,α(wm))f({{u}_{1}},\cdots,{{u}_{n}})=g({{w}_{1}},\cdots,{{w}_{m}})\Rightarrow f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}}))=g(\alpha({{w}_{1}}),\cdots,\alpha({{w}_{m}})).

Moreover, by UM,αVU\overset{M,\alpha}{\longleftrightarrow}V we mean that α\alpha is bijective, UM,αVU\xrightarrow{M,\alpha}V and VM,α1UV\xrightarrow{M,{{\alpha}^{-1}}}U; or equivalently, by UM,αVU\overset{M,\alpha}{\longleftrightarrow}V we mean that α\alpha is bijective and n,m+\forall n,m\in{{\mathbb{Z}}^{+}}, f,gMf,g\in M, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}}, and (w1,,wm)Um({{w}_{1}},\cdots,{{w}_{m}})\in{{U}^{m}},
f(u1,,un)=g(w1,,wm)f(α(u1),,α(un))=g(α(w1),,α(wm))f({{u}_{1}},\cdots,{{u}_{n}})=g(w_{1},\cdots,w_{m})\Leftrightarrow f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}}))=g(\alpha({{w}_{1}}),\cdots,\alpha({{w}_{m}})).

Remark.

Suppose UM,αVU\xrightarrow{M,\alpha}V. Then fM\forall f\in M of nn variables and (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}},
f(u1,,un)=f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}})=f({{u}_{1}},\cdots,{{u}_{n}}) \Rightarrow f(α(u1),,α(un))=f(α(u1),,α(un))f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}}))=f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})).
Hence by Convention 6.1.7, f(α(u1),,α(un))f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})) is well-defined if f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined.

Then Proposition 7.2.2 still holds but its proof changes as follows.

Proposition 14.2.2.

(Proposition 7.2.2) Let σ\sigma be a TT-morphism from a TT-space S1S_{1} to a TT-space S2S_{2}. Then AS1\forall A\subseteq{{S}_{1}}, AT,αS2A\xrightarrow{T,\alpha}{S}_{2}, where α:=σ|A\alpha:=\sigma{{|}_{A}}. In particular, S1T,σS2{{S}_{1}}\xrightarrow{T,\sigma}{S}_{2}.

Proof.

Let n,m+n,m\in{{\mathbb{Z}}^{+}}, let f,gTf,g\in T, let (a1,,an)An({{a}_{1}},\cdots,{{a}_{n}})\in{{A}^{n}}, and let (b1,,bm)Am({{b}_{1}},\cdots,{{b}_{m}})\in{{A}^{m}}. If f(a1,,an)=g(b1,,bm)f({{a}_{1}},\cdots,{{a}_{n}})=g({{b}_{1}},\cdots,{{b}_{m}}), by Definition 9.3.1, both f(σ(a1),,σ(an))f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) and g(σ(b1),,σ(bm))g(\sigma({{b}_{1}}),\cdots,\sigma({{b}_{m}})) are well-defined and

f(α(a1),,α(an))f(\alpha({{a}_{1}}),\cdots,\alpha({{a}_{n}}))

=f(σ(a1),,σ(an))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})) (since α=σ|A\alpha=\sigma{{|}_{A}})

=σ(f(a1,,an))=\sigma(f({{a}_{1}},\cdots,{{a}_{n}})) (by Definition 9.3.1)

=σ(g(b1,,bm))=\sigma(g({{b}_{1}},\cdots,{{b}_{m}})) (because f(a1,,an)=g(b1,,bm)f({{a}_{1}},\cdots,{{a}_{n}})=g({{b}_{1}},\cdots,{{b}_{m}}))

=g(σ(b1),,σ(bm))=g(\sigma({{b}_{1}}),\cdots,\sigma({{b}_{m}})) (by Definition 9.3.1)

=g(α(b1),,α(bm))=g(\alpha({{b}_{1}}),\cdots,\alpha({{b}_{m}})) (because α=σ|A\alpha=\sigma{{|}_{A}}).

Hence by Notation 14.2.1, AT,αS2A\xrightarrow{T,\alpha}S_{2}. ∎

Proposition 7.2.4 is generalized to the following, which also generalizes Proposition 14.1.3.

Proposition 14.2.3.

Let α:U(D)V(D)\alpha:U(\subseteq D)\to V(\subseteq D) be a map.

  1. (a)

    The following statements are equivalent:

    1. (i)

      UT,αVU\xrightarrow{T,\alpha}V.

    2. (ii)

      There exists a map σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} given by f(u1,,un)f(α(u1),,α(un))f({{u}_{1}},\cdots,{{u}_{n}})\mapsto f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})), n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fTf\in T such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined.

  2. (b)

    The map σ\sigma given in (ii) is a TT-morphism.

Proof.

For (a):

UT,αVU\xrightarrow{T,\alpha}V;
n,m+\Leftrightarrow\forall n,m\in{{\mathbb{Z}}^{+}}, fTf\in T of nn variables, gTg\in T of mm variables, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}}, and (w1,,wm)Um({{w}_{1}},\cdots,{{w}_{m}})\in{{U}^{m}},

f(u1,,un)=g(w1,,wm)f({{u}_{1}},\cdots,{{u}_{n}})=g({{w}_{1}},\cdots,{{w}_{m}})

implies

f(α(u1),,α(un))=g(α(w1),,α(wm));f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}}))=g(\alpha({{w}_{1}}),\cdots,\alpha({{w}_{m}}));

σ:UTVT\Leftrightarrow\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} given by

f(u1,,un)f(α(u1),,α(un)),f({{u}_{1}},\cdots,{{u}_{n}})\mapsto f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})),

n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fTf\in T such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined, is a well-defined map.

For (b):

Let (a1,,an)UTn(a_{1},\cdots,a_{n})\in\left\langle U\right\rangle_{T}^{n}. Then i=1,,n\forall i=1,\cdots,n, giT\exists{{g}_{i}}\in T of ni{{n}_{i}} variables and ziUni{{\text{z}}_{i}}\in{{U}^{{{n}_{i}}}} such that gi(zi)=ai{{g}_{i}}({{\text{z}}_{i}})={{a}_{i}}. For zi:=(u1,,uni){{\text{z}}_{i}}:=({{u}_{1}},\cdots,{{u}_{{{n}_{i}}}}), we denote (α(u1),,α(uni))(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{{{n}_{i}}}})) by α(zi)\alpha({{\text{z}}_{i}}) for brevity.

Let fTf\in T have nn variables. By Definition 9.1.2, f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) is a partial function from Dm{{D}^{m}} to DD given by (v1,,vn)f(g1(v1),,gn(vn))({{\text{v}}_{1}},\cdots,{{\text{v}}_{n}})\mapsto f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})), where (v1,,vn)=(x1,,xm)Dm({{\text{v}}_{1}},\cdots,{{\text{v}}_{n}})=({{x}_{1}},\cdots,{{x}_{m}})\in{{D}^{m}} and m=inim=\sum\nolimits_{i}{{{n}_{i}}}. By Definition 9.1.4, f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) is a restriction of some element of TT. Hence

f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined;
σ(f(a1,,an))\Leftrightarrow\sigma(f(a_{1},\cdots,a_{n}))

=σ(f(g1(z1),,gn(zn)))=\sigma(f({{g}_{1}}({{\text{z}}_{1}}),\cdots,{{g}_{n}}({{\text{z}}_{n}})))

=σ((f(g1,,gn))(z1,,zn))=\sigma((f\circ({{g}_{1}},\cdots,{{g}_{n}}))({{\text{z}}_{1}},\cdots,{{\text{z}}_{n}})) (by Definition 9.1.2)

=(f(g1,,gn))(α(z1),,α(zn))=(f\circ({{g}_{1}},\cdots,{{g}_{n}}))(\alpha({{\text{z}}_{1}}),\cdots,\alpha({{\text{z}}_{n}})) (by the definition of σ\sigma in (ii))

=f(g1(α(z1)),,gn(α(zn)))=f({{g}_{1}}(\alpha({{\text{z}}_{1}})),\cdots,{{g}_{n}}(\alpha({{\text{z}}_{n}}))) (by Definition 9.1.2)

=f(σ(g1(z1)),,σ(gn(zn)))=f(\sigma({{g}_{1}}({{\text{z}}_{1}})),\cdots,\sigma({{g}_{n}}({{\text{z}}_{n}}))) (by the definition of σ\sigma in (ii))

=f(σ(a1),,σ(an))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}))

is well-defined;
as desired for σ\sigma to be a TT-morphism (by Definition 9.3.1). ∎

Then Proposition 7.2.5 is generalized to the following.

Proposition 14.2.4.

Let α:U(D)V(D)\alpha:U(\subseteq D)\to V(\subseteq D) be a map.

  1. (a)

    The following statements are equivalent:

    1. (i)

      UT,αVU\overset{T,\alpha}{\longleftrightarrow}V.

    2. (ii)

      α\alpha is bijective and there exists a bijective map σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} given by f(u1,,un)f(α(u1),,α(un))f({{u}_{1}},\cdots,{{u}_{n}})\mapsto f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})), n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fTf\in T such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined.

  2. (b)

    The map σ\sigma given in (ii) is a TT-isomorphism.

Proof.

For (a):

UT,αVU\overset{T,\alpha}{\longleftrightarrow}V;

\Leftrightarrow α\alpha is bijective and n,m+\forall n,m\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}}, (w1,,wm)Um({{w}_{1}},\cdots,{{w}_{m}})\in{{U}^{m}} and f,gTf,g\in T
f(u1,,un)=g(w1,,wm)f(α(u1),,α(un))=g(α(w1),,α(wm))f({{u}_{1}},\cdots,{{u}_{n}})=g(w_{1},\cdots,w_{m})\Leftrightarrow f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}}))=g(\alpha({{w}_{1}}),\cdots,\alpha({{w}_{m}}));

\Leftrightarrow α\alpha is bijective and σ:UTVT\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}} given by

f(u1,,un)f(α(u1),,α(un)),f({{u}_{1}},\cdots,{{u}_{n}})\mapsto f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})),

n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fTf\in T such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined, is a well-defined bijective map.

For (b):

By (b) in Proposition 14.2.3, σ\sigma is a TT-morphism. Since σ\sigma is bijective, by Definition 2.4.1, it is a TT-isomorphism. ∎

Definition 7.2.7 is generalized to

Definition 14.2.5.

Let TT be a par-operator gen-semigroup on DD, let UDU\subseteq D, and let σ\sigma be a TT-morphism from UT{{\left\langle U\right\rangle}_{T}}. If there exists a map α:UD\alpha:U\to D such that σ(f(u1,,un))=f(α(u1),,α(un))\sigma(f({{u}_{1}},\cdots,{{u}_{n}}))=f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})), n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fTf\in T such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined, then we say that σ\sigma is constructible by α\alpha, or just say that σ\sigma is constructible for brevity.

Corollary 7.2.8 still holds, but its proof changes a little as follows.

Corollary 14.2.6.

(Corollary 7.2.8) Let UDU\subseteq D. If UUTU\subseteq{{\left\langle U\right\rangle}_{T}}, then any TT-morphism σ\sigma from UT{{\left\langle U\right\rangle}_{T}} is constructible by α:=σ|U\alpha:=\sigma{{|}_{U}}.

Proof.

Since UUTU\subseteq{{\left\langle U\right\rangle}_{T}}, by Definition 9.3.1,

σ(f(u1,,un))=f(σ(u1),,σ(un))=f(α(u1),,α(un)),\sigma(f({{u}_{1}},\cdots,{{u}_{n}}))=f(\sigma({{u}_{1}}),\cdots,\sigma({{u}_{n}}))=f(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})),

n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fTf\in T such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined. ∎

However, we could not generalize Theorem 7.2.9. Indeed, in Theorem 7.2.9, if TT were replaced by a par-operator gen-semigroup T=gT=\left\langle g\right\rangle (generated by Definition 9.1.5, where gg is a partial function from some Dn{{D}^{n}} to DD) and if we defined α\alpha in a way as in the proof of Theorem 7.2.9, then because any fTf\in T may have more than one variable, we could not show that α\alpha is a well-defined map.

14.3. A construction of θ\theta-morphisms

In this subsection, unless otherwise specified, θ\theta-morphisms are defined by Definition 9.5.5, and T1{{T}_{1}} and T2{{T}_{2}} are par-operator gen-semigroups.

Notation 7.3.1 is generalized to the following, which also generalizes Notation 14.2.1.

Notation 14.3.1.

Let D1D_{1} (resp. D2D_{2}) be a set, let M1M_{1} (resp. M2M_{2}) be a subset of n+{\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{}all partial functions from D1nD_{1}^{n} to D1D_{1}} (resp. a subset of n+{\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{}all partial functions from D2nD_{2}^{n} to D2D_{2}}), let θM1×M2\theta\subseteq{{M}_{1}}\times{{M}_{2}}, and let α:U(D1)V(D2)\alpha:U(\subseteq{D_{1}})\to V(\subseteq{{D}_{2}}) be a map.

By Uθ,αVU\xrightarrow{\theta,\alpha}V we mean that f,gDomθ\forall f,g\in\operatorname{Dom}\theta, n,m+n,m\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}}, and (w1,,wm)Um({{w}_{1}},\cdots,{{w}_{m}})\in{{U}^{m}}, f(u1,,un)=g(w1,,wm)f({{u}_{1}},\cdots,{{u}_{n}})=g({{w}_{1}},\cdots,{{w}_{m}}) implies

θ(f)(α(u1),,α(un))=θ(g)(α(w1),,α(wm)).\theta(f)(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}}))=\theta(g)(\alpha({{w}_{1}}),\cdots,\alpha({{w}_{m}})).

Moreover, by Uθ,αVU\overset{\theta,\alpha}{\longleftrightarrow}V we mean that α\alpha is bijective, Uθ,αVU\xrightarrow{\theta,\alpha}V and Vθ1,α1UV\xrightarrow{{{\theta}^{-1}},{{\alpha}^{-1}}}U, where θ1:={(h,f)|(f,h)θ}{{\theta}^{-1}}:=\{(h,f)\,|\,(f,h)\in\theta\}.

Remark.

If M1=M2=:M{{M}_{1}}={{M}_{2}}=:M and θ\theta is the identity map on MM, then Uθ,αVU\xrightarrow{\theta,\alpha}V is equivalent to UM,αVU\xrightarrow{M,\alpha}V (defined by Notation 14.2.1).

Proposition 7.3.2 would still hold if T1T_{1} and T2T_{2} in it were generalized to be par-operator gen-semigroups, but the proof would change as follows.

Proposition 14.3.2.

Let ϕ\phi be a θ\theta-morphism from a T1T_{1}-space S1S_{1} to a T2T_{2}-space S2S_{2}. Then AS1\forall A\subseteq{{S}_{1}}, Aθ,αS2A\xrightarrow{\theta,\alpha}S_{2}, where α:=ϕ|A\alpha:=\phi{{|}_{A}}. In particular, S1θ,ϕS2{{S}_{1}}\xrightarrow{\theta,\phi}S_{2}.

Proof.

By Definition 9.5.5, f,gDomθ\forall f,g\in\operatorname{Dom}\theta, (a1,,an)An(a_{1},\cdots,a_{n})\in{{A}^{n}} and (b1,,bm)Am({{b}_{1}},\cdots,{{b}_{m}})\in{{A}^{m}}, if f(a1,,an)=g(b1,,bm)f(a_{1},\cdots,a_{n})=g({{b}_{1}},\cdots,{{b}_{m}}), then

θ(f)(ϕ(a1),,ϕ(an))\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))

=ϕ(f(a1,,an))=\phi(f(a_{1},\cdots,a_{n})) (by Definition 9.5.5)

=ϕ(g(b1,,bm))=\phi(g({{b}_{1}},\cdots,{{b}_{m}})) (since f(a1,,an)=g(b1,,bm)f(a_{1},\cdots,a_{n})=g({{b}_{1}},\cdots,{{b}_{m}}))

=θ(g)(ϕ(b1),,ϕ(bm))=\theta(g)(\phi({{b}_{1}}),\cdots,\phi({{b}_{m}})) (by Definition 9.5.5),

and hence θ(f)(α(a1),,α(an))=θ(g)(α(b1),,α(bm))\theta(f)(\alpha({{a}_{1}}),\cdots,\alpha({{a}_{n}}))=\theta(g)(\alpha({{b}_{1}}),\cdots,\alpha({{b}_{m}})). By Notation 14.3.1, Aθ,αS2A\xrightarrow{\theta,\alpha}S_{2}. ∎

Definition 7.3.3 is generalized as follows.

Definition 14.3.3.

Let T1T_{1} and T2T_{2} be par-operator gen-semigroups on D1D_{1} and D2D_{2}, respectively, let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}}, and let AD2A\subseteq{{D}_{2}}. If f,g1,,gnDomθ\forall f,{{g}_{1}},\cdots,{{g}_{n}}\in\operatorname{Dom}\theta and zAm\text{z}\in{{A}^{m}} such that θ(f(g1,,gn))(z)\theta(f\circ({{g}_{1}},\cdots,{{g}_{n}}))(\text{z}) is well-defined,

θ(f(g1,,gn))(z)=(θ(f)(θ(g1),,θ(gn)))(z),\theta(f\circ({{g}_{1}},\cdots,{{g}_{n}}))(\text{z})=(\theta(f)\circ(\theta({{g}_{1}}),\cdots,\theta({{g}_{n}})))(\text{z}),

then we say that θ\theta is distributive over AA.

Propositions 7.3.4 and 14.2.3 are generalized as follows.

Proposition 14.3.4.

Let T1T_{1} and T2T_{2} be par-operator gen-semigroups on D1D_{1} and D2D_{2}, respectively, let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}} with Domθ=T1\operatorname{Dom}\theta={{T}_{1}}, and let α:U(D1)V(D2)\alpha:U(\subseteq{D_{1}})\to V(\subseteq{{D}_{2}}) be a map.

  1. (a)

    The following statements are equivalent:

    1. (i)

      Uθ,αVU\xrightarrow{\theta,\alpha}V.

    2. (ii)

      There exists a map ϕ:UT1VT2\phi:{{\left\langle U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}} given by

      f(u1,,un)θ(f)(α(u1),,α(un)),f({{u}_{1}},\cdots,{{u}_{n}})\mapsto\theta(f)(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})),

      n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fT1f\in{{T}_{1}} such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined.

  2. (b)

    Suppose that θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map and θ\theta is distributive over VV. Then the map ϕ\phi in (ii) is a θ\theta-morphism.

Proof.

For (a):

Uθ,αVU\xrightarrow{\theta,\alpha}V;
f,gDomθ(=T1)\Leftrightarrow\forall f,g\in\operatorname{Dom}\theta(={{T}_{1}}), (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and (w1,,wm)Um(w_{1},\cdots,w_{m})\in{{U}^{m}},

f(u1,,un)=g(w1,,wm)f({{u}_{1}},\cdots,{{u}_{n}})=g(w_{1},\cdots,w_{m})

implies

θ(f)(α(u1),,α(un))=θ(g)(α(v1),,α(vm));\theta(f)(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}}))=\theta(g)(\alpha({{v}_{1}}),\cdots,\alpha({{v}_{m}}));

ϕ:UT1VT2\Leftrightarrow\phi:{{\left\langle U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}} given by

f(u1,,un)θ(f)(α(u1),,α(un)),f({{u}_{1}},\cdots,{{u}_{n}})\mapsto\theta(f)(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})),

n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fT1f\in{{T}_{1}} such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined, is a well-defined map.

For (b):

Since condition (i) in Definition 9.5.5 is satisfied, it suffices show that condition (ii) in Definition 9.5.5 is satisfied.

Let fT1f\in{{T}_{1}} have nn variables and let (a1,,an)UT1n(a_{1},\cdots,a_{n})\in\left\langle U\right\rangle_{{{T}_{1}}}^{n}. Then i=1,,n\forall i=1,\cdots,n, giT1\exists{{g}_{i}}\in{{T}_{1}} of ni{{n}_{i}} variables and ziUni{{\text{z}}_{i}}\in{{U}^{{{n}_{i}}}} such that gi(zi)=ai{{g}_{i}}({{\text{z}}_{i}})={{a}_{i}}. For zi:=(u1,,uni){{\text{z}}_{i}}:=({{u}_{1}},\cdots,{{u}_{{{n}_{i}}}}), we denote (α(u1),,α(uni))(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{{{n}_{i}}}})) by α(zi)\alpha({{\text{z}}_{i}}) for brevity.

By Definition 9.1.2, f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) is a partial function from D1mD_{1}^{m} to D1{D_{1}} given by (v1,,vn)f(g1(v1),,gn(vn))({{\text{v}}_{1}},\cdots,{{\text{v}}_{n}})\mapsto f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})), where (v1,,vn)=(x1,,xm)D1m({{\text{v}}_{1}},\cdots,{{\text{v}}_{n}})=({{x}_{1}},\cdots,{{x}_{m}})\in D_{1}^{m} and m=inim=\sum\nolimits_{i}{{{n}_{i}}}. By Definition 9.1.4, f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) is a restriction of some element of T1{{T}_{1}}. Hence

f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined;
ϕ(f(a1,,an))\Leftrightarrow\phi(f(a_{1},\cdots,a_{n}))

=ϕ(f(g1(z1),,gn(zn)))=\phi(f({{g}_{1}}({{\text{z}}_{1}}),\cdots,{{g}_{n}}({{\text{z}}_{n}})))

=ϕ((f(g1,,gn))(z1,,zn))=\phi((f\circ({{g}_{1}},\cdots,{{g}_{n}}))({{\text{z}}_{1}},\cdots,{{\text{z}}_{n}})) (by Definition 9.1.2)

=θ(f(g1,,gn))(α(z1),,α(zn))=\theta(f\circ({{g}_{1}},\cdots,{{g}_{n}}))(\alpha({{\text{z}}_{1}}),\cdots,\alpha({{\text{z}}_{n}})) (by the definition of ϕ\phi in (ii))

=(θ(f)(θ(g1),,θ(gn)))(α(z1),,α(zn))=(\theta(f)\circ(\theta({{g}_{1}}),\cdots,\theta({{g}_{n}})))(\alpha({{\text{z}}_{1}}),\cdots,\alpha({{\text{z}}_{n}})) (since θ\theta is distributive over VV)

=θ(f)(θ(g1)(α(z1)),,θ(gn)(α(zn)))=\theta(f)(\theta({{g}_{1}})(\alpha({{\text{z}}_{1}})),\cdots,\theta({{g}_{n}})(\alpha({{\text{z}}_{n}}))) (by Definition 9.1.2)

=θ(f)(ϕ(g1(z1)),,ϕ(gn(zn)))=\theta(f)(\phi({{g}_{1}}({{\text{z}}_{1}})),\cdots,\phi({{g}_{n}}({{\text{z}}_{n}}))) (by the definition of ϕ\phi in (ii))

=θ(f)(ϕ(a1),,ϕ(an))=\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))

is well-defined;
as desired for ϕ\phi to be a θ\theta-morphism. ∎

The following generalizes both Propositions 7.3.5 and 14.2.4.

Proposition 14.3.5.

Let T1T_{1}, T2T_{2}, and α\alpha be defined as in Proposition 14.3.4. Let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}} such that Domθ=T1\operatorname{Dom}\theta={{T}_{1}} and Imθ=T2\operatorname{Im}\theta={{T}_{2}}.

  1. (a)

    The following statements are equivalent:

    1. (i)

      Uθ,αVU\overset{\theta,\alpha}{\longleftrightarrow}V.

    2. (ii)

      α\alpha is bijective, there exists a bijective map ϕ:UT1VT2\phi:{{\left\langle U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}} given by

      f(u1,,un)θ(f)(α(u1),,α(un)),f({{u}_{1}},\cdots,{{u}_{n}})\mapsto\theta(f)(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})),

      n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fT1f\in{{T}_{1}} such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined, and its inverse ϕ1:VT2UT1\phi^{-1}:{{\left\langle V\right\rangle}_{{{T}_{2}}}}\to{{\left\langle U\right\rangle}_{{{T}_{1}}}} can be given by

      g(v1,,vn)θ1(g)(α1(v1),,α1(vn)),g(v_{1},\cdots,v_{n})\mapsto\theta^{-1}(g)(\alpha^{-1}(v_{1}),\cdots,\alpha^{-1}(v_{n})),

      n+\forall n\in{{\mathbb{Z}}^{+}}, (v1,,vn)Vn(v_{1},\cdots,v_{n})\in V^{n} and gT2g\in{{T}_{2}} such that g(v1,,vn)g(v_{1},\cdots,v_{n}) is well-defined.

  2. (b)

    Suppose that θ\theta is distributive over VV and θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map. Then the bijective map ϕ\phi given in (ii) is a θ\theta-isomorphism.

Proof.

For (a):

Uθ,αVU\overset{\theta,\alpha}{\longleftrightarrow}V;

\Leftrightarrow α\alpha is bijective, Uθ,αVU\xrightarrow{\theta,\alpha}V and Vθ1,α1UV\xrightarrow{{{\theta}^{-1}},{{\alpha}^{-1}}}U (by Notation 14.3.1);

\Leftrightarrow α\alpha is bijective, ϕ:UT1VT2\phi:{{\left\langle U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}} given by

f(u1,,un)θ(f)(α(u1),,α(un)),f({{u}_{1}},\cdots,{{u}_{n}})\mapsto\theta(f)(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})),

n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fT1f\in{{T}_{1}} such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined, is a well-defined map, and ϕ:VT2UT1\phi^{\prime}:{{\left\langle V\right\rangle}_{{{T}_{2}}}}\to{{\left\langle U\right\rangle}_{{{T}_{1}}}} given by

g(v1,,vn)θ1(g)(α1(v1),,α1(vn)),g(v_{1},\cdots,v_{n})\mapsto\theta^{-1}(g)(\alpha^{-1}(v_{1}),\cdots,\alpha^{-1}(v_{n})),

n+\forall n\in{{\mathbb{Z}}^{+}}, (v1,,vn)Vn(v_{1},\cdots,v_{n})\in V^{n} and gT2g\in{{T}_{2}} such that g(v1,,vn)g(v_{1},\cdots,v_{n}) is well-defined, is also a well-defined map (by Proposition 14.3.4);

\Leftrightarrow α\alpha is bijective, there is a bijective map ϕ:UT1VT2\phi:{{\left\langle U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}} given by

f(u1,,un)θ(f)(α(u1),,α(un)),f({{u}_{1}},\cdots,{{u}_{n}})\mapsto\theta(f)(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})),

n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fT1f\in{{T}_{1}} such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined, and its inverse ϕ1:VT2UT1\phi^{-1}:{{\left\langle V\right\rangle}_{{{T}_{2}}}}\to{{\left\langle U\right\rangle}_{{{T}_{1}}}} can be given by

g(v1,,vn)θ1(g)(α1(v1),,α1(vn)),g(v_{1},\cdots,v_{n})\mapsto\theta^{-1}(g)(\alpha^{-1}(v_{1}),\cdots,\alpha^{-1}(v_{n})),

n+\forall n\in{{\mathbb{Z}}^{+}}, (v1,,vn)Vn(v_{1},\cdots,v_{n})\in V^{n} and gT2g\in{{T}_{2}} such that g(v1,,vn)g(v_{1},\cdots,v_{n}) is well-defined.

The third equivalence relation is explained as follows. The sufficiency (\Leftarrow) is obvious, so we only show the necessity (\Rightarrow). For this purpose, we show that both ϕϕ\phi^{\prime}\circ\phi and ϕϕ\phi\circ\phi^{\prime} are the identity map.

Let (f,g)θ(f,g)\in\theta and (u1,,un)Un(u_{1},\cdots,u_{n})\in U^{n} such that f(u1,,un)f(u_{1},\cdots,u_{n}) is well-defined. Then

ϕ(ϕ(f(u1,,un)))\phi^{\prime}(\phi(f(u_{1},\cdots,u_{n})))

=ϕ(θ(f)(α(u1),,α(un)))=\phi^{\prime}(\theta(f)(\alpha(u_{1}),\cdots,\alpha(u_{n}))) (by the definition of ϕ\phi)

=ϕ(g(α(u1),,α(un)))=\phi^{\prime}(g(\alpha(u_{1}),\cdots,\alpha(u_{n}))) (by Definition 9.5.3)

=θ1(g)(u1,,un)=\theta^{-1}(g)(u_{1},\cdots,u_{n}) (by the definition of ϕ\phi^{\prime})

=f(u1,,un)=f(u_{1},\cdots,u_{n}) (by (the θ1\theta^{-1} version of) Definition 9.5.3)

Then ϕϕ\phi^{\prime}\circ\phi is the identity map on UT1{{\left\langle U\right\rangle}_{{{T}_{1}}}} (because Domθ=T1\operatorname{Dom}\theta={{T}_{1}}).

Analogously, we can show that ϕϕ\phi\circ\phi^{\prime} is the identity map on VT2{{\left\langle V\right\rangle}_{{{T}_{2}}}}. Therefore, both ϕ\phi and ϕ\phi^{\prime} are bijective and ϕ\phi^{\prime} is the inverse of ϕ\phi.

For (b):

By (b) in Proposition 14.3.4, ϕ\phi is a θ\theta-morphism. Since ϕ\phi is bijective, ϕ\phi is a θ\theta-isomorphism. ∎

Definition 7.3.7 is generalized as follows, which also generalizes Definition 14.2.5.

Definition 14.3.6.

Let T1T_{1} and T2T_{2} be par-operator gen-semigroups on D1D_{1} and D2D_{2}, respectively, let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}}, let UD1U\subseteq D_{1} and let ϕ\phi be a θ\theta-morphism from UT1{\left\langle U\right\rangle}_{{{T}_{1}}} to a T2T_{2}-space. If Domθ=T1\operatorname{Dom}\theta={{T}_{1}} and there exists a map α:UD2\alpha:U\to{{D}_{2}} such that ϕ(f(u1,,un))=θ(f)(α(u1),,α(un))\phi(f({{u}_{1}},\cdots,{{u}_{n}}))=\theta(f)(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})), n+\forall n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fT1f\in{{T}_{1}} such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined, then we say that ϕ\phi is constructible by α\alpha, or just say that ϕ\phi is constructible for brevity.

Corollary 7.3.8 would still hold if T1T_{1} and T2T_{2} in it were generalized to be par-operator gen-semigroups, as shown below. The following also generalizes Corollary 14.2.6.

Corollary 14.3.7.

Let T1T_{1} and T2T_{2} be par-operator gen-semigroups on D1D_{1} and D2D_{2}, respectively, let θT1×T2\theta\subseteq{{T}_{1}}\times{{T}_{2}} and let UD1U\subseteq{D_{1}}. If Domθ=T1\operatorname{Dom}\theta={{T}_{1}} and UUT1U\subseteq{{\left\langle U\right\rangle}_{{{T}_{1}}}}, then any θ\theta-morphism ϕ\phi from UT1{{\left\langle U\right\rangle}_{{{T}_{1}}}} to a T2T_{2}-space is constructible by α:=ϕ|U\alpha:=\phi{{|}_{U}}.

Proof.

Since Domθ=T1\operatorname{Dom}\theta={{T}_{1}} and UUT1U\subseteq{{\left\langle U\right\rangle}_{{{T}_{1}}}}, by Definition 9.5.5, for all n+n\in{{\mathbb{Z}}^{+}}, (u1,,un)Un({{u}_{1}},\cdots,{{u}_{n}})\in{{U}^{n}} and fT1f\in{{T}_{1}} such that f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) is well-defined,

ϕ(f(u1,,un))=θ(f)(ϕ(u1),,ϕ(un))=θ(f)(α(u1),,α(un)).\phi(f({{u}_{1}},\cdots,{{u}_{n}}))=\theta(f)(\phi({{u}_{1}}),\cdots,\phi({{u}_{n}}))=\theta(f)(\alpha({{u}_{1}}),\cdots,\alpha({{u}_{n}})).

As we said at the end of Subsection 14.2, we could not generalize Theorem 7.2.9. For the same reason, we could not generalize Theorem 7.3.9, either.

For a par-operator gen-semigroup TT, we could not generalize results in Subsection 7.4 because now any fTf\in T may have more than one variable.

Part III. SOLVABILITY OF EQUATIONS

In Part III, we shall deviate from the topics of Parts I and II and study solvability of equations. A main goal of the classical Galois theory is to study the solvability by radicals of polynomial equations. In Section 15, we shall introduce a new understanding of solvability of polynomial equations, which involves a composition series of the Galois group of the polynomial. Analogously, for homogeneous linear differential equations, in Section 16, we shall introduce a solvability which involves a normal series of the differential Galois group of the differential equation. In Section 17, we shall generalize our results to “general” equations in terms of the theory developed in Parts I and II.

15. A solvability of polynomial equations

In this section, we shall show that for any separable polynomial p(x)p(x) over a field BB with a splitting field EBE\supsetneq B, there are m+m\in{\mathbb{Z}}^{+} polynomials p1(x),,pm(x)p_{1}(x),\cdots,p_{m}(x) such that the Galois group of each pi(x){{p}_{i}}(x) is simple and for any root of p(x)p(x) in EE, there is a formula which only involves rational functions over BB on the roots of p1(x),,pm(x)p_{1}(x),\cdots,p_{m}(x) in EE.

15.1. Canonical extensions and canonical solvability

Lemma 15.1.1.

Let E/BE/B be an algebraic field extension. Suppose B=EGal(E/B)B={{E}^{\operatorname{Gal}(E/B)}}, where EGal(E/B){{E}^{\operatorname{Gal}(E/B)}} denotes the fixed field of the Galois group Gal(E/B)\operatorname{Gal}(E/B), and G:=Gal(E/B)G:=\operatorname{Gal}(E/B) has a composition series G=G0G1Gm={1}G={{G}_{0}}\,\triangleright\,{{G}_{1}}\,\triangleright\,\cdots\,\triangleright\,{{G}_{m}}=\{1\}. Then there are intermediate fields B0,B1,,Bm{{B}_{0}},{{B}_{1}},\cdots,{{B}_{m}} such that B=B0B1Bm=EB={{B}_{0}}\subseteq{{B}_{1}}\subseteq\cdots\subseteq{{B}_{m}}=E, Bi/Bi1{{B}_{i}}/{{B}_{i-1}} is Galois, and Gal(Bi/Bi1)Gi1/Gi\operatorname{Gal}({{B}_{i}}/{{B}_{i-1}})\cong{{G}_{i-1}}/{{G}_{i}} is a simple group or the trivial group {1}\{1\}, i=1,,m\forall i=1,\cdots,m.

Proof.

i=0,,m\forall i=0,\cdots,m, let Bi=EGi{{B}_{i}}={{E}^{{{G}_{i}}}}. Then B=B0B1Bm=EB={{B}_{0}}\subseteq{{B}_{1}}\subseteq\cdots\subseteq{{B}_{m}}=E and i=0,,m\forall i=0,\cdots,m, Gal(E/Bi)=Gal(E/EGi)=Gi\operatorname{Gal}(E/{{B}_{i}})=\operatorname{Gal}(E/{{E}^{{{G}_{i}}}})={{G}_{i}}, and hence

G=Gal(E/B0)Gal(E/B1)Gal(E/Bm)={1}.G=\operatorname{Gal}(E/{{B}_{0}})\,\triangleright\,\operatorname{Gal}(E/{{B}_{1}})\,\triangleright\,\cdots\,\triangleright\,\operatorname{Gal}(E/{{B}_{m}})=\{1\}.

Since the normal series is a composition series, by the fundamental theorem of the classical Galois theory or infinite Galois theory, i=1,,m\forall i=1,\cdots,m, Bi/Bi1{{B}_{i}}/{{B}_{i-1}} is Galois and

Gal(Bi/Bi1)Gal(E/Bi1)/Gal(E/Bi)=Gi1/Gi\operatorname{Gal}({{B}_{i}}/{{B}_{i-1}})\cong\operatorname{Gal}(E/{{B}_{i-1}})/\operatorname{Gal}(E/{{B}_{i}})={{G}_{i-1}}/{{G}_{i}}

is a simple group or the trivial group. ∎

Lemma 15.1.1 implies that we may define a notion of solvability other than the solvability by radicals. In Subsection 15.2, it will be clear why we introduce the notions as follows.

Definition 15.1.2.

A field extension F/BF/B is called an irreducible Galois extension if F/BF/B is Galois and Gal(F/B)\operatorname{Gal}(F/B) is a simple group or the trivial group. A field extension E/BE/B is called a canonical ((field)) extension if there is a tower of fields B=B0B1Bm=EB={{B}_{0}}\subseteq{{B}_{1}}\subseteq\cdots\subseteq{{B}_{m}}=E, where m+m\in{{\mathbb{Z}}^{+}} and i=1,,m\forall i=1,\cdots,m, Bi/Bi1{{B}_{i}}/{{B}_{i-1}} is an irreducible Galois extension.

Definition 15.1.3.

A polynomial p(x)p(x) is said to be canonically solvable if there is a canonical field extension E/BE/B such that p(x)p(x) is over BB, p(x)p(x) is not over any proper subfield of BB, and EE is a splitting field of p(x)p(x).

Proposition 15.1.4.

Every separable polynomial is canonically solvable, and so is every polynomial over a field of characteristic 0.

Proof.

Let p(x)p(x) be a separable polynomial over a field BB with a splitting field EE such that p(x)p(x) is not over any proper subfield of BB. Then E/BE/B is Galois and B=EGal(E/B)B={{E}^{\operatorname{Gal}(E/B)}}. Since Gal(E/B)\operatorname{Gal}(E/B) is finite as the Galois group of a polynomial, it has a composition series. Then by Lemma 15.1.1 and Definition 15.1.3, p(x)p(x) is canonically solvable. ∎

15.2. Formulas for roots of separable polynomials

Now let’s explain why we introduced the notion of canonical solvability.

As is well-known, when we say that a polynomial over a field is solvable by radicals, we imply that there are formulas for the roots of the polynomial which may involve extraction of roots (an,n+\sqrt[n]{a},n\in{{\mathbb{Z}}^{+}}), rational functions, and no operation else. Thus actually we assume that solving equations in the form of Xn=a{X}^{n}=a (n+)(n\in{{\mathbb{Z}}^{+}}) is a basic operation. Analogous to this assumption, now we are assuming that solving a separable polynomial whose Galois group is simple is a basic operation. Said differently, we assume that the roots of any separable polynomial with a simple Galois group can be determined (by symbols, formulas, or values). Then theoretically, there exists a formula for any root of any separable polynomial, explained below by a process described in steps.

Step 15.2.1.

Given a separable polynomial p(x)p(x) over a field BB with a splitting field EE, by Proposition 15.1.4 and Lemma 15.1.1, there exist intermediate fields B0,B1,,Bm{{B}_{0}},{{B}_{1}},\cdots,{{B}_{m}} such that B=B0B1Bm=EB={{B}_{0}}\subseteq{{B}_{1}}\subseteq\cdots\subseteq{{B}_{m}}=E, Bi/Bi1{{B}_{i}}/{{B}_{i-1}} is Galois, and Gal(Bi/Bi1)Gi1/Gi\operatorname{Gal}({{B}_{i}}/{{B}_{i-1}})\simeq{{G}_{i-1}}/{{G}_{i}} is a simple group or the trivial group, i=1,,m(+)\forall i=1,\cdots,m(\in{{\mathbb{Z}}^{+}}). Obviously we only need to deal with the case where each Gal(Bi/Bi1)\operatorname{Gal}({{B}_{i}}/{{B}_{i-1}}) is a simple group, or equivalently, where each Bi/Bi1{{B}_{i}}/{{B}_{i-1}} is an irreducible Galois extension (Definition 15.1.2) and BiBi1{{B}_{i}}\neq{{B}_{i-1}}.

Step 15.2.2.

Since B1/B0{{B}_{1}}/{{B}_{0}} is an irreducible Galois extension and B1B0{{B}_{1}}\neq{{B}_{0}}, there is a separable polynomial p1(x)B0[x]{{p}_{1}}(x)\in{{B}_{0}}[x] such that B1{{B}_{1}} is the splitting field of p1(x){{p}_{1}}(x) in EE and the Galois group of p1(x){{p}_{1}}(x) is simple. Now we assume that every root of p1(x){{p}_{1}}(x) can be determined (by a symbol, a formula, or a value). Then B1{{B}_{1}}, which is a splitting field of p1(x){{p}_{1}}(x), can be generated over BB by the roots of p1(x){{p}_{1}}(x) in EE. Hence every element of B1{{B}_{1}} can be determined. Said precisely, bB1\forall b\in{{B}_{1}}, there is a formula for bb which only involves rational functions over BB on the roots of p1(x){{p}_{1}}(x) in EE.

Step 15.2.3.

Similarly, because B2/B1{{B}_{2}}/{{B}_{1}} is an irreducible Galois extension and B2B1{{B}_{2}}\neq{{B}_{1}}, there is a separable polynomial p2(x){{p}_{2}}(x) over B1{{B}_{1}} such that B2{{B}_{2}} is the splitting field of p2(x){{p}_{2}}(x) in EE and the Galois group of p2(x){{p}_{2}}(x) is simple. Again we assume that every root of p2(x){{p}_{2}}(x) in EE can be determined. Then B2{{B}_{2}}, which is a splitting field of p2(x){{p}_{2}}(x), can be generated over B1{{B}_{1}} by the roots of p2(x){{p}_{2}}(x) in EE. And as was shown in Step 15.2.2, B1{{B}_{1}} can be generated over BB by the roots of p1(x){{p}_{1}}(x) in EE. Hence by substitutions, B2{{B}_{2}} can be generated over BB by the roots of p1(x){{p}_{1}}(x) and p2(x){{p}_{2}}(x) in EE. Thus, bB2\forall b\in{{B}_{2}}, there is a formula for bb which only involves rational functions over BB on the roots of p1(x){{p}_{1}}(x) and p2(x){{p}_{2}}(x) in EE.

Step 15.2.4.

By induction, eventually for each element of Bm=E{{B}_{m}}=E, there is a formula which only involves rational functions over BB on the roots of p1(x),,pm(x)p_{1}(x),\cdots,p_{m}(x) in EE, where each pi(x){{p}_{i}}(x) is a separable polynomial over Bi1{{B}_{i-1}} whose splitting field in EE is Bi{{B}_{i}} and whose Galois group is simple. Since EE is a splitting field of p(x)p(x), for any root of p(x)p(x) in EE, there is a formula which only involves rational functions over BB on the roots of p1(x),,pm(x)p_{1}(x),\cdots,p_{m}(x) in EE.

In fact, in the above process, the operation of solving equation p(x)=0p(x)=0 is “reduced” to the operations of solving p1(x)=0,,pm(x)=0p_{1}(x)=0,\cdots,p_{m}(x)=0, where the Galois group of each pi(x){{p}_{i}}(x) is simple, which implies that the operation of solving pi(x)=0{{p}_{i}}(x)=0 is “irreducible”. This is why we use the terminology “canonical”.

The above process applies to any polynomial which is canonically solvable. Therefore, the following is obvious, which also explains why we introduced the notion of canonical solvability.

Corollary 15.2.5.

Let p(x)p(x) be a polynomial over a field BB with a splitting field EE such that EBE\supsetneq B. If p(x)p(x) is separable, or more generally, p(x)p(x) is canonically solvable, then there exist m(+)m(\in{{\mathbb{Z}}^{+}}) polynomials p1(x),,pm(x)p_{1}(x),\cdots,p_{m}(x) such that the Galois group of each pi(x){{p}_{i}}(x) is simple and for any root of p(x)p(x) in EE, there is a formula which only involves rational functions over BB on the roots of p1(x),,pm(x)p_{1}(x),\cdots,p_{m}(x) in EE.

16. A solvability of homogeneous linear differential equations

For homogeneous linear differential equations, we can do a similar thing as we just did for polynomial equations, although the process will be relatively complicated.

16.1. Normal Zariski-closed series and composition Zariski-closed series

In differential Galois theory, the Picard-Vessiot extension of a homogeneous linear differential equation is the analogue of a splitting field of a polynomial (see e.g. [1, 4, 9]). However, by the fundamental theorem of differential Galois theory, only Zariski-closed subgroups (not necessarily all subgroups) of a differential Galois group have a bijective correspondence with intermediate differential fields of a Picard-Vessiot extension. Hence we first give the following definition.

Definition 16.1.1.

Let G=G0G1Gm={1}(m+)G={{G}_{0}}\,\triangleright\,{{G}_{1}}\,\triangleright\,\cdots\,\triangleright\,{{G}_{m}}=\{1\}\,(m\in{{\mathbb{Z}}^{+}}) be a normal series where every Gi{{G}_{i}} is a Zariski-closed subgroup of a general linear group over a field. Then we call it a normal Zariski-closed series of GG. Moreover, if any possible proper refinement of this normal series is not a normal Zariski-closed one, then we call it a composition Zariski-closed series of GG.

Remark.

We do not require that a composition Zariski-closed series to be a composition series.

The following is comparable with Lemma 15.1.1.

Lemma 16.1.2.

Let a differential field extension BEB\subseteq E be a Picard-Vessiot extension such that the constant field of BB is algebraically closed. Suppose that the differential Galois group Gal(E/B)\operatorname{Gal}(E/B) has a normal Zariski-closed series

Gal(E/B)=:G0G1Gm={1}.\operatorname{Gal}(E/B)=:{G_{0}}\,\triangleright\,G_{1}\,\triangleright\,\cdots\,\triangleright\,{{G}_{m}}=\{1\}.

Then there are intermediate differential fields B0,B1,,Bm{{B}_{0}},{{B}_{1}},\cdots,{{B}_{m}} such that

B=B0B1Bm=E,B={{B}_{0}}\subseteq{{B}_{1}}\subseteq\cdots\subseteq{{B}_{m}}=E,

i=1,,m\forall i=1,\cdots,m, Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} is a Picard-Vessiot extension and Gal(Bi/Bi1)Gi1/Gi\operatorname{Gal}({{B}_{i}}/{{B}_{i-1}})\cong{{G}_{i-1}}/{{G}_{i}}, and i=0,,m\forall i=0,\cdots,m, Gal(E/Bi)=Gi\operatorname{Gal}(E/{{B}_{i}})={{G}_{i}}.

Proof.

Let CC be the constant field of BB, which is algebraically closed. Hence we can apply the fundamental theorem of differential Galois theory (see e.g. [1, 4, 9]) for any Picard-Vessiot extension of BB.

i=0,,m\forall i=0,\cdots,m, let Bi=EGi{{B}_{i}}={{E}^{{{G}_{i}}}}. Then

B=EGal(E/B)=EG0=B0B1Bm=EB={{E}^{\operatorname{Gal}(E/B)}}={{E}^{{{G}_{0}}}}={{B}_{0}}\subseteq{{B}_{1}}\subseteq\cdots\subseteq{{B}_{m}}=E

and, by the fundamental theorem, each Gal(E/Bi)=Gal(E/EGi)=Gi\operatorname{Gal}(E/{{B}_{i}})=\operatorname{Gal}(E/{{E}^{{{G}_{i}}}})={{G}_{i}} because each Gi{{G}_{i}} is a Zariski-closed subgroup of Gal(E/B)\operatorname{Gal}(E/B). Hence

Gal(E/B)=Gal(E/B0)Gal(E/B1)Gal(E/Bm)={1}.\operatorname{Gal}(E/B)=\operatorname{Gal}(E/{{B}_{0}})\,\triangleright\,\operatorname{Gal}(E/{{B}_{1}})\,\triangleright\,\cdots\,\triangleright\,\operatorname{Gal}(E/{{B}_{m}})=\{1\}.

Moreover, i=0,,m\forall i=0,\cdots,m, Bi=EGi{{B}_{i}}={{E}^{{{G}_{i}}}} is a differential field with BBiEB\subseteq{{B}_{i}}\subseteq E. And since the constant field CC of B=B0B={{B}_{0}} is algebraically closed, by the fundamental theorem, B0B1{{B}_{0}}\subseteq{{B}_{1}} is a Picard-Vessiot extension because Gal(E/B0)Gal(E/B1)\operatorname{Gal}(E/{{B}_{0}})\,\triangleright\,\operatorname{Gal}(E/{{B}_{1}}). Hence by the definition of Picard-Vessiot extensions (see e.g. [1]), CC is also the constant field of B1{{B}_{1}}. Again by the fundamental theorem, B1B2{{B}_{1}}\subseteq{{B}_{2}} is also a Picard-Vessiot extension because Gal(E/B1)Gal(E/B2)\operatorname{Gal}(E/{{B}_{1}})\,\triangleright\,\operatorname{Gal}(E/{{B}_{2}}). By induction, eventually each Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} is a Picard-Vessiot extension and each Gal(Bi/Bi1)Gal(E/Bi1)/Gal(E/Bi)=Gi1/Gi\operatorname{Gal}({{B}_{i}}/{{B}_{i-1}})\cong\operatorname{Gal}(E/{{B}_{i-1}})/\operatorname{Gal}(E/{{B}_{i}})={{G}_{i-1}}/{{G}_{i}} by the fundamental theorem. ∎

Because Gal(E/B)Gal(E/E)={1}\operatorname{Gal}(E/B)\,\triangleright\,\operatorname{Gal}(E/E)=\{1\}, Gal(E/B)\operatorname{Gal}(E/B) always has a normal Zariski-closed series. Thus we can employ Lemma 16.1.2 to any Picard-Vessiot extension whenever the constant field is algebraically closed.

16.2. Canonical extensions and canonical solvability

Let L(Y)=0L(Y)=0 be a homogeneous linear differential equation over BB for which BEB\subseteq E is a Picard-Vessiot extension with an algebraically closed field of constants. Then Lemma 16.1.2 implies that L(Y)=0L(Y)=0 can be “decomposed” into homogeneous linear differential equations L1(Y)=0,,Lm(Y)=0{{L}_{1}}(Y)=0,\cdots,{{L}_{m}}(Y)=0 such that i=1,,m\forall i=1,\cdots,m, Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} is a Picard-Vessiot extension for Li(Y)=0{{L}_{i}}(Y)=0, and hence we can solve L(Y)=0L(Y)=0 by solving L1(Y)=0,,Lm(Y)=0{{L}_{1}}(Y)=0,\cdots,{{L}_{m}}(Y)=0. We shall explain this process in Subsection 16.3. In light of this idea, we define a notion of solvability other than solvability by quadratures.

Definition 16.2.1.

A Picard-Vessiot extension BEB\subseteq E is said to be irreducible if for any KK with BKEB\subsetneq K\subsetneq E, KK is not a Picard-Vessiot extension of BB. Otherwise it is said to be reducible.

Remark.

Trivially, B=EB=E is an irreducible Picard-Vessiot extension.

Then by the fundamental theorem of differential Galois theory, the following, which characterizes reducible Picard-Vessiot extensions, is obvious.

Corollary 16.2.2.

Let BEB\subseteq E be a Picard-Vessiot extension with the constant field of BB being algebraically closed. Then BEB\subseteq E is reducible if and only if Gal(E/B)\operatorname{Gal}(E/B) has a proper nontrivial normal Zariski-closed subgroup.

The following two definitions are comparable with Definitions 15.1.2 and 15.1.3, respectively.

Definition 16.2.3.

A differential field extension BEB\subseteq E is called a canonical extension if there is a tower of differential fields B=B0B1Bm=EB={{B}_{0}}\subseteq{{B}_{1}}\subseteq\cdots\subseteq{{B}_{m}}=E, where m+m\in{{\mathbb{Z}}^{+}}, such that i=1,,m\forall i=1,\cdots,m, Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} is an irreducible Picard-Vessiot extension.

Definition 16.2.4.

Let L(Y)=0L(Y)=0 be a homogeneous linear differential equation. If there is a Picard-Vessiot extension BEB\subseteq E for L(Y)=0L(Y)=0 such that

  1. (i)

    the constant field of BB is algebraically closed;

  2. (ii)

    there does not exist any proper differential subfield B{B}^{\prime} of BB such that L(Y)=0L(Y)=0 is over B{B}^{\prime} and the constant field of B{B}^{\prime} is algebraically closed; and

  3. (iii)

    BEB\subseteq E is a canonical extension;

then L(Y)=0L(Y)=0 is said to be canonically solvable.

The following characterizes canonical solvability in terms of a composition Zariski-closed series of the Galois group.

Proposition 16.2.5.

Let L(Y)=0L(Y)=0 be a homogeneous linear differential equation. Then L(Y)=0L(Y)=0 is canonically solvable if and only if there is a Picard-Vessiot extension BEB\subseteq E for L(Y)=0L(Y)=0 such that conditions (i) and (ii) in Definition 16.2.4 are satisfied and Gal(E/B)\operatorname{Gal}(E/B) has a composition Zariski-closed series.

Proof.

Assume that there is a Picard-Vessiot extension BEB\subseteq E for L(Y)=0L(Y)=0 such that conditions (i) and (ii) in Definition 16.2.4 are satisfied. Then by Definition 16.2.4, it is sufficient to show that the following two statements are equivalent.

(a) BEB\subseteq E is a canonical extension.

(b) Gal(E/B)\operatorname{Gal}(E/B) has a composition Zariski-closed series.
1. (a) \Rightarrow (b):

Suppose that BEB\subseteq E is a canonical extension. By Definition 16.2.3, there is a tower of differential fields B=B0B1Bm=EB={{B}_{0}}\subseteq{{B}_{1}}\subseteq\cdots\subseteq{{B}_{m}}=E where each Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} is an irreducible Picard-Vessiot extension. Hence by the definition of Picard-Vessiot extensions, the constant field of every Bi{{B}_{i}} is the same, which we assume to be CC. Because condition (i) in Definition 16.2.4 is assumed to be satisfied, CC is algebraically closed, and hence we can apply the fundamental theorem of differential Galois theory to any Picard-Vessiot extension of any Bi{{B}_{i}} as follows.

(Gal(E/B)=)Gal(E/B0)Gal(E/B1)(\operatorname{Gal}(E/B)=)\operatorname{Gal}(E/{{B}_{0}})\,\triangleright\,\operatorname{Gal}(E/{{B}_{1}}) because both B0E{{B}_{0}}\subseteq E and B0B1{{B}_{0}}\subseteq{{B}_{1}} are Picard-Vessiot extensions. Analogously, since L(Y)=0L(Y)=0 can be viewed as defined over B1{{B}_{1}}, B1E{{B}_{1}}\subseteq E is a Picard-Vessiot extension, and hence Gal(E/B1)Gal(E/B2)\operatorname{Gal}(E/{{B}_{1}})\,\triangleright\,\operatorname{Gal}(E/{{B}_{2}}) because B1B2{{B}_{1}}\subseteq{{B}_{2}} is also Picard-Vessiot. Applying this argument for each Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}}, eventually we obtain

(16.1) Gal(E/B)=Gal(E/B0)Gal(E/B1)Gal(E/Bm)={1}.\operatorname{Gal}(E/B)=\operatorname{Gal}(E/{{B}_{0}})\,\triangleright\,\operatorname{Gal}(E/{{B}_{1}})\,\triangleright\,\cdots\,\triangleright\,\operatorname{Gal}(E/{{B}_{m}})=\{1\}.

Because CC is an algebraically closed field, each Gal(E/Bi)\operatorname{Gal}(E/{{B}_{i}}) is a Zariski-closed subgroup of the general linear group GL(n,C)\operatorname{GL}(n,C), where nn is the order of L(Y)L(Y). Thus by Definition 16.1.1, (16.1) is a normal Zariski-closed series of Gal(E/B)\operatorname{Gal}(E/B).

Suppose that (16.1) has a proper refinement which is still normal Zariski-closed. Then i{1,,m}\exists i\in\{1,\cdots,m\} and a Zariski-closed subgroup HH of GL(n,C)\operatorname{GL}(n,C) such that Gal(E/Bi1)HGal(E/Bi)\operatorname{Gal}(E/{{B}_{i-1}})\,\triangleright\,H\,\triangleright\,\operatorname{Gal}(E/{{B}_{i}}) with Gal(E/Bi1)HGal(E/Bi)\operatorname{Gal}(E/{{B}_{i-1}})\neq H\neq\operatorname{Gal}(E/{{B}_{i}}). Hence by the fundamental theorem of differential Galois theory, H=Gal(E/EH)H=\operatorname{Gal}(E/E^{H}), Bi1EHBi{{B}_{i-1}}\subsetneq{{E}^{H}}\subsetneq{{B}_{i}}, and EH{{E}^{H}} is a Picard-Vessiot extension of Bi1{{B}_{i-1}}. But Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} is an irreducible Picard-Vessiot extension, a contradiction.

Therefore, any possible proper refinement of (16.1) is no longer a normal Zariski-closed one. By Definition 16.1.1, (16.1) is a composition Zariski-closed series of Gal(E/B)\operatorname{Gal}(E/B).
2. (b) \Rightarrow (a):

Suppose that the differential Galois group Gal(E/B)\operatorname{Gal}(E/B) has a composition Zariski-closed series: Gal(E/B)=:G0G1Gm={1}\operatorname{Gal}(E/B)=:{{G}_{0}}\,\triangleright\,{{G}_{1}}\,\triangleright\,\cdots\,\triangleright\,{{G}_{m}}=\{1\}.

By Lemma 16.1.2, there are intermediate differential fields B0,B1,,Bm{{B}_{0}},{{B}_{1}},\cdots,{{B}_{m}} such that B=B0B1Bm=EB={{B}_{0}}\subseteq{{B}_{1}}\subseteq\cdots\subseteq{{B}_{m}}=E, where Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}}, i=1,,m\forall i=1,\cdots,m, is a Picard-Vessiot extension and i=0,,m\forall i=0,\cdots,m, Gal(E/Bi)=Gi\operatorname{Gal}(E/{{B}_{i}})={{G}_{i}}. Thus,

(16.2) Gal(E/B)=Gal(E/B0)Gal(E/B1)Gal(E/Bm)={1}\operatorname{Gal}(E/B)=\operatorname{Gal}(E/{{B}_{0}})\,\triangleright\,\operatorname{Gal}(E/{{B}_{1}})\,\triangleright\,\cdots\,\triangleright\,\operatorname{Gal}(E/{{B}_{m}})=\{1\}

is a composition Zariski-closed series.

Besides, by the definition of Picard-Vessiot extensions, all Bi{{B}_{i}} share the same constant field, which we assume to be CC. Because condition (i) in Definition 16.2.4 is assumed to be satisfied, CC is algebraically closed. Hence we can apply the fundamental theorem of differential Galois theory to any Picard-Vessiot extension of Bi{{B}_{i}}, i=0,,m\forall i=0,\cdots,m.

Suppose that there exists a reducible Picard-Vessiot extension Bi1Bi{{B}_{i-1}}\subsetneq{{B}_{i}}, where i{1,,m}i\in\{1,\cdots,m\}. Then by Definition 16.2.1, K\exists K with Bi1KBi{{B}_{i-1}}\subsetneq K\subsetneq{{B}_{i}} such that KK is a Picard-Vessiot extension of Bi1{{B}_{i-1}}. Hence CC is also the constant field of KK. By the fundamental theorem, Gal(E/Bi1)Gal(E/K)\operatorname{Gal}(E/{{B}_{i-1}})\,\triangleright\,\operatorname{Gal}(E/K), where Gal(E/K)\operatorname{Gal}(E/K) is also Zariski-closed. Moreover, since Bi1Bi{{B}_{i-1}}\subsetneq{{B}_{i}} is a Picard-Vessiot extension, so is KBiK\subsetneq{{B}_{i}}. Then again by the fundamental theorem, Gal(E/K)Gal(E/Bi)\operatorname{Gal}(E/K)\,\triangleright\,\operatorname{Gal}(E/{{B}_{i}}). Therefore, Gal(E/Bi1)Gal(E/K)Gal(E/Bi)\operatorname{Gal}(E/{{B}_{i-1}})\,\triangleright\,\operatorname{Gal}(E/K)\,\triangleright\,\operatorname{Gal}(E/{{B}_{i}}). And Gal(E/Bi1)Gal(E/K)Gal(E/Bi)\operatorname{Gal}(E/{{B}_{i-1}})\neq\operatorname{Gal}(E/K)\neq\operatorname{Gal}(E/{{B}_{i}}) because Bi1KBi{{B}_{i-1}}\neq K\neq{{B}_{i}}. But (16.2) is a composition Zariski-closed series, a contradiction.

Thus, every (Picard-Vessiot extension) Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} of B=B0B1Bm=EB={{B}_{0}}\subseteq{{B}_{1}}\subseteq\cdots\subseteq{{B}_{m}}=E is irreducible. By Definition 16.2.3, BEB\subseteq E is a canonical extension. ∎

16.3. Formulas for solutions of homogeneous linear differential equations

Now we show that in a sense we may develop formulas for the solutions of any homogeneous linear differential equation by the following process described in steps. To simplify our descriptions, we introduce

Terminology 16.3.1.

Let BB(U)B\subseteq B(U) be a differential field extension generated by UU. Then by differential field operations over BB on UU we mean rational functions over BB on UU and the derivation function of B(U)B(U).

Step 16.3.2.

Let L(Y)=0L(Y)=0 be a homogeneous linear differential equation for which BEB\subseteq E is a Picard-Vessiot extension such that conditions (i) and (ii) in Definition 16.2.4 are satisfied. Let CC be the algebraically closed field of constants of BEB\subseteq E.

Because Gal(E/B)Gal(E/E)={1}\operatorname{Gal}(E/B)\,\triangleright\,\operatorname{Gal}(E/E)=\{1\}, there must be a normal Zariski-closed series of Gal(E/B)\operatorname{Gal}(E/B) in the form Gal(E/B)=:G0G1Gm={1}\operatorname{Gal}(E/B)=:{{G}_{0}}\,\triangleright\,{{G}_{1}}\,\triangleright\,\cdots\,\triangleright\,{{G}_{m}}=\{1\}. By Lemma 16.1.2, there are intermediate differential fields B0,B1,,Bm{{B}_{0}},{{B}_{1}},\cdots,{{B}_{m}} such that B=B0B1Bm=EB={{B}_{0}}\subseteq{{B}_{1}}\subseteq\cdots\subseteq{{B}_{m}}=E with every Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} being a Picard-Vessiot extension. Thus there exist homogeneous linear differential equations L1(Y)=0,,Lm(Y)=0{{L}_{1}}(Y)=0,\cdots,{{L}_{m}}(Y)=0 such that i=1,,m\forall i=1,\cdots,m, Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} is a Picard-Vessiot extension for Li(Y)=0{{L}_{i}}(Y)=0.

Moreover, if Gal(E/B)\operatorname{Gal}(E/B) has a composition Zariski-closed series, then by Proposition 16.2.5, L(Y)=0L(Y)=0 is canonically solvable, and hence by Definitions 16.2.3 and 16.2.4, in such case we may assume that i=1,,m\forall i=1,\cdots,m, Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} is an irreducible Picard-Vessiot extension.

Step 16.3.3.

Now B0B1{{B}_{0}}\subseteq{{B}_{1}} is a Picard-Vessiot extension for L1(Y)=0{{L}_{1}}(Y)=0. Just as the solutions of Y=a{Y}^{\prime}=a are assumed to be adx\int{a}dx, now the solutions of L1(Y)=0{{L}_{1}}(Y)=0 in EE are assumed to be determined (by symbols, formulas, or values). Then because B1{{B}_{1}} can be generated over B0=B{{B}_{0}}=B by the solutions of L1(Y)=0{{L}_{1}}(Y)=0 in EE, every element of B1{{B}_{1}} can be determined. Said precisely, bB1\forall b\in{{B}_{1}}, there is a formula for bb which only involves differential field operations over BB on the solutions of L1(Y)=0{{L}_{1}}(Y)=0 in EE (cf. Terminology 16.3.1).

Step 16.3.4.

Analogously, because B1B2{{B}_{1}}\subseteq{{B}_{2}} is a Picard-Vessiot extension for L2(Y)=0{{L}_{2}}(Y)=0, B2{{B}_{2}} can be generated over B1{{B}_{1}} by the solutions of L2(Y)=0{{L}_{2}}(Y)=0 in EE. Hence bB2\forall b\in{{B}_{2}}, there is a formula for bb which only involves differential field operations over B1{{B}_{1}} on the solutions of L2(Y)=0{{L}_{2}}(Y)=0 in EE. As was shown in Step 16.3.3, every element of B1{{B}_{1}} can be obtained by a formula which only involves differential field operations over BB on the solutions of L1(Y)=0{{L}_{1}}(Y)=0 in EE. Hence by substitutions, every element of B2{{B}_{2}} can be determined by a formula which only involves differential field operations over BB on the solutions of L1(Y)=0{{L}_{1}}(Y)=0 and L2(Y)=0{{L}_{2}}(Y)=0 in EE.

Step 16.3.5.

By induction, eventually every element of Bm=E{{B}_{m}}=E can be determined by a formula which only involves differential field operations over BB on the solutions of equations L1(Y)=0,,Lm(Y)=0{{L}_{1}}(Y)=0,\cdots,{{L}_{m}}(Y)=0 in EE. Therefore, for each solution of L(Y)=0L(Y)=0 in EE, there is a formula which only involves differential field operations over BB on the solutions of equations L1(Y)=0,,Lm(Y)=0{{L}_{1}}(Y)=0,\cdots,{{L}_{m}}(Y)=0 in EE.

In fact, in the above process, the operation of solving equation L(Y)=0L(Y)=0 is “decomposed” into the operations of solving L1(Y)=0,,Lm(Y)=0{{L}_{1}}(Y)=0,\cdots,{{L}_{m}}(Y)=0. Moreover, Proposition 16.2.5 tells us that in the case where Gal(E/B)\operatorname{Gal}(E/B) has a composition Zariski-closed series, i=1,,m\forall i=1,\cdots,m, Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} is an irreducible Picard-Vessiot extension, which implies that any further “decomposition” of any Li(Y)=0{{L}_{i}}(Y)=0, i=1,,m\forall i=1,\cdots,m, would be trivial. This explains why in this case we say that L(Y)=0L(Y)=0 is canonically solvable.

The following summarizes this subsection.

Corollary 16.3.6.

Let L(Y)=0L(Y)=0 be a homogeneous linear differential equation for which BEB\subseteq E is a Picard-Vessiot extension such that the constant field is algebraically closed.

Suppose that Gal(E/B)=:G0G1Gm={1}\operatorname{Gal}(E/B)=:{{G}_{0}}\,\triangleright\,{{G}_{1}}\,\triangleright\,\cdots\,\triangleright\,{{G}_{m}}=\{1\}, where m+m\in{{\mathbb{Z}}^{+}}. Then there exist intermediate differential fields (B=)B0,B1,,Bm(=E)(B=){{B}_{0}},{{B}_{1}},\cdots,{{B}_{m}}(=E) such that Gal(Bi/Bi1)Gi1/Gi\operatorname{Gal}({{B}_{i}}/{{B}_{i-1}})\cong{{G}_{i-1}}/{{G}_{i}}, i=1,,m\forall i=1,\cdots,m, and homogeneous linear differential equations L1(Y)=0,,Lm(Y)=0{{L}_{1}}(Y)=0,\cdots,{{L}_{m}}(Y)=0 such that i=1,,m\forall i=1,\cdots,m, Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} is a Picard-Vessiot extension for Li(Y)=0{{L}_{i}}(Y)=0. And for any solution of L(Y)=0L(Y)=0 in E, there is a formula which only involves differential field operations over BB on the solutions of equations L1(Y)=0,,Lm(Y)=0{{L}_{1}}(Y)=0,\cdots,{{L}_{m}}(Y)=0 in EE ((cf. Terminology 16.3.1)).

Moreover, if Gal(E/B)G1Gm={1}\operatorname{Gal}(E/B)\,\triangleright\,{{G}_{1}}\,\triangleright\,\cdots\,\triangleright\,{{G}_{m}}=\{1\} is a composition Zariski-closed series ((cf. Definition 16.1.1)), then i=1,,m\forall i=1,\cdots,m, the Picard-Vessiot extension Bi1Bi{{B}_{i-1}}\subseteq{{B}_{i}} for Li(Y)=0{{L}_{i}}(Y)=0 is irreducible ((cf. Definition 16.2.1)).

17. A possible strategy for equation solving

Now for “general” equations (with unknowns), we shall generalize our results (in the two preceding sections) in terms of the theory developed in Parts I and II. Because this section is only a sketch of some ideas, it is not developed in a rigorous way.

17.1. Splitting TT-spaces, Galois TT-groups and Galois TT-monoids of equations

To simplify our descriptions, we introduce

Terminology 17.1.1.

Let TT be a par-operator gen-semigroup on a set DD and let BB be a set. If UD\forall U\subseteq D and aUTa\in{{\left\langle U\right\rangle}_{T}}, there is a formula for aa which only involves some given operations on elements of BB and UU, then we say that T is ((defined on D)D) over BB.

Remark.

Note that we do not require BDB\subseteq D.

For instance, the par-operator gen-semigroups defined in Examples 2.1.2, 8.1.3, 9.1.6 and 9.1.7 are all defined over BB.

Notation 17.1.2.

Let QQ be an equation (with unknowns) and let DD be a set. We denote by UQDU_{Q}^{D} the set of all solutions of QQ in DD.

Example 17.1.3.

Let F/BF/B be a field extension and let TT be the par-operator gen-semigroup on FF defined in Example 9.1.6. Let p(x)B[x]p(x)\in B[x] such that pp splits over FF and let UU be the set of roots of pp in FF. Then UQFT=UT{{\left\langle U_{Q}^{F}\right\rangle}_{T}}={{\left\langle U\right\rangle}_{T}} is a splitting field of pp over BB, where QQ denotes the equation p(x)=0p(x)=0.

Example 17.1.4.

Let F/BF/B be a differential field extension such that every constant of FF lies in BB. Let TT be the par-operator gen-semigroup on FF defined in Example 9.1.7. Let L(Y)=0L(Y)=0 be a homogeneous linear differential equation over BB with UU being a fundamental set of solutions of it in FF. Then BUQFT(=UT)B\subseteq{{\left\langle U_{Q}^{F}\right\rangle}_{T}}(={{\left\langle U\right\rangle}_{T}}) is a Picard-Vessiot extension (for LL), where QQ denotes the equation L(Y)=0L(Y)=0.

Inspired by the above two examples, we generalize the concepts of splitting fields and Picard-Vessiot extensions as follows.

Definition 17.1.5.

Let QQ be an equation (with unknowns) and let TT be a par-operator gen-semigroup on a set DD. With UQDU_{Q}^{D} defined by Notation 17.1.2, we call UQDT{{\left\langle U_{Q}^{D}\right\rangle}_{T}} the splitting TT-space of QQ. Moreover, if TT is defined over BB (cf. Terminology 17.1.1), then we call UQDT{{\left\langle U_{Q}^{D}\right\rangle}_{T}} the splitting TT-space of QQ over BB.

Remark.
  1. (1)

    If QQ has no solution in DD, then of course UQDT=UQD={{\left\langle U_{Q}^{D}\right\rangle}_{T}}=U_{Q}^{D}=\emptyset.

  2. (2)

    We use the terminology “splitting” to be consistent with the classical Galois theory although maybe nothing of QQ “split” like a polynomial.

  3. (3)

    It is possible that UQDUQDTU_{Q}^{D}\nsubseteq{{\left\langle U_{Q}^{D}\right\rangle}_{T}} and/or BUQDTB\nsubseteq{{\left\langle U_{Q}^{D}\right\rangle}_{T}}.

  4. (4)

    UQDUQDTU_{Q}^{D}\subseteq{{\left\langle U_{Q}^{D}\right\rangle}_{T}} if IdT\operatorname{Id}\in T.

Moreover, we generalize the notions of Galois groups of polynomials and differential Galois groups of homogeneous linear differential equations as follows.

Definition 17.1.6.

Let QQ be an equation (with unknowns) and let TT be a par-operator gen-semigroup on a set DD over a set BB (cf. Terminology 17.1.1). If BUQDTB\subseteq{{\left\langle U_{Q}^{D}\right\rangle}_{T}}, then the Galois TT-group of QQ and the Galois TT-monoid of QQ are GGrT(UQDT/B)\operatorname{GGr}_{T}({{\left\langle U_{Q}^{D}\right\rangle}_{T}}/B) and GMnT(UQDT/B)\operatorname{GMn}_{T}({{\left\langle U_{Q}^{D}\right\rangle}_{T}}/B) (cf. Definition 2.6.1), respectively.

In Example 17.1.3, we can tell from Proposition 9.3.4 that GGrT(UQFT/B)=GMnT(UQFT/B)\operatorname{GGr}_{T}({{\left\langle U_{Q}^{F}\right\rangle}_{T}}/B)=\operatorname{GMn}_{T}({{\left\langle U_{Q}^{F}\right\rangle}_{T}}/B) and they are the Galois group Gal(UT/B)\operatorname{Gal}({{\left\langle U\right\rangle}_{T}}/B) of p(x)p(x) over BB, where UU is the set of roots of pp in FF.

And in Example 17.1.4, it is not hard to tell from Proposition 9.3.6 that GGrT(UQFT/B)=GMnT(UQFT/B)\operatorname{GGr}_{T}({{\left\langle U_{Q}^{F}\right\rangle}_{T}}/B)=\operatorname{GMn}_{T}({{\left\langle U_{Q}^{F}\right\rangle}_{T}}/B) and they are the differential Galois group Gal(UT/B)\operatorname{Gal}({{\left\langle U\right\rangle}_{T}}/B) of L(Y)=0L(Y)=0 over BB, where UU is a fundamental set of solutions of L(Y)=0L(Y)=0 in FF.

17.2. Formulas for elements of splitting TT-spaces of “general” equations

Generalizing the strategies used in Subsections 15.2 and 16.3, the following shows that we can determine the splitting TT-space of an equation provided that some conditions are satisfied.

Proposition 17.2.1.

Given an equation QQ, we are required to determine UQDT{{\left\langle U_{Q}^{D}\right\rangle}_{T}}, where TT is a par-operator gen-semigroup on a set DD over a set BB.

If there exist m(+)m(\in{{\mathbb{Z}}^{+}}) equations Q1,,QmQ_{1},\cdots,Q_{m}, where every UQiDU_{{{Q}_{i}}}^{D} is nonempty and is assumed to be determined (by a symbol, a formula, or a value), and mm par-operator gen-semigroups T1,,TmT_{1},\cdots,T_{m} on DD such that the following two conditions are satisfied,

  1. (i)

    i=1,,m\forall i=1,\cdots,m, Ti{{T}_{i}} is defined over Bi1{{B}_{i-1}}, where B0:=B{{B}_{0}}:=B and i=1,,(m1)\forall i=1,\cdots,(m-1), Bi:=UQiDTi{{B}_{i}}:={{\left\langle U_{{{Q}_{i}}}^{D}\right\rangle}_{{{T}_{i}}}} and all given operations on BiB_{i} are also given as operations on BB (cf. Terminology 17.1.1);

  2. (ii)

    UQDTUQmDTm{{\left\langle U_{Q}^{D}\right\rangle}_{T}}\subseteq{{\left\langle U_{{{Q}_{m}}}^{D}\right\rangle}_{{{T}_{m}}}};

then for each element of UQDT{{\left\langle U_{Q}^{D}\right\rangle}_{T}}, there is a formula which only involves the given operations on BB and UQ1D,,UQmDU_{Q_{1}}^{D},\cdots,U_{Q_{m}}^{D}.

Proof.

Since UQ1DT1=B1{{\left\langle U_{{{Q}_{1}}}^{D}\right\rangle}_{{{T}_{1}}}}={{B}_{1}} and T1{{T}_{1}} is defined over B0=B{{B}_{0}}=B (condition (i)), by Terminology 17.1.1, bB1\forall b\in{{B}_{1}}, there is a formula for bb which only involves the given operations on BB and UQ1DU_{{{Q}_{1}}}^{D}.

Analogously, because UQ2DT2=B2{{\left\langle U_{{{Q}_{2}}}^{D}\right\rangle}_{{{T}_{2}}}}={{B}_{2}} and T2{{T}_{2}} is defined over B1{{B}_{1}}, bB2\forall b\in{{B}_{2}}, there is a formula for bb which only involves the given operations on B1{{B}_{1}} and UQ2DU_{{{Q}_{2}}}^{D}. And as was shown above, for each element of B1{{B}_{1}}, there is a formula which only involves the given operations on BB and UQ1DU_{{{Q}_{1}}}^{D}. Hence by substitutions and the assumption that all given operations on B1B_{1} are also given operations on BB, any bB2b\in{{B}_{2}} can be determined by a formula which only involves the given operations on BB, UQ1DU_{{{Q}_{1}}}^{D}, and UQ2DU_{{{Q}_{2}}}^{D}.

By induction, eventually every element of UQmDTm{{\left\langle U_{{{Q}_{m}}}^{D}\right\rangle}_{{{T}_{m}}}} can be determined by a formula which only involves the given operations on BB and UQ1D,,UQmDU_{Q_{1}}^{D},\cdots,U_{Q_{m}}^{D}. Thus, aUQDTUQmDTm\forall a\in{{\left\langle U_{Q}^{D}\right\rangle}_{T}}\subseteq{{\left\langle U_{{{Q}_{m}}}^{D}\right\rangle}_{{{T}_{m}}}}(condition (ii)), there is a formula for aa which only involves the given operations on BB and UQ1D,,UQmDU_{Q_{1}}^{D},\cdots,U_{Q_{m}}^{D}. ∎

The strategy described by Proposition 17.2.1 is actually “decomposing” QQ into Q1,,Qm{{Q}_{1}},\cdots,{{Q}_{m}} each of which the solutions are assumed to be known. This is just the strategy used in Sections 15 and 16.

Note that in Sections 15 and 16, the “decomposition” of an equation QQ into Q1,,Qm{{Q}_{1}},\cdots,{{Q}_{m}} depends on a normal series of the Galois TT-group of QQ (cf. Lemmas 15.1.1 and 16.1.2). However, for any other equation QQ, we do not know whether a similar situation arises, i.e. whether there is a connection between the “decomposition” of QQ described in Proposition 17.2.1 and a normal series of the Galois TT-group of QQ (defined by Definition 17.1.6).

Moreover, we do not know whether there exists any case where the “decomposition” of QQ is related to the Galois TT-monoid of QQ (defined by Definition 17.1.6) rather than the Galois TT-group of QQ.

Of course, the possible strategy for equation solving described in this section is just a very rough sketch and much more work is needed.

Part IV. OTHER TOPICS AND FUTURE RESEARCH

In Part IV, we shall briefly introduce some more results which we think to be important or deserve deeper study in the future.

18. Dualities of operator semigroups

Let TT be an operator semigroup and let SS be a TT-space. Then fT\forall f\in T, σEndT(S)\sigma\in\operatorname{End}_{T}(S) and aSa\in S, σ(f(a))=f(σ(a))\sigma(f(a))=f(\sigma(a)). The symmetry in this equation implies that TT and EndT(S)\operatorname{End}_{T}(S) may in a way exchange their roles with each other. Indeed, there exists a duality between EndT(S)\operatorname{End}_{T}(S) and the maximum operator semigroup on DD which “accommodates” EndT(S)\operatorname{End}_{T}(S), as explained below.

In the following two results, for any set FF of functions from DD to DD and SDS\subseteq D, we may denote by F|SF{{|}_{S}} the set {(f|S)|fF}\{(f{{|}_{S}})\,|\,f\in F\}.

Theorem 18.0.1.

Let TT be an operator semigroup on a set DD, let SS be a TT-space, let T=EndT(S)T^{*}=\operatorname{End}_{T}(S), and let

TEnd(S)={f:DD|Im(f|S)S,σ(f(a))=f(σ(a)),aS{{T}_{\operatorname{End}(S)}}=\{f:D\to D\,|\,\operatorname{Im}(f{{|}_{S}})\subseteq S,\sigma(f(a))=f(\sigma(a)),\forall a\in S and σEndT(S)}\sigma\in\operatorname{End}_{T}(S)\}.

  1. (a)

    TT^{*} is an operator semigroup on SS, SS is a TT^{*}-space, and TEnd(S){{T}_{\operatorname{End}(S)}} is an operator semigroup on DD.

  2. (b)

    EndT(S)T|S\operatorname{End}_{T^{*}}(S)\supseteq T{{|}_{S}}, where the equation holds if and only if T|S=TEnd(S)|ST{{|}_{S}}={{T}_{\operatorname{End}(S)}}{{|}_{S}}.

Remark.
  1. (1)

    Obviously TTEnd(S)T\subseteq{{T}_{\operatorname{End}(S)}}, and intuitively, TEnd(S){{T}_{\operatorname{End}(S)}} can be interpreted as the maximum operator semigroup on DD which “accommodates” EndT(S)\operatorname{End}_{T}(S) (i.e. σEndT(S)\forall\sigma\in\operatorname{End}_{T}(S), σ\sigma would still be a TT-endomorphism of SS if TT were extended to TEnd(S){{T}_{\operatorname{End}(S)}}).

  2. (2)

    Since T=EndT(S)T^{*}=\operatorname{End}_{T}(S), if T|S=TEnd(S)|ST{{|}_{S}}={{T}_{\operatorname{End}(S)}}{{|}_{S}}, then (b) implies EndEndT(S)(S)=EndT(S)=T|S\operatorname{End}_{\operatorname{End}_{T}(S)}(S)=\operatorname{End}_{T^{*}}(S)=T|_{S}. That is, EndEndT(S)(S)=T|S\operatorname{End}_{\operatorname{End}_{T}(S)}(S)=T|_{S} when TT reaches its “maximum” TEnd(S){T}_{\operatorname{End}(S)}. This is why we said that there exists a duality between EndT(S)\operatorname{End}_{T}(S) and TEnd(S){T}_{\operatorname{End}(S)}.

Proof.

For (a):

It follows from Proposition 2.3.8 and Definition 2.1.1 that T=EndT(S)T^{*}=\operatorname{End}_{T}(S) is an operator semigroup on SS, and hence STS{{\left\langle S\right\rangle}_{T^{*}}}\subseteq S. Since Id on SS lies in EndT(S)\operatorname{End}_{T}(S), ST=S{{\left\langle S\right\rangle}_{T^{*}}}=S. Hence the TT-space SS is also a TT^{*}-space.

To show that TEnd(S){{T}_{\operatorname{End}(S)}} is an operator semigroup on DD, we only need to show that it is closed under composition. Let h,gTEnd(S)h,g\in{{T}_{\operatorname{End}(S)}}, σEndT(S)\sigma\in\operatorname{End}_{T}(S) and aSa\in S. Then from the definition of TEnd(S){{T}_{\operatorname{End}(S)}}, we can tell that Im((hg)|S)S\operatorname{Im}((h\circ g){{|}_{S}})\subseteq S and

σ((hg)(a))=σ(h(g(a)))=h(σ(g(a)))=h(g(σ(a)))=(hg)(σ(a)).\sigma((h\circ g)(a))=\sigma(h(g(a)))=h(\sigma(g(a)))=h(g(\sigma(a)))=(h\circ g)(\sigma(a)).

Hence hgTEnd(S)h\circ g\in{{T}_{\operatorname{End}(S)}}, and thus TEnd(S){{T}_{\operatorname{End}(S)}} is an operator semigroup on DD.

For (b):

By Definition 2.3.1, fT\forall f\in T and σEndT(S)=T\sigma\in\operatorname{End}_{T}(S)=T^{*}, ff commutes with σ\sigma on SS, i.e. aS,σ(f(a))=f(σ(a))\forall a\in S,\sigma(f(a))=f(\sigma(a)), and hence f|Sf{{|}_{S}} is a TT^{*}-endomorphism (cf. Definition 2.3.7) of the TT^{*}-space SS (since Im(f|S)S\operatorname{Im}(f{{|}_{S}})\subseteq S by Proposition 2.2.7). Therefore, T|SEndT(S)T{{|}_{S}}\subseteq\operatorname{End}_{T^{*}}(S).

We are going to prove that EndT(S)=T|S\operatorname{End}_{T^{*}}(S)=T{{|}_{S}} if and only if TEnd(S)|S=T|S{{T}_{\operatorname{End}(S)}}{{|}_{S}}=T{{|}_{S}}.

Suppose EndT(S)=T|S\operatorname{End}_{T^{*}}(S)=T{{|}_{S}}. T|STEnd(S)|ST{{|}_{S}}\subseteq{{T}_{\operatorname{End}(S)}}{{|}_{S}} because TTEnd(S)T\subseteq{{T}_{\operatorname{End}(S)}}. To show TEnd(S)|ST|S{{T}_{\operatorname{End}(S)}}{{|}_{S}}\subseteq T{{|}_{S}}, let fTEnd(S)f\in{{T}_{\operatorname{End}(S)}}. Then by the definition of TEnd(S){{T}_{\operatorname{End}(S)}}, Im(f|S)S\operatorname{Im}(f{{|}_{S}})\subseteq S and σ(f(a))=f(σ(a))\sigma(f(a))=f(\sigma(a)), aS\forall a\in S and σT=EndT(S)\sigma\in T^{*}=\operatorname{End}_{T}(S), and thus f|SEndT(S)=T|Sf{{|}_{S}}\in\operatorname{End}_{T^{*}}(S)=T{{|}_{S}}. Hence TEnd(S)|ST|S{{T}_{\operatorname{End}(S)}}{{|}_{S}}\subseteq T{{|}_{S}}. Thus, TEnd(S)|S=T|S{{T}_{\operatorname{End}(S)}}{{|}_{S}}=T{{|}_{S}}.

Conversely, suppose TEnd(S)|S=T|S{{T}_{\operatorname{End}(S)}}{{|}_{S}}=T{{|}_{S}}. We already showed T|SEndT(S)T{{|}_{S}}\subseteq\operatorname{End}_{T^{*}}(S). To show EndT(S)T|S\operatorname{End}_{T^{*}}(S)\subseteq T{{|}_{S}}, let f:DDf:D\to D such that f|SEndT(S)f{{|}_{S}}\in\operatorname{End}_{T^{*}}(S). Then Im(f|S)S\operatorname{Im}(f{{|}_{S}})\subseteq S and σ(f(a))=f(σ(a)),aS\sigma(f(a))=f(\sigma(a)),\forall a\in S and σT=EndT(S)\sigma\in T^{*}=\operatorname{End}_{T}(S). It follows that fTEnd(S)f\in{{T}_{\operatorname{End}(S)}}, and thus f|STEnd(S)|S=T|Sf{{|}_{S}}\in{{T}_{\operatorname{End}(S)}}{{|}_{S}}=T{{|}_{S}}. Hence EndT(S)T|S\operatorname{End}_{T^{*}}(S)\subseteq T{{|}_{S}}. Therefore, EndT(S)=T|S\operatorname{End}_{T^{*}}(S)=T{{|}_{S}}. ∎

Analogously, there exists a duality between AutT(S)\operatorname{Aut}_{T}(S) and the maximum operator semigroup which “accommodates” AutT(S)\operatorname{Aut}_{T}(S), though an additional condition is required.

Theorem 18.0.2.

Let TT be an operator semigroup on a set DD, let SS be a TT-space, let T#=AutT(S){{T}^{\#}}=\operatorname{Aut}_{T}(S), and let

TAut(S)={f:DD|Im(f|S)S,σ(f(a))=f(σ(a)){{T}_{\operatorname{Aut}(S)}}=\{f:D\to D\,|\,\operatorname{Im}(f{{|}_{S}})\subseteq S,\sigma(f(a))=f(\sigma(a)), aS\forall a\in S and σAutT(S)}\sigma\in\operatorname{Aut}_{T}(S)\}.

  1. (a)

    T#{{T}^{\#}} is an operator semigroup on SS, SS is a T#-{{T}^{\#}}\text{-}space and TAut(S){{T}_{\operatorname{Aut}(S)}} is an operator semigroup on DD.

  2. (b)

    If fT\forall f\in T, f|Sf{{|}_{S}} is bijective onto SS, then AutT#(S)T|S\operatorname{Aut}_{{{T}^{\#}}}(S)\supseteq T{{|}_{S}}

  3. (c)

    Suppose that fTAut(S)\forall f\in{{T}_{\operatorname{Aut}(S)}}, f|Sf{{|}_{S}} is bijective onto SS. Then AutT#(S)T|S\operatorname{Aut}_{{{T}^{\#}}}(S)\supseteq T{{|}_{S}} and the equality holds if and only if T|S=TAut(S)|ST{{|}_{S}}={{T}_{\operatorname{Aut}(S)}}{{|}_{S}}.

Remark.
  1. (1)

    Apparently TTAut(S)T\subseteq{{T}_{\operatorname{Aut}(S)}}, and intuitively, TAut(S){{T}_{\operatorname{Aut}(S)}} can be interpreted as the maximum operator semigroup on DD which “accommodates” AutT(S)\operatorname{Aut}_{T}(S) (i.e. σAutT(S)\forall\sigma\in\operatorname{Aut}_{T}(S), σ\sigma would still be a TT-automorphism of SS if TT were extended to TAut(S){{T}_{\operatorname{Aut}(S)}}).

  2. (2)

    Since T#=AutT(S){{T}^{\#}}=\operatorname{Aut}_{T}(S), if T|S=TAut(S)|ST{{|}_{S}}={{T}_{\operatorname{Aut}(S)}}{{|}_{S}} and fTAut(S)\forall f\in{{T}_{\operatorname{Aut}(S)}}, f|Sf{{|}_{S}} is bijective onto SS, then (c) implies AutAutT(S)(S)=AutT#(S)=T|S=TAut(S)|S\operatorname{Aut}_{\operatorname{Aut}_{T}(S)}(S)=\operatorname{Aut}_{{{T}^{\#}}}(S)=T|_{S}={{T}_{\operatorname{Aut}(S)}}{{|}_{S}}. This is why we said that there exists a duality between AutT(S)\operatorname{Aut}_{T}(S) and (T)TAut(S)(T\subseteq){{T}_{\operatorname{Aut}(S)}}.

Proof.

For (a):

It follows from Proposition 2.4.3 and Definition 2.1.1 that T#=AutT(S){{T}^{\#}}=\operatorname{Aut}_{T}(S) is an operator semigroup on SS, and hence ST#S{{\left\langle S\right\rangle}_{{{T}^{\#}}}}\subseteq S. Because Id on SS lies in AutT(S)\operatorname{Aut}_{T}(S), ST#=S{{\left\langle S\right\rangle}_{{{T}^{\#}}}}=S. Hence the TT-space SS is also a T#-{{T}^{\#}}\text{-}space.

To show that TAut(S){{T}_{\operatorname{Aut}(S)}} is an operator semigroup on DD, it suffices to show that it is closed under composition. Let h,gTAut(S)h,g\in{{T}_{\operatorname{Aut}(S)}}, σAutT(S)\sigma\in\operatorname{Aut}_{T}(S) and aSa\in S. Then from the definition of TAut(S){{T}_{\operatorname{Aut}(S)}}, we can tell that Im((hg)|S)S\operatorname{Im}((h\circ g){{|}_{S}})\subseteq S and

σ((hg)(a))=σ(h(g(a)))=h(σ(g(a)))=h(g(σ(a)))=(hg)(σ(a)).\sigma((h\circ g)(a))=\sigma(h(g(a)))=h(\sigma(g(a)))=h(g(\sigma(a)))=(h\circ g)(\sigma(a)).

Hence hgTAut(S)h\circ g\in{{T}_{\operatorname{Aut}(S)}}. Thus, TAut(S){{T}_{\operatorname{Aut}(S)}} is an operator semigroup on DD.

For (b):

Assume that fT\forall f\in T, f|Sf{{|}_{S}} is bijective onto SS. By Definition 2.3.1, fT\forall f\in T and σAutT(S)=T#\sigma\in\operatorname{Aut}_{T}(S)={{T}^{\#}}, ff commutes with σ\sigma on SS, i.e. aS\forall a\in S, σ(f(a))=f(σ(a))\sigma(f(a))=f(\sigma(a)), and hence f|Sf{{|}_{S}} is a T#-{{T}^{\#}}\text{-}automorphism of the T#-{{T}^{\#}}\text{-}space SS (since f|Sf{{|}_{S}} is bijective onto S). Therefore, AutT#(S)T|S\operatorname{Aut}_{{{T}^{\#}}}(S)\supseteq T{{|}_{S}}.

For (c):

Assume that fTAut(S)\forall f\in{{T}_{\operatorname{Aut}(S)}}, f|Sf{{|}_{S}} is bijective onto SS. Since TTAut(S)T\subseteq{{T}_{\operatorname{Aut}(S)}}, fT\forall f\in T, f|Sf{{|}_{S}} is bijective onto SS. Then by (b), AutT#(S)T|S\operatorname{Aut}_{{{T}^{\#}}}(S)\supseteq T{{|}_{S}}.

We are going to prove that AutT#(S)=T|S\operatorname{Aut}_{{{T}^{\#}}}(S)=T{{|}_{S}} if and only if TAut(S)|S=T|S{{T}_{\operatorname{Aut}(S)}}{{|}_{S}}=T{{|}_{S}}.

Suppose AutT#(S)=T|S\operatorname{Aut}_{{{T}^{\#}}}(S)=T{{|}_{S}}. T|STAut(S)|ST{{|}_{S}}\subseteq{{T}_{\operatorname{Aut}(S)}}{{|}_{S}} because TTAut(S)T\subseteq{{T}_{\operatorname{Aut}(S)}}. To show TAut(S)|ST|S{{T}_{\operatorname{Aut}(S)}}{{|}_{S}}\subseteq T{{|}_{S}}, let fTAut(S)f\in{{T}_{\operatorname{Aut}(S)}}. Then by the definition of TAut(S){{T}_{\operatorname{Aut}(S)}}, σ(f(a))=f(σ(a))\sigma(f(a))=f(\sigma(a)), aS\forall a\in S and σT#=AutT(S)\sigma\in{{T}^{\#}}=\operatorname{Aut}_{T}(S), and hence f|SAutT#(S)=T|Sf{{|}_{S}}\in\operatorname{Aut}_{{{T}^{\#}}}(S)=T{{|}_{S}} (because f|Sf{{|}_{S}} is bijective onto S). Hence TAut(S)|ST|S{{T}_{\operatorname{Aut}(S)}}{{|}_{S}}\subseteq T{{|}_{S}}. Thus, TAut(S)|S=T|S{{T}_{\operatorname{Aut}(S)}}{{|}_{S}}=T{{|}_{S}}.

Conversely, suppose TAut(S)|S=T|S{{T}_{\operatorname{Aut}(S)}}{{|}_{S}}=T{{|}_{S}}. We already showed AutT#(S)T|S\operatorname{Aut}_{{{T}^{\#}}}(S)\supseteq T{{|}_{S}}. To show AutT#(S)T|S\operatorname{Aut}_{{{T}^{\#}}}(S)\subseteq T{{|}_{S}}, let f:DDf:D\to D such that f|SAutT#(S)f{{|}_{S}}\in\operatorname{Aut}_{{{T}^{\#}}}(S). Then Im(f|S)S\operatorname{Im}(f{{|}_{S}})\subseteq S and σ(f(a))=f(σ(a)),aS\sigma(f(a))=f(\sigma(a)),\forall a\in S and σT#=AutT(S)\sigma\in{{T}^{\#}}=\operatorname{Aut}_{T}(S). It follows that fTAut(S)f\in{{T}_{\operatorname{Aut}(S)}}, and hence f|STAut(S)|S=T|Sf{{|}_{S}}\in{{T}_{\operatorname{Aut}(S)}}{{|}_{S}}=T{{|}_{S}}. Thus, AutT#(S)T|S\operatorname{Aut}_{{{T}^{\#}}}(S)\subseteq T{{|}_{S}}. Therefore, AutT#(S)=T|S\operatorname{Aut}_{{{T}^{\#}}}(S)=T{{|}_{S}}. ∎

In Theorem 18.0.1, if TT were generalized to be a (par-)operator gen-semigroup, then statement (a) would still hold (though generally it would be false that TTEnd(S)T\subseteq{{T}_{\operatorname{End}(S)}} because TEnd(S){T}_{\operatorname{End}(S)} is always an operator semigroup). However, since the elements of EndT(S){}_{T^{*}}(S) are all unary functions but the elements of a (par-)operator gen-semigroup are possibly not, statement (b) in Theorem 18.0.1 would no longer hold.

Analogously, in Theorem 18.0.2, if TT were generalized to be a (par-)operator gen-semigroup, then statement (a) would still hold (though generally it would be false that TTAut(S)T\subseteq{{T}_{\operatorname{Aut}(S)}} because TAut(S){T}_{\operatorname{Aut}(S)} is always an operator semigroup), but statements (b) and (c) would no longer hold.

19. Fixed sets and transitive actions of EndT(S)\operatorname{End}_{T}(S) and AutT(S)\operatorname{Aut}_{T}(S)

In this section we develop, among other results, two analogues of the well-known fact that the Galois group of an irreducible polynomial acts transitively on its roots.

In this section, TT is always a par-operator gen-semigroup unless otherwise specified.

In the classical Galois theory, the fixed field of Aut(F)\operatorname{Aut}(F), where FF is a field, is denoted by FAut(F){{F}^{\operatorname{Aut}(F)}}. To study SEndT(S){{S}^{\operatorname{End}_{T}(S)}} and SAutT(S){{S}^{\operatorname{Aut}_{T}(S)}}, which generalize the notion of FAut(F){{F}^{\operatorname{Aut}(F)}}, we first give

Notation 19.0.1.

Let ADA\subseteq D, where DD is the domain of TT. Let

CAEnd={uA|[u)TA={u}},C_{A}^{\operatorname{End}}=\{u\in A\,|\,{{[u)}_{T}}\bigcap A=\{u\}\},

i.e. CAEndC_{A}^{\operatorname{End}} is the set of uAu\in A where uu is the only vAv\in A such that uTvu\xrightarrow{T}v. And let

CAAut={uA|[u]TA={u}},C_{A}^{\operatorname{Aut}}=\{u\in A\,|\,{{[u]}_{T}}\bigcap A=\{u\}\},

i.e. CAAutC_{A}^{\operatorname{Aut}} is the set of uAu\in A where uu is the only vAv\in A such that uTvu\overset{T}{\longleftrightarrow}v.

Remark.

For an operator semigroup TT, see Notation 7.1.1 for the notations involved. More generally, for a par-operator gen-semigroup TT, see Notation 14.1.1 for the notations involved.

For example, let TT, FF and BB be defined as in Proposition 7.1.2, then it is not hard to tell that CFEnd=CFAut=BC_{F}^{\operatorname{End}}=C_{F}^{\operatorname{Aut}}=B.

19.1. Fixed sets and transitive actions of EndT(S)\operatorname{End}_{T}(S)

The following explains why we have “End” in notation CAEndC_{A}^{\operatorname{End}}.

Lemma 19.1.1.

Let SS be a TT-space. Then CSEndSEndT(S)C_{S}^{\operatorname{End}}\subseteq{{S}^{\operatorname{End}_{T}(S)}}, and hence EndT(S)=GMnT(S/CSEnd)\operatorname{End}_{T}(S)=\operatorname{GMn}_{T}(S/C_{S}^{\operatorname{End}}).

Proof.

Let σEndT(S)\sigma\in\operatorname{End}_{T}(S). By Proposition 14.1.2, uS\forall u\in S, uTσ(u)u\xrightarrow{T}\sigma(u). Let uCSEndu\in C_{S}^{\operatorname{End}}. Then by Notation 19.0.1, u=σ(u)u=\sigma(u), and hence uSσu\in{{S}^{\sigma}}. It follows that uSEndT(S)u\in{{S}^{\operatorname{End}_{T}(S)}}. Thus CSEndSEndT(S)C_{S}^{\operatorname{End}}\subseteq{{S}^{\operatorname{End}_{T}(S)}}.

Hence by Propositions 2.6.5 and 2.6.6, EndT(S)GMnT(S/SEndT(S))GMnT(S/CSEnd)\operatorname{End}_{T}(S)\subseteq\operatorname{GMn}_{T}(S/{{S}^{\operatorname{End}_{T}(S)}})\subseteq\operatorname{GMn}_{T}(S/C_{S}^{\operatorname{End}}). Since EndT(S)GMnT(S/CSEnd)\operatorname{End}_{T}(S)\supseteq\operatorname{GMn}_{T}(S/C_{S}^{\operatorname{End}}), EndT(S)=GMnT(S/CSEnd)\operatorname{End}_{T}(S)=\operatorname{GMn}_{T}(S/C_{S}^{\operatorname{End}}). ∎

Recall a well-known fact in the classical Galois theory: the Galois group of an irreducible polynomial acts transitively on its roots. By Proposition 7.1.2, the following is an analogue of this fact.

Theorem 19.1.2.

Let SS be a TT-space. Suppose that the identity function lies in TT and aS\CSEnd\forall a\in S\backslash C_{S}^{\operatorname{End}}, aT=S{{\left\langle a\right\rangle}_{T}}=S. Then u,vS\forall u,v\in S with uTvu\xrightarrow{T}v, σGMnT(S/CSEnd)(=EndT(S))\exists\sigma\in\operatorname{GMn}_{T}(S/C_{S}^{\operatorname{End}})(=\operatorname{End}_{T}(S)) such that σ(u)=v\sigma(u)=v.

Proof.

Let u,vSu,v\in S with uTvu\xrightarrow{T}v. There are two cases as follows.

Suppose uS\CSEndu\in S\backslash C_{S}^{\operatorname{End}}. It is assumed that aT=S{{\left\langle a\right\rangle}_{T}}=S, aS\CSEnd\forall a\in S\backslash C_{S}^{\operatorname{End}}. Hence uT=S{{\left\langle u\right\rangle}_{T}}=S. By Proposition 14.1.3, the map σ:uTvT\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}} given by f(un)f(vn)f({{u}^{n}})\mapsto f({{v}^{n}}), n+\forall n\in{{\mathbb{Z}}^{+}} and fTf\in T of nn variables such that f(un)f({{u}^{n}}) is well-defined, is a TT-morphism from SS to itself (because vTS\left\langle v\right\rangle_{T}\subseteq S by Proposition 10.1.1). Then by Lemma 19.1.1, σEndT(S)=GMnT(S/CSEnd)\sigma\in\operatorname{End}_{T}(S)=\operatorname{GMn}_{T}(S/C_{S}^{\operatorname{End}}). Moreover, if let f=Idf=\operatorname{Id}, then by the definition of σ\sigma, σ(u)=σ(Id(u))=Id(v)=v\sigma(u)=\sigma(\operatorname{Id}(u))=\operatorname{Id}(v)=v, as desired.

Suppose uCSEndu\in C_{S}^{\operatorname{End}}. Then by Notation 19.0.1, v=uv=u because uTvu\xrightarrow{T}v. If let σ\sigma be the identity function on SS, then apparently σGMnT(S/CSEnd)\sigma\in\operatorname{GMn}_{T}(S/C_{S}^{\operatorname{End}}) and σ(u)=u=v\sigma(u)=u=v, as desired. ∎

Combining the above two results, we have

Corollary 19.1.3.

Let SS be a TT-space. Suppose that the identity function lies in TT and aS\CSEnd\forall a\in S\backslash C_{S}^{\operatorname{End}}, aT=S{{\left\langle a\right\rangle}_{T}}=S. Then CSEnd=SEndT(S)=SGMnT(S/CSEnd)C_{S}^{\operatorname{End}}={{S}^{\operatorname{End}_{T}(S)}}={{S}^{\operatorname{GMn}_{T}(S/C_{S}^{\operatorname{End}})}}.

Proof.

Let uSEndT(S)u\in{{S}^{\operatorname{End}_{T}(S)}}. Assume uCSEndu\notin C_{S}^{\operatorname{End}}, i.e. vS\exists v\in S such that vuv\neq u and uTvu\xrightarrow{T}v. Then by Theorem 19.1.2, σGMnT(S/CSEnd)\exists\sigma\in\operatorname{GMn}_{T}(S/C_{S}^{\operatorname{End}}) such that σ(u)=v\sigma(u)=v, which is contrary to uSEndT(S)u\in{{S}^{\operatorname{End}_{T}(S)}}. Hence SEndT(S)CSEnd{{S}^{\operatorname{End}_{T}(S)}}\subseteq C_{S}^{\operatorname{End}}.

By Lemma 19.1.1, CSEndSEndT(S)=SGMnT(S/CSEnd)C_{S}^{\operatorname{End}}\subseteq{{S}^{\operatorname{End}_{T}(S)}}={{S}^{\operatorname{GMn}_{T}(S/C_{S}^{\operatorname{End}})}}. Therefore, CSEnd=SEndT(S)=SGMnT(S/CSEnd)C_{S}^{\operatorname{End}}={{S}^{\operatorname{End}_{T}(S)}}={{S}^{\operatorname{GMn}_{T}(S/C_{S}^{\operatorname{End}})}}. ∎

19.2. Fixed sets and transitive actions of AutT(S)\operatorname{Aut}_{T}(S)

For CSAutC_{S}^{\operatorname{Aut}} (cf. Notation 19.0.1), the analogue of Lemma 19.1.1 is

Lemma 19.2.1.

Let SS be a TT-space. Then CSAutSAutT(S)C_{S}^{\operatorname{Aut}}\subseteq{{S}^{\operatorname{Aut}_{T}(S)}}, and hence AutT(S)=GGrT(S/CSAut)\operatorname{Aut}_{T}(S)=\operatorname{GGr}_{T}(S/C_{S}^{\operatorname{Aut}}).

Proof.

Recall that Proposition 7.1.4 still holds for TT being a par-operator gen-semigroup (cf. Subsection 14.1). Let σAutT(S)\sigma\in\operatorname{Aut}_{T}(S). Then by Proposition 7.1.4, uS\forall u\in S, uTσ(u)u\overset{T}{\longleftrightarrow}\sigma(u). Let uCSAutu\in C_{S}^{\operatorname{Aut}}. Then by Notation 19.0.1, u=σ(u)u=\sigma(u), and hence uSσu\in{{S}^{\sigma}}. It follows that uSAutT(S)u\in{{S}^{\operatorname{Aut}_{T}(S)}}. Hence CSAutSAutT(S)C_{S}^{\operatorname{Aut}}\subseteq{{S}^{\operatorname{Aut}_{T}(S)}}.

Then by Propositions 2.6.5 and 2.6.6, AutT(S)GGrT(S/SAutT(S))GGrT(S/CSAut)\operatorname{Aut}_{T}(S)\subseteq\operatorname{GGr}_{T}(S/{{S}^{\operatorname{Aut}_{T}(S)}})\subseteq\operatorname{GGr}_{T}(S/C_{S}^{\operatorname{Aut}}). Since AutT(S)GGrT(S/CSAut)\operatorname{Aut}_{T}(S)\supseteq\operatorname{GGr}_{T}(S/C_{S}^{\operatorname{Aut}}), AutT(S)=GGrT(S/CSAut)\operatorname{Aut}_{T}(S)=\operatorname{GGr}_{T}(S/C_{S}^{\operatorname{Aut}}). ∎

And for CSAutC_{S}^{\operatorname{Aut}}, the analogue of Theorem 19.1.2 is as follows, which, by Proposition 7.1.2, is also an analogue of the well-known fact that the Galois group of an irreducible polynomial acts transitively on its roots.

Theorem 19.2.2.

Let SS be a TT-space. Suppose that the identity function lies in TT and aS\CSAut\forall a\in S\backslash C_{S}^{\operatorname{Aut}}, aT=S{{\left\langle a\right\rangle}_{T}}=S. Then u,vS\forall u,v\in S with uTvu\overset{T}{\longleftrightarrow}v, σGGrT(S/CSAut)(=AutT(S))\exists\sigma\in\operatorname{GGr}_{T}(S/C_{S}^{\operatorname{Aut}})(=\operatorname{Aut}_{T}(S)) such that σ(u)=v\sigma(u)=v.

Proof.

Let u,vSu,v\in S with uTvu\overset{T}{\longleftrightarrow}v. There are two cases as follows.

Suppose uS\CSAutu\in S\backslash C_{S}^{\operatorname{Aut}}. It is assumed that aT=S{{\left\langle a\right\rangle}_{T}}=S, aS\CSAut\forall a\in S\backslash C_{S}^{\operatorname{Aut}}. Hence uT=S{{\left\langle u\right\rangle}_{T}}=S. By Corollary 14.1.4, the map σ:uTvT\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}} given by f(un)f(vn)f({{u}^{n}})\mapsto f({{v}^{n}}), n+\forall n\in{{\mathbb{Z}}^{+}} and fTf\in T of nn variables such that f(un)f({{u}^{n}}) is well-defined, is a TT-isomorphism from SS to itself (because vTS\left\langle v\right\rangle_{T}\subseteq S by Proposition 10.1.1). Then by Lemma 19.2.1, σAutT(S)=GGrT(S/CSAut)\sigma\in\operatorname{Aut}_{T}(S)=\operatorname{GGr}_{T}(S/C_{S}^{\operatorname{Aut}}). Moreover, if let f=Idf=\operatorname{Id}, then by the definition of σ\sigma, σ(u)=σ(Id(u))=Id(v)=v\sigma(u)=\sigma(\operatorname{Id}(u))=\operatorname{Id}(v)=v, as desired.

Suppose uCSAutu\in C_{S}^{\operatorname{Aut}}. Then by Notation 19.0.1, v=uv=u because uTvu\overset{T}{\longleftrightarrow}v. If let σ\sigma be the identity function on SS, then σGGrT(S/CSAut)\sigma\in\operatorname{GGr}_{T}(S/C_{S}^{\operatorname{Aut}}) and σ(u)=u=v\sigma(u)=u=v, as desired. ∎

Combining the above two results, we have

Corollary 19.2.3.

Let SS be a TT-space. Suppose that the identity function lies in TT and aS\CSAut\forall a\in S\backslash C_{S}^{\operatorname{Aut}}, aT=S{{\left\langle a\right\rangle}_{T}}=S. Then CSAut=SAutT(S)=SGGrT(S/CSAut)C_{S}^{\operatorname{Aut}}={{S}^{\operatorname{Aut}_{T}(S)}}={{S}^{\operatorname{GGr}_{T}(S/C_{S}^{\operatorname{Aut}})}}.

Proof.

Let uSAutT(S)u\in{{S}^{\operatorname{Aut}_{T}(S)}}. Assume uCSAutu\notin C_{S}^{\operatorname{Aut}}, i.e. vS\exists v\in S such that vuv\neq u and uTvu\overset{T}{\longleftrightarrow}v. Then by Theorem 19.2.2, σGGrT(S/CSAut)\exists\sigma\in\operatorname{GGr}_{T}(S/C_{S}^{\operatorname{Aut}}) such that σ(u)=v\sigma(u)=v, which is contrary to uSAutT(S)u\in{{S}^{\operatorname{Aut}_{T}(S)}}. Hence SAutT(S)CSAut{{S}^{\operatorname{Aut}_{T}(S)}}\subseteq C_{S}^{\operatorname{Aut}}.

By Lemma 19.2.1, CSAutSAutT(S)=SGGrT(S/CSAut)C_{S}^{\operatorname{Aut}}\subseteq{{S}^{\operatorname{Aut}_{T}(S)}}={{S}^{\operatorname{GGr}_{T}(S/C_{S}^{\operatorname{Aut}})}}. Therefore, CSAut=SAutT(S)=SGGrT(S/CSAut)C_{S}^{\operatorname{Aut}}={{S}^{\operatorname{Aut}_{T}(S)}}={{S}^{\operatorname{GGr}_{T}(S/C_{S}^{\operatorname{Aut}})}}. ∎

20. Two questions on Galois TT-extensions and normal subgroups of Galois TT-groups

In this section, TT is always a par-operator gen-semigroup unless otherwise specified.

As was promised at the end of Section 3, this section focuses on the following.

Problem 20.0.1.

Let SS be a TT-space and let BKSB\subseteq K\subseteq S. If B=SGGrT(S/B)B={{S}^{\operatorname{GGr}_{T}(S/B)}} and K=SGGrT(S/K)K={{S}^{\operatorname{GGr}_{T}(S/K)}}, then

  1. (1)

    when does B=KGGrT(K/B)B={{K}^{\operatorname{GGr}_{T}(K/B)}}?

  2. (2)

    when do GGrT(S/K)GGrT(S/B)\operatorname{GGr}_{T}(S/K)\,\triangleleft\,\operatorname{GGr}_{T}(S/B) and

    GGrT(S/B)/GGrT(S/K)GGrT(K/B)?\operatorname{GGr}_{T}(S/B)/\operatorname{GGr}_{T}(S/K)\cong\operatorname{GGr}_{T}(K/B)?

In particular, if let BKSB\subseteq K\subseteq S be fields such that SS is a Galois extension of BB, then, as is well-known, KK is Galois over BB (i.e. B=KGal(K/B)B={{K}^{\operatorname{Gal}(K/B)}}) if and only if Gal(S/K)\operatorname{Gal}(S/K) is normal in Gal(S/B)\operatorname{Gal}(S/B) (that is, (1) is true if and only if (2) is true). However, for a general TT-space, the answer to Problem 20.0.1 seems to be not so simple. We can only give some sufficient conditions for the two questions in Problem 20.0.1. Nevertheless, one still can see that some, if not all, results and notions introduced in this section have analogues in well-known Galois theories.

20.1. HH-stable subsets of a TT-space

Definition 20.1.1.

Let SS be a TT-space with KSK\subseteq S and HEndT(S)H\subseteq\operatorname{End}_{T}(S). If σH\forall\sigma\in H and aKa\in K, σ(a)K,\sigma(a)\in K, then we say that KK is stable under HH or KK is H-stable.

Remark.

Another choice of this terminology is HH-invariant.

Lemma 20.1.2.

Let SS be a TT-space with BSB\subseteq S. Let HGGrT(S/B)H\,\triangleleft\,\operatorname{GGr}_{T}(S/B). Then SH{{S}^{H}} is GGrT(S/B)\operatorname{GGr}_{T}(S/B)-stable.

Proof.

Suppose that σGGrT(S/B)\exists\sigma\in\operatorname{GGr}_{T}(S/B) and aSHa\in{{S}^{H}} such that σ(a)SH\sigma(a)\notin{{S}^{H}}. Then σ(a)S\SH\sigma(a)\in S\backslash{{S}^{H}} and aSH\Ba\in{{S}^{H}}\backslash B (otherwise aBa\in B and σ(a)=aSH\sigma(a)=a\in S^{H}). Assume that τH\forall\tau\in H, τ(σ(a))=σ(a)\tau(\sigma(a))=\sigma(a), then σ(a)SH\sigma(a)\in{{S}^{H}}, a contradiction. It follows that there is τaH{{\tau}_{a}}\in H such that τa(σ(a))σ(a){{\tau}_{a}}(\sigma(a))\neq\sigma(a). However, since HGGrT(S/B)H\,\triangleleft\,\operatorname{GGr}_{T}(S/B), σ1τaσH{{\sigma}^{-1}}{{\tau}_{a}}\sigma\in H. Then σ1(τa(σ(a)))=a{{\sigma}^{-1}}({{\tau}_{a}}(\sigma(a)))=a because aSHa\in{{S}^{H}}, and hence τa(σ(a))=σ(a){{\tau}_{a}}(\sigma(a))=\sigma(a), a contradiction again. ∎

Lemma 20.1.3.

Let SS be a TT-space with BKSB\subseteq K\subseteq S. Suppose that K=SGGrT(S/K)K={{S}^{\operatorname{GGr}_{T}(S/K)}}, KK is GGrT(S/B)\operatorname{GGr}_{T}(S/B)-stable, and every element of GGrT(K/B)\operatorname{GGr}_{T}(K/B) can be extended to an element of GGrT(S/B)\operatorname{GGr}_{T}(S/B). Then SGGrT(S/B)=KGGrT(K/B){{S}^{\operatorname{GGr}_{T}(S/B)}}={{K}^{\operatorname{GGr}_{T}(K/B)}}.

Proof.

K=SGGrT(S/K)K={{S}^{\operatorname{GGr}_{T}(S/K)}} implies that no element of S\KS\backslash K is fixed under the action of GGrT(S/K)(GGrT(S/B))\operatorname{GGr}_{T}(S/K)(\subseteq\operatorname{GGr}_{T}(S/B)). Then no element of S\KS\backslash K is fixed under the action of GGrT(S/B)\operatorname{GGr}_{T}(S/B), and hence SGGrT(S/B)=KGGrT(S/B){{S}^{\operatorname{GGr}_{T}(S/B)}}={{K}^{\operatorname{GGr}_{T}(S/B)}} (cf. Definition 2.5.1 for KGGrT(S/B){{K}^{\operatorname{GGr}_{T}(S/B)}}).

KK is GGrT(S/B)\operatorname{GGr}_{T}(S/B)-stable, and hence σGGrT(S/B)(AutTS),\forall\sigma\in\operatorname{GGr}_{T}(S/B)(\subseteq\operatorname{Aut}_{T}S), σ|KGGrT(K/B)\sigma{{|}_{K}}\in\operatorname{GGr}_{T}(K/B). Hence GGrT(S/B)|KGGrT(K/B)\operatorname{GGr}_{T}(S/B){{|}_{K}}\subseteq\operatorname{GGr}_{T}(K/B), and therefore KGGrT(S/B)KGGrT(K/B){{K}^{\operatorname{GGr}_{T}(S/B)}}\supseteq{{K}^{\operatorname{GGr}_{T}(K/B)}}.

On the other hand, since every element of GGrT(K/B)\operatorname{GGr}_{T}(K/B) can be extended to an element of GGrT(S/B)\operatorname{GGr}_{T}(S/B), KGGrT(K/B)KGGrT(S/B){{K}^{\operatorname{GGr}_{T}(K/B)}}\supseteq{{K}^{\operatorname{GGr}_{T}(S/B)}}. Thus KGGrT(K/B)=KGGrT(S/B){{K}^{\operatorname{GGr}_{T}(K/B)}}={{K}^{\operatorname{GGr}_{T}(S/B)}}.

Therefore, SGGrT(S/B)=KGGrT(S/B)=KGGrT(K/B){{S}^{\operatorname{GGr}_{T}(S/B)}}={{K}^{\operatorname{GGr}_{T}(S/B)}}={{K}^{\operatorname{GGr}_{T}(K/B)}}. ∎

Now we obtain our first answer to the first question in Problem 20.0.1 as follows, where an “extra” condition is that “every element of GGrT(K/B)\operatorname{GGr}_{T}(K/B) can be extended to an element of GGrT(S/B)\operatorname{GGr}_{T}(S/B)”.

Corollary 20.1.4.

Let SS be a TT-space with BKSB\subseteq K\subseteq S. Suppose that B=SGGrT(S/B)B={{S}^{\operatorname{GGr}_{T}(S/B)}}, K=SGGrT(S/K)K={{S}^{\operatorname{GGr}_{T}(S/K)}}, GGrT(S/K)GGrT(S/B)\operatorname{GGr}_{T}(S/K)\,\triangleleft\,\operatorname{GGr}_{T}(S/B), and every element of GGrT(K/B)\operatorname{GGr}_{T}(K/B) can be extended to an element of GGrT(S/B)\operatorname{GGr}_{T}(S/B). Then B=KGGrT(K/B)B={{K}^{\operatorname{GGr}_{T}(K/B)}}.

Proof.

By Lemma 20.1.2, KK(=SGGrT(S/K)={{S}^{\operatorname{GGr}_{T}(S/K)}}) is GGrT(S/B)\operatorname{GGr}_{T}(S/B)-stable. Then by Lemma 20.1.3, B=SGGrT(S/B)=KGGrT(K/B)B={{S}^{\operatorname{GGr}_{T}(S/B)}}={{K}^{\operatorname{GGr}_{T}(K/B)}}. ∎

A converse of Lemma 20.1.2 is as follows, which is our first answer to the second question in Problem 20.0.1.

Proposition 20.1.5.

Let SS be a TT-space with BKSB\subseteq K\subseteq S. Suppose that KK is GGrT(S/B)\operatorname{GGr}_{T}(S/B)-stable. Then there exists a group homomorphism γ:GGrT(S/B)GGrT(K/B)\gamma:\operatorname{GGr}_{T}(S/B)\to\operatorname{GGr}_{T}(K/B) such that kerγ=GGrT(S/K)\ker\gamma=\operatorname{GGr}_{T}(S/K), and so GGrT(S/K)GGrT(S/B)\operatorname{GGr}_{T}(S/K)\,\triangleleft\,\operatorname{GGr}_{T}(S/B) and GGrT(S/B)/GGrT(S/K)GGrT(K/B)\operatorname{GGr}_{T}(S/B)/\operatorname{GGr}_{T}(S/K)\cong\operatorname{GGr}_{T}(K/B).

Proof.

Let γ:GGrT(S/B)GGrT(K/B)\gamma:\operatorname{GGr}_{T}(S/B)\to\operatorname{GGr}_{T}(K/B) be given by γ(σ)=σ|K\gamma(\sigma)=\sigma{{|}_{K}}.

Since KK is GGrT(S/B)-\operatorname{GGr}_{T}(S/B)\text{-}stable, σGGrT(S/B)\forall\sigma\in\operatorname{GGr}_{T}(S/B), σ(K)K\sigma(K)\subseteq K. And σ(K)=K\sigma(K)=K because σGGrT(S/B)AutT(S)\sigma\in\operatorname{GGr}_{T}(S/B)\subseteq\operatorname{Aut}_{T}(S). Hence γ(σ)=σ|KGGrT(K/B)\gamma(\sigma)=\sigma{{|}_{K}}\in\operatorname{GGr}_{T}(K/B), which implies that γ\gamma is a well-defined map.

Moreover, if σ1,σ2GGrT(S/B){{\sigma}_{1}},{{\sigma}_{2}}\in\operatorname{GGr}_{T}(S/B), then (σ1|K)(σ2|K)=(σ1σ2)|K({{\sigma}_{1}}{{|}_{K}})\circ({{\sigma}_{2}}{{|}_{K}})=({{\sigma}_{1}}\circ{{\sigma}_{2}}){{|}_{K}}, and so γ(σ1)γ(σ2)=γ(σ1σ2)\gamma({{\sigma}_{1}})\circ\gamma({{\sigma}_{2}})=\gamma({{\sigma}_{1}}\circ{{\sigma}_{2}}), which implies that γ\gamma is a group homomorphism.

Finally, because γ\gamma is defined as the restriction to KK, kerγ=GGrT(S/K)\ker\gamma=\operatorname{GGr}_{T}(S/K): ∎

Combining Proposition 20.1.5 with Corollary 20.1.4, we immediately obtain our second answer to the first question in Problem 20.0.1.

Corollary 20.1.6.

Let SS be a TT-space with BKSB\subseteq K\subseteq S. If B=SGGrT(S/B)B={{S}^{\operatorname{GGr}_{T}(S/B)}}, K=SGGrT(S/K)K={{S}^{\operatorname{GGr}_{T}(S/K)}}, KK is GGrT(S/B)\operatorname{GGr}_{T}(S/B)-stable, and each element of GGrT(K/B)\operatorname{GGr}_{T}(K/B) can be extended to an element of GGrT(S/B)\operatorname{GGr}_{T}(S/B), then B=KGGrT(K/B)B={{K}^{\operatorname{GGr}_{T}(K/B)}}.

20.2. Normal TT-subsets of a TT-space

Definition 20.2.1.

Let KK be a subset of a TT-space SS. If aK\forall a\in K, [a]TSK{{[a]}_{T}}\bigcap S\subseteq K (cf. Notation 14.1.1 for [a]T{{[a]}_{T}}), then we call KK a normal TT-subset of SS and write KSK\,\triangleleft\,S or SKS\,\triangleright\,K.

Remark.

We use this terminology because there is a correspondence between normal TT-subsets of SS and normal subgroups of a Galois TT-group of SS, which will be shown in Corollary 20.2.3 and Proposition 20.2.4.

Any normal TT-subset of a TT-space SS is AutT(S)\operatorname{Aut}_{T}(S)-stable, as shown below.

Lemma 20.2.2.

Let SS be a TT-space. Suppose that KK is a normal TT-subset of SS. Then KK is stable under AutT(S)\operatorname{Aut}_{T}(S).

Proof.

Recall that Proposition 7.1.4 still holds for TT being a par-operator gen-semigroup (cf. Subsection 14.1). Hence by Proposition 7.1.4, aTσ(a)a\overset{T}{\longleftrightarrow}\sigma(a), aS\forall a\in S and σAutT(S)\sigma\in\operatorname{Aut}_{T}(S).

Let aKa\in K and σAutT(S)\sigma\in\operatorname{Aut}_{T}(S). Then aTσ(a)a\overset{T}{\longleftrightarrow}\sigma(a), and thus σ(a)[a]TSK\sigma(a)\in{{[a]}_{T}}\bigcap S\subseteq K because KSK\,\triangleleft\,S. Hence by Definition 20.1.1, KK is stable under AutT(S)\operatorname{Aut}_{T}(S). ∎

The combination of Lemma 20.2.2 and Proposition 20.1.5 yields the following, which is obvious because GGrT(S/B)AutT(S)\operatorname{GGr}_{T}(S/B)\subseteq\operatorname{Aut}_{T}(S). It is our second answer to the second question in Problem 20.0.1.

Corollary 20.2.3.

Let SS be a TT-space with BKSB\subseteq K\subseteq S. Suppose that KK is a normal TT-subset of SS. Then there exists a group homomorphism γ:GGrT(S/B)GGrT(K/B)\gamma:\operatorname{GGr}_{T}(S/B)\to\operatorname{GGr}_{T}(K/B) such that kerγ=GGrT(S/K)\ker\gamma=\operatorname{GGr}_{T}(S/K), and hence GGrT(S/K)GGrT(S/B)\operatorname{GGr}_{T}(S/K)\,\triangleleft\,\operatorname{GGr}_{T}(S/B) and GGrT(S/B)/GGrT(S/K)GGrT(K/B)\operatorname{GGr}_{T}(S/B)/\operatorname{GGr}_{T}(S/K)\cong\operatorname{GGr}_{T}(K/B).

A converse of Corollary 20.2.3 is as follows, which also explains the terminology “normal TT-subsets”.

Proposition 20.2.4.

Let SS be a TT-space. Suppose that the identity function lies in TT, aS\CSAut\forall a\in S\backslash C_{S}^{\operatorname{Aut}}, aT=S{{\left\langle a\right\rangle}_{T}}=S, and HAutT(S)H\,\triangleleft\,\operatorname{Aut}_{T}(S). Then SHS{{S}^{H}}\,\triangleleft\,S.

Proof.

Let uSHu\in{{S}^{H}} and v[u]TSv\in{{[u]}_{T}}\bigcap S. By Theorem 19.2.2, σAutT(S)\exists\sigma\in\operatorname{Aut}_{T}(S) such that σ(u)=v\sigma(u)=v. By Lemma 20.1.2 (with B=B=\emptyset), SH{{S}^{H}} is AutT(S)\operatorname{Aut}_{T}(S)-stable, and hence vSHv\in{{S}^{H}}. Thus uSH\forall u\in{{S}^{H}}, [u]TSSH{{[u]}_{T}}\bigcap S\subseteq{{S}^{H}}. Hence by Definition 20.2.1, SHS{{S}^{H}}\,\triangleleft\,S. ∎

Finally, combining Corollary 20.2.3 with Corollary 20.1.4, we immediately obtain our third answer to the first question in Problem 20.0.1:

Corollary 20.2.5.

Let SS be a TT-space with BKSB\subseteq K\subseteq S. If B=SGGrT(S/B)B={{S}^{\operatorname{GGr}_{T}(S/B)}}, K=SGGrT(S/K)K={{S}^{\operatorname{GGr}_{T}(S/K)}}, KSK\,\triangleleft\,S, and each element of GGrT(K/B)\operatorname{GGr}_{T}(K/B) can be extended to an element of GGrT(S/B)\operatorname{GGr}_{T}(S/B), then B=KGGrT(K/B)B={{K}^{\operatorname{GGr}_{T}(K/B)}}.

21. \mathcal{F}-transcendental elements and \mathcal{F}-transcendental subsets

In this section, the algebraic notions of transcendental elements over a field and purely transcendental field extensions are generalized for formal partial functions (cf. Definition 9.3.2). To simplify our descriptions, we introduce

Definition 21.0.1.

Let ff and gg be formal partial functions of nn variables. If for each set DD,

  1. (i)

    DD is a domain of ff if and only if DD is a domain of gg; and

  2. (ii)

    (u1,,un)Dn\forall({{u}_{1}},\cdots,{{u}_{n}})\in{{D}^{n}}, neither f(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}}) nor g(u1,,un)g({{u}_{1}},\cdots,{{u}_{n}}) is well-defined or f(u1,,un)=g(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}})=g({{u}_{1}},\cdots,{{u}_{n}});

then we say that ff is equivalent to gg and write f=gf=g.

Moreover, throughout this section, unless otherwise specified, DD always stands for a set and \mathcal{F} denotes a set of formal partial functions.

21.1. \mathcal{F}-transcendental elements

To generalize the concept of transcendental elements in field theory, we give

Definition 21.1.1.

Let n+n\in{{\mathbb{Z}}^{+}} and let (u1,,un)Dn({{u}_{1}},\cdots,{{u}_{n}})\in{{D}^{n}}. If f(u1,,un)=g(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}})=g({{u}_{1}},\cdots,{{u}_{n}}) implies f=gf=g, f,g\forall f,g\in\mathcal{F}, then we say that (u1,,un)({{u}_{1}},\cdots,{{u}_{n}}) is \mathcal{F}-transcendental.

The above definition of transcendental elements coincides with that in field theory, as shown below.

Proposition 21.1.2.

Let BB be a subfield of a field FF and let \mathcal{F} be the polynomial ring B[x]B[x]. Then aF\forall a\in F, aa is transcendental over BB if and only if aa is \mathcal{F}-transcendental.

Proof.

Let aFa\in F. Then

aa is algebraic over BB;
\Leftrightarrow\exists nonzero p(x)B[x]p(x)\in B[x] such that p(a)=0p(a)=0;
f,g(=B[x])\Leftrightarrow\exists f,g\in\mathcal{F}(=B[x]) such that fgf\neq g and f(a)=g(a)f(a)=g(a);
a\Leftrightarrow a is not \mathcal{F}-transcendental. ∎

21.2. \mathcal{F}-transcendental subsets

Definition 21.1.1 may be generalized in more than one way. One of them is as follows.

Definition 21.2.1.

Let UDU\subseteq D. If n+\forall n\in{{\mathbb{Z}}^{+}} and {u1,,un}U\{{{u}_{1}},\cdots,{{u}_{n}}\}\subseteq U, (u1,,un)({{u}_{1}},\cdots,{{u}_{n}}) is \mathcal{F}-transcendental, i.e. f(u1,,un)=g(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}})=g({{u}_{1}},\cdots,{{u}_{n}}) implies f=gf=g, f,g\forall f,g\in\mathcal{F}, then we say that UU is \mathcal{F}-transcendental.

Remark.

By {u1,,un}U\{{{u}_{1}},\cdots,{{u}_{n}}\}\subseteq U we imply that 1i<jn\forall 1\leq i<j\leq n, uiuju_{i}\neq u_{j}.

The following shows that the above definition of \mathcal{F}-transcendental subsets of DD coincides with the concept of purely transcendental field extension in field theory.

Proposition 21.2.2.

Let BB be a subfield of a field FF, let

=n+Frac(B[x1,,xn]),\mathcal{F}\text{=}\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\text{Frac}(B[{{x}_{1}},\cdots,{{x}_{n}}])},

let TT be the par-operator gen-semigroup on FF defined in Example 9.1.6, and let nonempty UFU\subseteq F. Then UT{{\left\langle U\right\rangle}_{T}} is a purely transcendental field over BB if and only if UT=B{{\left\langle U\right\rangle}_{T}}=B or UU is \mathcal{F}-transcendental.

Proof.

Since UT=n+{B(u1,,un)|u1,,unU}=B(U){{\left\langle U\right\rangle}_{T}}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\{B({{u}_{1}},\cdots,{{u}_{n}})|{{u}_{1}},\cdots,{{u}_{n}}\in U\}}=B(U), the field UT{{\left\langle U\right\rangle}_{T}} is purely transcendental over BB if and only if UT=B{{\left\langle U\right\rangle}_{T}}=B or UU is algebraically independent over BB. Hence we only need to show that UU is \mathcal{F}-transcendental if and only if UU is algebraically independent over BB.

UU is \mathcal{F}-transcendental ;
n+\Leftrightarrow\forall n\in{{\mathbb{Z}}^{+}}, {u1,,un}U\{{{u}_{1}},\cdots,{{u}_{n}}\}\subseteq U and f,gf,g\in\mathcal{F},

f(u1,,un)=g(u1,,un)f({{u}_{1}},\cdots,{{u}_{n}})=g({{u}_{1}},\cdots,{{u}_{n}}) implies f=gf=g;

n+\Leftrightarrow\forall n\in{{\mathbb{Z}}^{+}}, {u1,,un}U\{{{u}_{1}},\cdots,{{u}_{n}}\}\subseteq U and f,gf,g\in\mathcal{F},

f(u1,,un)g(u1,,un)=0f({{u}_{1}},\cdots,{{u}_{n}})-g({{u}_{1}},\cdots,{{u}_{n}})=0 implies that fgf-g is a zero polynomial;

n+\Leftrightarrow\forall n\in{{\mathbb{Z}}^{+}}, {u1,,un}U\{{{u}_{1}},\cdots,{{u}_{n}}\}\subseteq U and hh\in\mathcal{F},

h(u1,,un)=0h({{u}_{1}},\cdots,{{u}_{n}})=0 implies that hh is a zero polynomial;

U\Leftrightarrow U is algebraically independent over BB. ∎

22. A generalized first isomorphism theorem

We shall derive a theorem which generalizes the first isomorphism theorems for groups, rings, and modules.

Notation 22.0.1.

In this section, a θ\theta-morphism is always defined by Definition 9.5.5. Moreover, ϕ\phi is always a θ\theta-morphism from a TT-space SS to a TT^{\prime}-space SS^{\prime}.

22.1. “Quotient space” QϕQ_{\phi}

Notation 22.1.1.

For ϕ\phi, we denote by QϕQ_{\phi} the set {ϕ1(z)|zImϕ}\{\phi^{-1}(z)\,|\,z\in\operatorname{Im}\phi\}.

Remark.

Obviously, QϕQ_{\phi} is a partition of SS. In fact, we may regard QϕQ_{\phi} as the “quotient space” induced by ϕ\phi (cf. Corollary 22.1.9 and Example 22.1.10).

To define a par-operator gen-semigroup (corresponding to ϕ\phi) on QϕQ_{\phi}, which we shall denote by TϕT_{\phi}^{*} (Notation 22.1.6), we define the elements fϕf_{\phi}^{*} (or ff^{*} for brevity) of TϕT_{\phi}^{*} as follows.

Notation 22.1.2.

Let nn-variable fTf\in T. Then by fϕf_{\phi}^{*} or ff^{*} (Qϕ××Qϕ\subseteq Q_{\phi}\times\cdots\times Q_{\phi}) we denote the relation

{(ϕ1(ϕ(x1)),,ϕ1(ϕ(xn)),ϕ1(ϕ(f(x1,,xn))))|(x1,,xn)Sn\{(\phi^{-1}(\phi(x_{1})),\cdots,\phi^{-1}(\phi(x_{n})),\phi^{-1}(\phi(f(x_{1},\cdots,x_{n}))))\,|\,(x_{1},\cdots,x_{n})\in S^{n} and f(x1,,xn)f(x_{1},\cdots,x_{n}) is well-defined}.

Remark.

If f(x1,,xn)f(x_{1},\cdots,x_{n}) is well-defined, by Proposition 10.1.1, f(x1,,xn)Sf(x_{1},\cdots,x_{n})\in S, and hence ϕ(f(x1,,xn))\phi(f(x_{1},\cdots,x_{n})) is well-defined.

To better understand the notation, we give

Example 22.1.3.

Let ϕ\phi be a group homomorphism from a group SS to a group SS^{\prime} (cf. Proposition 8.5.11). Let ee be the identity element of SS. Then ϕ1(ϕ(e))\phi^{-1}(\phi(e)) is the kernel of ϕ\phi, xS\forall x\in S, ϕ1(ϕ(x))\phi^{-1}(\phi(x)) is a coset of the kernel, and QϕQ_{\phi} is the set of all cosets of the kernel. Moreover, we shall see that fϕf_{\phi}^{*} is a group operation among elements of QϕQ_{\phi}.

Proposition 22.1.4.

Let nn-variable fTf\in T. Then fϕf_{\phi}^{*} is a partial function from QϕnQ_{\phi}^{n} to QϕQ_{\phi}, where QϕnQ_{\phi}^{n} denotes the cartesian product of nn copies of QϕQ_{\phi}.

Proof.

It suffices to show that (C1,,Cn)Qϕn\forall(C_{1},\cdots,C_{n})\in Q_{\phi}^{n}, the set {C|(C1,,Cn,C)fϕ}\{C\,|\,(C_{1},\cdots,C_{n},C)\in f_{\phi}^{*}\} has at most one element.

For this purpose, suppose that (a1,,an),(b1,,bn)Sn(a_{1},\cdots,a_{n}),(b_{1},\cdots,b_{n})\in S^{n} such that both f(a1,,an)f(a_{1},\cdots,a_{n}) and f(b1,,bn)f(b_{1},\cdots,b_{n}) are well-defined and ϕ(ai)=ϕ(bi),i=1,,n\phi(a_{i})=\phi(b_{i}),\forall i=1,\cdots,n. By Notation 22.1.2, it suffices to show ϕ(f(a1,,an))=ϕ(f(b1,,bn))\phi(f(a_{1},\cdots,a_{n}))=\phi(f(b_{1},\cdots,b_{n})).

Since ϕ\phi is a θ\theta-morphism, by Definition 9.5.5,

ϕ(f(a1,,an))\displaystyle\phi(f(a_{1},\cdots,a_{n})) =θ(f)(ϕ(a1),,ϕ(an))\displaystyle=\theta(f)(\phi(a_{1}),\cdots,\phi(a_{n}))
=θ(f)(ϕ(b1),,ϕ(bn))\displaystyle=\theta(f)(\phi(b_{1}),\cdots,\phi(b_{n}))
=ϕ(f(b1,,bn)),\displaystyle=\phi(f(b_{1},\cdots,b_{n})),

as desired. ∎

Then the following is obvious.

Corollary 22.1.5.

Let nn-variable fTf\in T. Then fϕ:QϕnQϕf_{\phi}^{*}:Q_{\phi}^{n}\to Q_{\phi} given by

(ϕ1(ϕ(x1)),,ϕ1(ϕ(xn)))ϕ1(ϕ(f(x1,,xn))),(\phi^{-1}(\phi(x_{1})),\cdots,\phi^{-1}(\phi(x_{n})))\mapsto\phi^{-1}(\phi(f(x_{1},\cdots,x_{n}))),

(x1,,xn)Sn\forall(x_{1},\cdots,x_{n})\in S^{n} such that f(x1,,xn)f(x_{1},\cdots,x_{n}) is well-defined, is a well-defined partial function.

Notation 22.1.6.

We denote by TϕT_{\phi}^{*} the set {fϕ|fT}\{f_{\phi}^{*}\,|\,f\in T\}, where fϕ:QϕnQϕf_{\phi}^{*}:Q_{\phi}^{n}\to Q_{\phi} is given in Corollary 22.1.5.

Proposition 22.1.7.

TϕT_{\phi}^{*} is a par-operator gen-semigroup on QϕQ_{\phi}.

Remark.

If the proposition holds, then QϕTϕQϕ{{\left\langle Q_{\phi}\right\rangle}_{T_{\phi}^{*}}}\subseteq Q_{\phi}, and thus by Definition 2.2.8, QϕQ_{\phi} is a quasi-TϕT_{\phi}^{*}-space.

Proof.

By Corollary 22.1.5, fϕTϕ\forall f_{\phi}^{*}\in T_{\phi}^{*}, fϕf_{\phi}^{*} is a partial function from some QϕnQ_{\phi}^{n} to QϕQ_{\phi}.

Let fTf\in T have nn variables, and i=1,,n\forall i=1,\cdots,n, let giT{g}_{i}\in T have nin_{i} variables. To show that TϕT_{\phi}^{*} is a par-operator gen-semigroup on QϕQ_{\phi}, by Definition 9.1.4, we only need to show that the composite f(g1,,gn)f^{*}\circ({{g}_{1}^{*}},\cdots,{{g}_{n}^{*}}) is a restriction of some element of TϕT_{\phi}^{*}. For this purpose, since f(g1,,gn)Tf\circ({{g}_{1}},\cdots,{{g}_{n}})\in T and hence (f(g1,,gn))Tϕ(f\circ({{g}_{1}},\cdots,{{g}_{n}}))^{*}\in T_{\phi}^{*}, it suffices to show

(22.1) f(g1,,gn)=(f(g1,,gn)).f^{*}\circ({{g}_{1}^{*}},\cdots,{{g}_{n}^{*}})=(f\circ({{g}_{1}},\cdots,{{g}_{n}}))^{*}.

Let DD be the domain of TT. Then by Definition 9.1.2, f(g1,,gn)f\circ({{g}_{1}},\cdots,{{g}_{n}}) is the partial function from Dm{{D}^{m}} to DD given by

(x1,,xm)f(g1(v1),,gn(vn)),({{x}_{1}},\cdots,{{x}_{m}})\mapsto f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})),

where m:=inim:=\sum\nolimits_{i}{{{n}_{i}}}, vi:=(xmi+1,,xmi+ni)Dni,i=1,,n{{\text{v}}_{i}}:=({{x}_{{{m}_{i}}+1}},\cdots,{{x}_{{{m}_{i}}+{{n}_{i}}}})\in{{D}^{{{n}_{i}}}},\forall i=1,\cdots,n, m1:=0{{m}_{1}}:=0 and mi:=k=1i1nk,i=2,,n{{m}_{i}}:=\sum\nolimits_{k=1}^{i-1}{{{n}_{k}}},\forall i=2,\cdots,n.

Then by Corollary 22.1.5, (f(g1,,gn))(f\circ({{g}_{1}},\cdots,{{g}_{n}}))^{*} is the partial function from QϕmQ_{\phi}^{m} to QϕQ_{\phi} given by

(ϕ1(ϕ(x1)),,ϕ1(ϕ(xm)))ϕ1(ϕ(f(g1(v1),,gn(vn)))),(\phi^{-1}(\phi(x_{1})),\cdots,\phi^{-1}(\phi(x_{m})))\mapsto\phi^{-1}(\phi(f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})))),

(x1,,xm)Sm\forall(x_{1},\cdots,x_{m})\in S^{m} such that f(g1(v1),,gn(vn))f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})) is well-defined.

To prove Equation (22.1), now we are showing that f(g1,,gn)f^{*}\circ({{g}_{1}^{*}},\cdots,{{g}_{n}^{*}}) can be given exactly the same as above.

By Definition 9.1.2, f(g1,,gn)f^{*}\circ({{g}_{1}^{*}},\cdots,{{g}_{n}^{*}}) is the partial function from QϕmQ_{\phi}^{m} to QϕQ_{\phi} given by

(C1,,Cm)f(g1(V1),,gn(Vn)),({C_{1}},\cdots,{C_{m}})\mapsto f^{*}({{g}_{1}^{*}}({{\text{V}}_{1}}),\cdots,{{g}_{n}^{*}}({{\text{V}}_{n}})),

where Vi:=(Cmi+1,,Cmi+ni)(Qϕ)ni,i=1,,n{{\text{V}}_{i}}:=({C_{{{m}_{i}}+1}},\cdots,{C_{{{m}_{i}}+{{n}_{i}}}})\in{{(Q_{\phi})}^{n_{i}}},\forall i=1,\cdots,n.

By Corollary 22.1.5, i=1,,n\forall i=1,\cdots,n, gig_{i}^{*} can be given by

(ϕ1(ϕ(xmi+1)),,ϕ1(ϕ(xmi+ni)))ϕ1(ϕ(gi(vi))),(\phi^{-1}(\phi({{x}_{{{m}_{i}}+1}})),\cdots,\phi^{-1}(\phi({{x}_{{{m}_{i}}+{{n}_{i}}}})))\mapsto\phi^{-1}(\phi(g_{i}(\text{v}_{i}))),

viSni\forall\text{v}_{i}\in S^{n_{i}} such that gi(vi)g_{i}(\text{v}_{i}) is well-defined. Combining this with the preceding paragraph, we can tell that f(g1,,gn)f^{*}\circ({{g}_{1}^{*}},\cdots,{{g}_{n}^{*}}) can be given by

(ϕ1(ϕ(x1)),,ϕ1(ϕ(xm)))f(ϕ1(ϕ(g1(v1))),,ϕ1(ϕ(gn(vn)))),(\phi^{-1}(\phi(x_{1})),\cdots,\phi^{-1}(\phi(x_{m})))\mapsto f^{*}(\phi^{-1}(\phi(g_{1}(\text{v}_{1}))),\cdots,\phi^{-1}(\phi(g_{n}(\text{v}_{n})))),

(x1,,xm)Sm\forall(x_{1},\cdots,x_{m})\in S^{m} such that f(ϕ1(ϕ(g1(v1))),,ϕ1(ϕ(gn(vn))))f^{*}(\phi^{-1}(\phi(g_{1}(\text{v}_{1}))),\cdots,\phi^{-1}(\phi(g_{n}(\text{v}_{n})))) is well-defined.

By Corollary 22.1.5, a well-defined f(ϕ1(ϕ(g1(v1))),,ϕ1(ϕ(gn(vn))))f^{*}(\phi^{-1}(\phi(g_{1}(\text{v}_{1}))),\cdots,\phi^{-1}(\phi(g_{n}(\text{v}_{n})))) is given the value ϕ1(ϕ(f(g1(v1),,gn(vn))))\phi^{-1}(\phi(f(g_{1}(\text{v}_{1}),\cdots,g_{n}(\text{v}_{n})))) with f(g1(v1),,gn(vn))f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})) being well-defined. Combining this with the preceding paragraph, we see that f(g1,,gn)f^{*}\circ({{g}_{1}^{*}},\cdots,{{g}_{n}^{*}}) can be given by

(ϕ1(ϕ(x1)),,ϕ1(ϕ(xm)))ϕ1(ϕ(f(g1(v1),,gn(vn)))),(\phi^{-1}(\phi(x_{1})),\cdots,\phi^{-1}(\phi(x_{m})))\mapsto\phi^{-1}(\phi(f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})))),

(x1,,xm)Sm\forall(x_{1},\cdots,x_{m})\in S^{m} such that f(g1(v1),,gn(vn))f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})) is well-defined.

Since we showed that (f(g1,,gn))(f\circ({{g}_{1}},\cdots,{{g}_{n}}))^{*} can be given exactly the same as above, Equation (22.1) holds, as desired. ∎

Notation 22.1.8.

By Tϕ#T_{\phi}^{\#} we denote the set {\{Id on Qϕ}TϕQ_{\phi}\}\bigcup T_{\phi}^{*}.

Corollary 22.1.9.

Tϕ#T_{\phi}^{\#} is a par-operator gen-semigroup on QϕQ_{\phi} and QϕQ_{\phi} is a Tϕ#T_{\phi}^{\#}-space. Specifically, Qϕ=QϕTϕ#Q_{\phi}={{\left\langle Q_{\phi}\right\rangle}_{T_{\phi}^{\#}}}.

Proof.

By Proposition 22.1.7, TϕT_{\phi}^{*} is a par-operator gen-semigroup on QϕQ_{\phi}, so is Tϕ#(={T_{\phi}^{\#}(=\{Id on Qϕ}Tϕ)Q_{\phi}\}\bigcup T_{\phi}^{*}). Hence QϕQϕTϕ#Q_{\phi}\supseteq{{\left\langle Q_{\phi}\right\rangle}_{T_{\phi}^{\#}}}. Since Id Tϕ#\in T_{\phi}^{\#}, QϕQϕTϕ#Q_{\phi}\subseteq{{\left\langle Q_{\phi}\right\rangle}_{T_{\phi}^{\#}}}. Thus, Qϕ=QϕTϕ#Q_{\phi}={{\left\langle Q_{\phi}\right\rangle}_{T_{\phi}^{\#}}}. ∎

QϕQ_{\phi} in Corollary 22.1.9 is actually a “quotient space” of SS corresponding to ϕ\phi. To see this, we give

Example 22.1.10.

Let ϕ\phi be the same as in Example 22.1.3. Let KK be the kernel of ϕ\phi. Then the Tϕ#T_{\phi}^{\#}-space QϕQ_{\phi} is the quotient group S/KS/K, where Tϕ#T_{\phi}^{\#} is the set of all group operations among (a finite number of) elements of S/KS/K. Moreover, ϕ\phi^{*} defined as follows is the group isomorphism from S/KS/K to Imϕ\operatorname{Im}\phi induced by ϕ\phi.

22.2. A generalized first isomorphism theorem

Notation 22.2.1.

Let ϕ:QϕImϕ\phi^{*}:Q_{\phi}\to\operatorname{Im}\phi be given by ϕ1(z)z\phi^{-1}(z)\mapsto z and let θ={(f,g)|(f,g)θ}\theta^{*}=\{(f^{*},g)\,|\,(f,g)\in\theta\}.

Then θTϕ#×T\theta^{*}\subseteq{T_{\phi}^{\#}}\times{T^{\prime}}. We are going to show that ϕ\phi^{*} is a θ\theta^{*}-isomorphism from the Tϕ#T_{\phi}^{\#}-space QϕQ_{\phi} to Imϕ\operatorname{Im}\phi if Imϕ\operatorname{Im}\phi is a TT^{\prime}-space. But first, we need to show that ϕ\phi^{*} is a θ\theta^{*}-morphism, which requires a lemma as follows.

Lemma 22.2.2.

Suppose that θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map. Then θ|Imϕ\theta^{*}{{|}_{\operatorname{Im}\phi^{*}}} is also a map. Moreover, θ(f)|Imϕ=θ(f)|Imϕ,fDomθ\theta^{*}(f^{*})|_{\operatorname{Im}\phi^{*}}=\theta(f)|_{\operatorname{Im}\phi},\forall f\in\operatorname{Dom}\theta.

Proof.

Apparently Imϕ=Imϕ\operatorname{Im}\phi^{*}=\operatorname{Im}\phi. Thus, to show that θ|Imϕ\theta^{*}{{|}_{\operatorname{Im}\phi^{*}}} is a map, it suffices to show that θ|Imϕ\theta^{*}{{|}_{\operatorname{Im}\phi}} is a map.

For this purpose, let f1,f2Tf_{1},f_{2}\in T such that f1=f2f_{1}^{*}=f_{2}^{*}. Let (f1,g1),(f2,g2)θ(f_{1},g_{1}),(f_{2},g_{2})\in\theta, and hence (f1,g1),(f2,g2)θ(f_{1}^{*},g_{1}),(f_{2}^{*},g_{2})\in\theta^{*}. Let n+n\in{{\mathbb{Z}}^{+}} and (a1,,an),(b1,,bn)Sn(a_{1},\cdots,a_{n}),(b_{1},\cdots,b_{n})\in S^{n} such that ϕ(ai)=ϕ(bi),i=1,,n\phi(a_{i})=\phi(b_{i}),\forall i=1,\cdots,n.

Then to show that θ|Imϕ\theta^{*}{{|}_{\operatorname{Im}\phi}} is a map, by Proposition 9.5.2, it suffices to show that neither g1(ϕ(a1),,ϕ(an))g_{1}(\phi(a_{1}),\cdots,\phi(a_{n})) nor g2(ϕ(b1),,ϕ(bn))g_{2}(\phi(b_{1}),\cdots,\phi(b_{n})) is well-defined or g1(ϕ(a1),,ϕ(an))=g2(ϕ(b1),,ϕ(bn))g_{1}(\phi(a_{1}),\cdots,\phi(a_{n}))=g_{2}(\phi(b_{1}),\cdots,\phi(b_{n})):
g1(ϕ(a1),,ϕ(an))g_{1}(\phi(a_{1}),\cdots,\phi(a_{n})) is well-defined;
\Leftrightarrow θ(f1)(ϕ(a1),,ϕ(an))\theta(f_{1})(\phi(a_{1}),\cdots,\phi(a_{n})) is well-defined (since (f1,g1)θ(f_{1},g_{1})\in\theta and θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map);
\Leftrightarrow f1(a1,,an)f_{1}(a_{1},\cdots,a_{n}) is well-defined (by (ii) in Definition 9.5.5);
\Leftrightarrow f1(ϕ1(ϕ(a1)),,ϕ1(ϕ(an)))f_{1}^{*}(\phi^{-1}(\phi(a_{1})),\cdots,\phi^{-1}(\phi(a_{n}))) is well-defined (by Corollary 22.1.5);
\Leftrightarrow f2(ϕ1(ϕ(b1)),,ϕ1(ϕ(bn)))f_{2}^{*}(\phi^{-1}(\phi(b_{1})),\cdots,\phi^{-1}(\phi(b_{n}))) is well-defined (since f1=f2f_{1}^{*}=f_{2}^{*} and ϕ(ai)=ϕ(bi),i=1,,n\phi(a_{i})=\phi(b_{i}),\forall i=1,\cdots,n);
\Leftrightarrow f2(b1,,bn)f_{2}(b_{1},\cdots,b_{n}) is well-defined (by Corollary 22.1.5);
\Leftrightarrow θ(f2)(ϕ(b1),,ϕ(bn))\theta(f_{2})(\phi(b_{1}),\cdots,\phi(b_{n})) is well-defined (by (ii) in Definition 9.5.5);
\Leftrightarrow g2(ϕ(b1),,ϕ(bn))g_{2}(\phi(b_{1}),\cdots,\phi(b_{n})) is well-defined (since (f2,g2)θ(f_{2},g_{2})\in\theta and θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map).

And in the case where the above equivalent conditions are satisfied,

g1(ϕ(a1),,ϕ(an))\displaystyle g_{1}(\phi(a_{1}),\cdots,\phi(a_{n})) =θ(f1)(ϕ(a1),,ϕ(an))\displaystyle=\theta(f_{1})(\phi(a_{1}),\cdots,\phi(a_{n}))
(by (ii) in Definition 9.5.5)\displaystyle(\text{by (ii) in Definition \ref{10.4.2}}) =ϕ(f1(a1,,an))\displaystyle=\phi(f_{1}(a_{1},\cdots,a_{n}))
(by the definition of ϕ)\displaystyle(\text{by the definition of }\phi^{*}) =ϕ(ϕ1(ϕ(f1(a1,,an))))\displaystyle=\phi^{*}(\phi^{-1}(\phi(f_{1}(a_{1},\cdots,a_{n}))))
(by Corollary 22.1.5)\displaystyle(\text{by Corollary \ref{24.5}}) =ϕ(f1(ϕ1(ϕ(a1)),,ϕ1(ϕ(an))))\displaystyle=\phi^{*}(f_{1}^{*}(\phi^{-1}(\phi(a_{1})),\cdots,\phi^{-1}(\phi(a_{n}))))
(since f1=f2 and ϕ(ai)=ϕ(bi),i=1,,n)\displaystyle(\text{since }f_{1}^{*}=f_{2}^{*}\text{ and }\phi(a_{i})=\phi(b_{i}),\forall i=1,\cdots,n) =ϕ(f2(ϕ1(ϕ(b1)),,ϕ1(ϕ(bn))))\displaystyle=\phi^{*}(f_{2}^{*}(\phi^{-1}(\phi(b_{1})),\cdots,\phi^{-1}(\phi(b_{n}))))
(by Corollary 22.1.5)\displaystyle(\text{by Corollary \ref{24.5}}) =ϕ(ϕ1(ϕ(f2(b1,,bn))))\displaystyle=\phi^{*}(\phi^{-1}(\phi(f_{2}(b_{1},\cdots,b_{n}))))
(by the definition of ϕ)\displaystyle(\text{by the definition of }\phi^{*}) =ϕ(f2(b1,,bn))\displaystyle=\phi(f_{2}(b_{1},\cdots,b_{n}))
(by (ii) in Definition 9.5.5)\displaystyle(\text{by (ii) in Definition \ref{10.4.2}}) =θ(f2)(ϕ(b1),,ϕ(bn))\displaystyle=\theta(f_{2})(\phi(b_{1}),\cdots,\phi(b_{n}))
=g2(ϕ(b1),,ϕ(bn)),\displaystyle=g_{2}(\phi(b_{1}),\cdots,\phi(b_{n})),

as desired.

Therefore, θ|Imϕ\theta^{*}{{|}_{\operatorname{Im}\phi}} is a map, so is θ|Imϕ\theta^{*}{{|}_{\operatorname{Im}\phi^{*}}}.

Moreover, let (f,g)θ(f,g)\in\theta. Then (f,g)θ(f^{*},g)\in\theta^{*}. Since both θ|Imϕ\theta^{*}{{|}_{\operatorname{Im}\phi^{*}}} and θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} are maps,

θ(f)|Imϕ=g|Imϕ=g|Imϕ=θ(f)|Imϕ.\theta^{*}(f^{*})|_{\operatorname{Im}\phi^{*}}=g|_{\operatorname{Im}\phi^{*}}=g|_{\operatorname{Im}\phi}=\theta(f)|_{\operatorname{Im}\phi}.

Proposition 22.2.3.

ϕ\phi^{*} is a θ\theta^{*}-morphism from the Tϕ#T_{\phi}^{\#}-space QϕQ_{\phi} to the TT^{\prime}-space SS^{\prime}.

Proof.

Since ϕ\phi is a θ\theta-morphism from SS to a TT^{\prime}-space, by Definition 9.5.5,

  1. (i)

    θ|Imϕ\theta{{|}_{\operatorname{Im}\phi}} is a map and

  2. (ii)

    n+\forall n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in{{S}^{n}} and fDomθf\in\operatorname{Dom}\theta, neither f(a1,,an)f(a_{1},\cdots,a_{n}) nor θ(f)(ϕ(a1),,ϕ(an))\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})) is well-defined or

    ϕ(f(a1,,an))=θ(f)(ϕ(a1),,ϕ(an)).\phi(f(a_{1},\cdots,a_{n}))=\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).

Then by Lemma 22.2.2, θ|Imϕ\theta^{*}{{|}_{\operatorname{Im}\phi^{*}}} is a map and θ(f)|Imϕ=θ(f)|Imϕ,fDomθ\theta^{*}(f^{*})|_{\operatorname{Im}\phi^{*}}=\theta(f)|_{\operatorname{Im}\phi},\forall f\in\operatorname{Dom}\theta. Hence to show that ϕ\phi^{*} is a θ\theta^{*}-morphism, again by Definition 9.5.5, we only need to show that n+\forall n\in{{\mathbb{Z}}^{+}}, (C1,,Cn)Qϕn(C_{1},\cdots,C_{n})\in{Q_{\phi}^{n}} and fDomθf^{*}\in\operatorname{Dom}\theta^{*}, neither f(C1,,Cn)f^{*}(C_{1},\cdots,C_{n}) nor θ(f)(ϕ(C1),,ϕ(Cn))\theta^{*}(f^{*})(\phi^{*}({C_{1}}),\cdots,\phi^{*}({C_{n}})) is well-defined or

ϕ(f(C1,,Cn))=θ(f)(ϕ(C1),,ϕ(Cn)).\phi^{*}(f^{*}(C_{1},\cdots,C_{n}))=\theta^{*}(f^{*})(\phi^{*}({C_{1}}),\cdots,\phi^{*}({C_{n}})).

For this purpose, let n+n\in{{\mathbb{Z}}^{+}}, (a1,,an)Sn(a_{1},\cdots,a_{n})\in S^{n} and fDomθf\in\operatorname{Dom}\theta.

Then it suffices to show that neither f(ϕ1(ϕ(a1)),,ϕ1(ϕ(an)))f^{*}(\phi^{-1}(\phi(a_{1})),\cdots,\phi^{-1}(\phi(a_{n}))) nor θ(f)(ϕ(a1),,ϕ(an))\theta^{*}(f^{*})(\phi(a_{1}),\cdots,\phi(a_{n})) is well-defined or

ϕ(f(ϕ1(ϕ(a1)),,ϕ1(ϕ(an))))=θ(f)(ϕ(a1),,ϕ(an)):\phi^{*}(f^{*}(\phi^{-1}(\phi(a_{1})),\cdots,\phi^{-1}(\phi(a_{n}))))=\theta^{*}(f^{*})(\phi(a_{1}),\cdots,\phi(a_{n})):

f(ϕ1(ϕ(a1)),,ϕ1(ϕ(an)))f^{*}(\phi^{-1}(\phi(a_{1})),\cdots,\phi^{-1}(\phi(a_{n}))) is well-defined;
\Leftrightarrow f(a1,,an)f(a_{1},\cdots,a_{n}) is well-defined (by Corollary 22.1.5);
\Leftrightarrow θ(f)(ϕ(a1),,ϕ(an))\theta(f)(\phi(a_{1}),\cdots,\phi(a_{n})) is well-defined (by (ii) in Definition 9.5.5);
\Leftrightarrow θ(f)(ϕ(a1),,ϕ(an))\theta^{*}(f^{*})(\phi(a_{1}),\cdots,\phi(a_{n})) is well-defined (since θ(f)|Imϕ=θ(f)|Imϕ\theta^{*}(f^{*})|_{\operatorname{Im}\phi^{*}}=\theta(f)|_{\operatorname{Im}\phi}).

And in the case where the above equivalent conditions are satisfied,
ϕ(f(ϕ1(ϕ(a1)),,ϕ1(ϕ(an))))\phi^{*}(f^{*}(\phi^{-1}(\phi(a_{1})),\cdots,\phi^{-1}(\phi(a_{n}))))
=ϕ(ϕ1(ϕ(f(a1,,an))))=\phi^{*}(\phi^{-1}(\phi(f(a_{1},\cdots,a_{n})))) (by Corollary 22.1.5)
=ϕ(f(a1,,an))=\phi(f(a_{1},\cdots,a_{n})) (by the definition of ϕ\phi^{*})
=θ(f)(ϕ(a1),,ϕ(an))=\theta(f)(\phi(a_{1}),\cdots,\phi(a_{n})) (by (ii) in Definition 9.5.5)
=θ(f)(ϕ(a1),,ϕ(an))=\theta^{*}(f^{*})(\phi(a_{1}),\cdots,\phi(a_{n})) (because θ(f)|Imϕ=θ(f)|Imϕ\theta^{*}(f^{*})|_{\operatorname{Im}\phi^{*}}=\theta(f)|_{\operatorname{Im}\phi}),
as desired. ∎

Then we have the main theorem of this section as follows.

Theorem 22.2.4.

Suppose that Imϕ\operatorname{Im}\phi is a TT^{\prime}-space. Then ϕ\phi^{*} is a θ\theta^{*}-isomorphism from the Tϕ#T_{\phi}^{\#}-space QϕQ_{\phi} to the TT^{\prime}-space Imϕ\operatorname{Im}\phi.

Proof.

Since Imϕ\operatorname{Im}\phi is a TT^{\prime}-space, ϕ\phi is also a θ\theta-morphism to the TT^{\prime}-space Imϕ\operatorname{Im}\phi. Then by Proposition 22.2.3, ϕ\phi^{*} is a θ\theta^{*}-morphism from the Tϕ#T_{\phi}^{\#}-space QϕQ_{\phi} to the TT^{\prime}-space Imϕ\operatorname{Im}\phi. Moreover, because ϕ\phi^{*} is bijective, ϕ\phi^{*} is a θ\theta^{*}-isomorphism from the Tϕ#T_{\phi}^{\#}-space QϕQ_{\phi} to the TT^{\prime}-space Imϕ\operatorname{Im}\phi. ∎

Then by Proposition 10.2.2, the following is immediate.

Corollary 22.2.5.

If Imθ=T\operatorname{Im}\theta=T^{\prime} and IdT\operatorname{Id}\in T^{\prime}, then Imϕ\operatorname{Im}\phi is a TT^{\prime}-space and ϕ\phi^{*} is a θ\theta^{*}-isomorphism from the Tϕ#T_{\phi}^{\#}-space QϕQ_{\phi} to the TT^{\prime}-space Imϕ\operatorname{Im}\phi.

From the corollary together with Propositions 8.5.11, 8.5.8 and 8.5.9, it is not hard to deduce the first isomorphism theorems for groups, rings, and modules (cf. Example 22.1.10).

23. On topological spaces

In Subsection 23.1, we introduce some notions and properties of topological spaces which are related to our theory. To show an application, in Subsection 23.2, we employ Corollaries 3.2.5 and 3.2.6 and obtain the Galois correspondences on topological spaces.

23.1. Basic notions and properties

Most results in this subsection will not be employed in Subsection 23.2, but they may be useful for future research on topological spaces.

To simplify our descriptions, we define some notations for Section 23 as follows.

Notation 23.1.1.

Unless otherwise specified, XX and YY always denote topological spaces. Let 𝒫(X)\mathcal{P}(X) and 𝒫(Y)\mathcal{P}(Y) denote the power sets of XX and YY, respectively. Moreover,

T:=TX:={IdXT:={{T}_{X}}:=\{\text{I}{{\text{d}}_{X}}, ClX:𝒫(X)𝒫(X)\text{C}{{\text{l}}_{X}}:\mathcal{P}(X)\to\mathcal{P}(X) given by AXAX¯}{{A}_{X}}\mapsto\overline{{{A}_{X}}}\}

and

TY:={IdY{{T}_{Y}}:=\{\text{I}{{\text{d}}_{Y}}, ClY:𝒫(Y)𝒫(Y)\text{C}{{\text{l}}_{Y}}:\mathcal{P}(Y)\to\mathcal{P}(Y) given by AYAY¯}{{A}_{Y}}\mapsto\overline{{{A}_{Y}}}\},

where IdX\text{I}{{\text{d}}_{X}} and IdY\text{I}{{\text{d}}_{Y}} denote the identity functions on 𝒫(X)\mathcal{P}(X) and 𝒫(Y)\mathcal{P}(Y), respectively.

Remark.

By Definition 2.1.1, it is easy to see that T(=TX)T(={{T}_{X}}) and TY{{T}_{Y}} are operator semigroups on 𝒫(X)\mathcal{P}(X) and 𝒫(Y)\mathcal{P}(Y), respectively.

Then we have

Proposition 23.1.2.

𝒫(X)\mathcal{P}(X) is a TT-space and

EndT(𝒫(X))={σ:𝒫(X)𝒫(X)|σ(A¯)=σ(A)¯,A𝒫(X)}\operatorname{End}_{T}(\mathcal{P}(X))=\{\sigma:\mathcal{P}(X)\to\mathcal{P}(X)\,|\,\sigma(\overline{A})=\overline{\sigma(A)},\forall A\in\mathcal{P}(X)\}.

Proof.

By Definition 2.2.1, 𝒫(X)\mathcal{P}(X) is a TT-space because 𝒫(X)T=P(X){{\left\langle\mathcal{P}(X)\right\rangle}_{T}}=P(X).

Since T={IdXT=\{\text{I}{{\text{d}}_{X}}, ClX:𝒫(X)𝒫(X)\text{C}{{\text{l}}_{X}}:\mathcal{P}(X)\to\mathcal{P}(X) given by AXAX¯}{{A}_{X}}\mapsto\overline{{{A}_{X}}}\}, by Definition 2.3.1, a map σ:𝒫(X)𝒫(X)\sigma:\mathcal{P}(X)\to\mathcal{P}(X) is a TT-morphism if and only if

(σ(ClX(A))=)σ(A¯)=σ(A)¯(=ClX(σ(A))),A𝒫(X)(\sigma(\text{C}{{\text{l}}_{X}}(A))=)\sigma(\overline{A})=\overline{\sigma(A)}(=\text{C}{{\text{l}}_{X}}(\sigma(A))),\forall A\in\mathcal{P}(X).

Thus,

EndT(𝒫(X))={σ:𝒫(X)𝒫(X)|σ(A¯)=σ(A)¯,A𝒫(X)}.\operatorname{End}_{T}(\mathcal{P}(X))=\{\sigma:\mathcal{P}(X)\to\mathcal{P}(X)\,|\,\sigma(\overline{A})=\overline{\sigma(A)},\forall A\in\mathcal{P}(X)\}.

For brevity, we take out a part from Proposition 6.1.3 as follows.

Definition 23.1.3.

Let f:XYf:X\to Y be a map. Then we define its induced map f:𝒫(X)𝒫(Y)f^{*}:\mathcal{P}(X)\to\mathcal{P}(Y) by

  1. (1)

    A𝒫(X)\forall A\in\mathcal{P}(X) that is closed in XX, let f(A)=f(A)¯f^{*}(A)=\overline{f(A)}, and

  2. (2)

    A𝒫(X)\forall A\in\mathcal{P}(X) that is not closed in XX, let f(A)=f(A)f^{*}(A)=f(A),

where f(A)f(A) denotes the set {f(x)|xA}\{f(x)\,|\,x\in A\} for convenience.

Remark.

In Section 23, unless otherwise specified, ff^{*} is defined by Definition 23.1.3 whenever ff is a map between topological spaces.

With Definition 23.1.3, we can restate Proposition 2.3.5 as follows.

Proposition 23.1.4.

A map f:XXf:X\to X is continuous if and only if its induced map f:𝒫(X)𝒫(X)f^{*}:\mathcal{P}(X)\to\mathcal{P}(X) is a TT-morphism.

To facilitate our following discussion, we give

Notation 23.1.5.

Let F(X,Y)F(X,Y) denote the set of all functions from XX to YY and let F(X,Y)F^{*}(X,Y) denote the set {f|fF(X,Y)}\{f^{*}\,|\,f\in F(X,Y)\}. Let F(X)=F(X,X)F(X)=F(X,X) and let F(X)=F(X,X)F^{*}(X)=F^{*}(X,X). Moreover, let C(X)C(X) denote the set of all continuous functions from XX to XX, and let C(X)C^{*}(X) denote the set {f|fC(X)}\{f^{*}\,|\,f\in C(X)\}.

Then by Proposition 23.1.4, the following is obvious.

Corollary 23.1.6.

F(X)EndT(𝒫(X))=C(X)F^{*}(X)\bigcap\operatorname{End}_{T}(\mathcal{P}(X))=C^{*}(X).

It would be desirable if F(X)EndT(𝒫(X))=EndT(𝒫(X))F^{*}(X)\bigcap\operatorname{End}_{T}(\mathcal{P}(X))=\operatorname{End}_{T}(\mathcal{P}(X)). However, it is possible that F(X)EndT(𝒫(X))EndT(𝒫(X))F^{*}(X)\bigcap\operatorname{End}_{T}(\mathcal{P}(X))\subsetneq\operatorname{End}_{T}(\mathcal{P}(X)) (and so C(X)EndT(𝒫(X))C^{*}(X)\subsetneq\operatorname{End}_{T}(\mathcal{P}(X))). That is, there may exist a TT-morphism σ\sigma from 𝒫(X)\mathcal{P}(X) to 𝒫(X)\mathcal{P}(X) such that σ\sigma is not the induced map of any map from XX to XX, as shown below.

Example 23.1.7.

Let XX be any topological space with the discrete topology. Then it follows from Proposition 23.1.2 that any map from 𝒫(X)\mathcal{P}(X) to 𝒫(X)\mathcal{P}(X) is a TT-morphism. Let σ:𝒫(X)𝒫(X)\sigma:\mathcal{P}(X)\to\mathcal{P}(X) be given by A,A𝒫(X)A\mapsto\emptyset,\forall A\in\mathcal{P}(X). If XX is nonempty, then by Definition 23.1.3, it is not hard to see that σF(X)\sigma\notin F^{*}(X).

Before going further, we define

Notation 23.1.8.

We denote by IX,Y{{I}_{X,Y}} the map from F(X,Y)F(X,Y) to F(X,Y)F^{*}(X,Y) given by fff\mapsto f^{*}. Besides, let IX=IX,X{{I}_{X}}={{I}_{X,X}}.

Obviously IX,Y{{I}_{X,Y}} is surjective. It would be desirable if IX,Y{{I}_{X,Y}} were always injective. However, there are cases where IX,Y{{I}_{X,Y}} is not injective, as shown below.

Example 23.1.9.

Let X={1}X=\{1\} and Y={2,3}Y=\{2,3\}. Let 𝒯X={,X}{{\mathcal{T}}_{X}}=\{\emptyset,X\} and 𝒯Y={,Y}{{\mathcal{T}}_{Y}}=\{\emptyset,Y\} be the (trivial) topologies on XX and YY, respectively. Let two maps f,g:XYf,g:X\to Y be given by f(1)=2f(1)=2 and g(1)=3g(1)=3, respectively. Then F(X,Y)={f,g}F(X,Y)=\{f,g\}. And by Definition 23.1.3, f()==g()f^{*}(\emptyset)=\emptyset=g^{*}(\emptyset) and f({1})={f(1)}¯={2}¯={2,3}={3}¯={g(1)}¯=g({1})f^{*}(\{1\})=\overline{\{f(1)\}}=\overline{\{2\}}=\{2,3\}=\overline{\{3\}}=\overline{\{g(1)\}}=g^{*}(\{1\}). Thus f=gf^{*}=g^{*}. Hence the map IX,Y{{I}_{X,Y}} from F(X,Y)F(X,Y) to F(X,Y)={f(=g)}F^{*}(X,Y)=\{f^{*}(=g^{*})\} is not injective.

Surprisingly, there are two important cases where IX,Y{{I}_{X,Y}} is injective, as shown by the following two results.

Proposition 23.1.10.

Suppose that every finite point set in YY is closed, i.e. YY satisfies the T1{{T}_{1}} axiom. Then IX,Y{{I}_{X,Y}} is injective.

Proof.

Assume that IX,Y{{I}_{X,Y}} is not injective. Then there are two different maps f,g:XYf,g:X\to Y such that f=gf^{*}=g^{*}. Hence xX\exists x\in X such that f(x)g(x)f(x)\neq g(x).

From Definition 23.1.3, we can tell that {x}\{x\} is closed in XX (otherwise f({x})={f(x)}{g(x)}=g({x})f^{*}(\{x\})=\{f(x)\}\neq\{g(x)\}=g^{*}(\{x\}), which is contrary to the assumption f=gf^{*}=g^{*}). Then again by Definition 23.1.3, {f(x)}¯=f({x})=g({x})={g(x)}¯\overline{\{f(x)\}}=f^{*}(\{x\})=g^{*}(\{x\})=\overline{\{g(x)\}}. However, since every finite point set in YY is closed, {f(x)}¯={f(x)}\overline{\{f(x)\}}=\{f(x)\} and {g(x)}¯={g(x)}\overline{\{g(x)\}}=\{g(x)\}. Thus f(x)=g(x)f(x)=g(x), which is contrary to f(x)g(x)f(x)\neq g(x). ∎

Proposition 23.1.11.

IX{{I}_{X}} is injective.

Proof.

Assume that IX{{I}_{X}} is not injective. Then there are two different maps f,g:XXf,g:X\to X such that f=gf^{*}=g^{*}. Hence xX\exists x\in X such that f(x)g(x)f(x)\neq g(x). Then by the same argument as in the proof of Proposition 23.1.10, {x}\{x\} is closed in XX and {f(x)}¯={g(x)}¯\overline{\{f(x)\}}=\overline{\{g(x)\}}.

It follows that {f(x)}¯={g(x)}¯{f(x),g(x)}\overline{\{f(x)\}}=\overline{\{g(x)\}}\supseteq\{f(x),g(x)\}. Hence {f(x)}¯{f(x)}\overline{\{f(x)\}}\neq\{f(x)\} and {g(x)}¯{g(x)}\overline{\{g(x)\}}\neq\{g(x)\} (because f(x)g(x)f(x)\neq g(x)), and thus f(x)xf(x)\neq x (otherwise {f(x)}¯={x}¯={x}={f(x)}\overline{\{f(x)\}}=\overline{\{x\}}=\{x\}=\{f(x)\}, a contradiction). Analogously, g(x)xg(x)\neq x.

Then {x,f(x)}¯{x,f(x),g(x)}{x,f(x)}\overline{\{x,f(x)\}}\supseteq\{x,f(x),g(x)\}\supsetneq\{x,f(x)\} because {f(x)}¯{f(x),g(x)}\overline{\{f(x)\}}\supseteq\{f(x),g(x)\} and f(x)g(x)xf(x)\neq g(x)\neq x. Hence {x,f(x)}¯{x,f(x)}\overline{\{x,f(x)\}}\neq\{x,f(x)\}; that is, {x,f(x)}\{x,f(x)\} is not closed in XX. Hence by Definition 23.1.3,

f({x,f(x)})={f(x),f(f(x))}f^{*}(\{x,f(x)\})=\{f(x),f(f(x))\} and g({x,f(x)})={g(x),g(f(x))}g^{*}(\{x,f(x)\})=\{g(x),g(f(x))\}.

Because it is assumed that f=gf^{*}=g^{*}, {f(x),f(f(x))}={g(x),g(f(x))}\{f(x),f(f(x))\}=\{g(x),g(f(x))\}.

Since f(x)g(x)f(x)\neq g(x), we can tell from the above equation that f(f(x))=g(x)f(x)=g(f(x))f(f(x))=g(x)\neq f(x)=g(f(x)). Then because it was shown that {f(x)}¯{f(x)}\overline{\{f(x)\}}\neq\{f(x)\}, by Definition 23.1.3, f({f(x)})={f(f(x))}{g(f(x))}=g({f(x)})f^{*}(\{f(x)\})=\{f(f(x))\}\neq\{g(f(x))\}=g^{*}(\{f(x)\}), which is contrary to the assumption that f=gf^{*}=g^{*}. ∎

Moreover, there is an important case where IX{{I}_{X}} commutes with composition of functions, as shown below.

Lemma 23.1.12.

Let ff and gg be continuous maps from XX to XX. Suppose that gg is injective. Then (fg)=fg(f\circ g)^{*}=f^{*}\circ g^{*}.

Proof.

It suffices to show that (fg)(A)=f(g(A)),A𝒫(X)(f\circ g)^{*}(A)=f^{*}(g^{*}(A)),\forall A\in\mathcal{P}(X). There are two cases as follows.

(1) A=A¯A=\overline{A}:

By Definition 23.1.3,

(fg)(A)=(fg)(A)¯=f(g(A))¯(f\circ g)^{*}(A)=\overline{(f\circ g)(A)}=\overline{f(g(A))}

and

f(g(A))=f(g(A)¯)=f(g(A)¯)¯.f^{*}(g^{*}(A))=f^{*}(\overline{g(A)})=\overline{f(\overline{g(A)})}.

Then it suffices to show f(g(A)¯)¯=f(g(A))¯\overline{f(\overline{g(A)})}=\overline{f(g(A))}.

Since ff is continuous, by e.g. [6], f(g(A)¯)f(g(A))¯f(\overline{g(A)})\subseteq\overline{f(g(A))}. Hence f(g(A)¯)¯f(g(A))¯\overline{f(\overline{g(A)})}\subseteq\overline{f(g(A))}. On the other hand, f(g(A)¯)f(g(A))f(\overline{g(A)})\supseteq f(g(A)), and hence f(g(A)¯)¯f(g(A))¯\overline{f(\overline{g(A)})}\supseteq\overline{f(g(A))}. Therefore, f(g(A)¯)¯=f(g(A))¯\overline{f(\overline{g(A)})}=\overline{f(g(A))}, as desired.

(2) AA¯A\neq\overline{A}:

Assume that g(A)g(A) is closed. Since gg is continuous, g1(g(A)){{g}^{-1}}(g(A)) is also closed. However, since gg is injective, g1(g(A))=A{{g}^{-1}}(g(A))=A. Hence g1(g(A)){{g}^{-1}}(g(A)) is not closed (because AA¯A\neq\overline{A}), a contradiction. Thus g(A)g(A) is not closed. Then by Definition 23.1.3,

f(g(A))=f(g(A))=f(g(A))=(fg)(A)=(fg)(A),f^{*}(g^{*}(A))=f^{*}(g(A))=f(g(A))=(f\circ g)(A)=(f\circ g)^{*}(A),

as desired. ∎

Notation 23.1.13.

We denote by Hom(X)\text{Hom}(X) the set of all homeomorphisms from XX to XX. Let Hom(X)={f|fHom(X)}\text{Hom}^{*}(X)=\{f^{*}\,|\,f\in\text{Hom}(X)\}.

Then we immediately have

Corollary 23.1.14.

Hom(X)\emph{Hom}(X) constitutes a group, which we still denote by Hom(X)\emph{Hom}(X), with composition of functions as the binary operation.

Hom(X)\text{Hom}^{*}(X) also constitutes a group, as shown below.

Proposition 23.1.15.

With composition of functions as the binary operation, Hom(X)\emph{Hom}^{*}(X) constitutes a subgroup, which we still denote by Hom(X)\emph{Hom}^{*}(X), of AutT(𝒫(X))\operatorname{Aut}_{T}(\mathcal{P}(X)) ((defined in Proposition 2.4.3)).

Remark.

Recall that TT was defined in Notation 23.1.1.

Proof.

It follows from Lemma 23.1.12 that Hom(X)\text{Hom}^{*}(X) is closed under composition of functions as the binary operation; that is, f,gHom(X)\forall f^{*},g^{*}\in\text{Hom}^{*}(X), fgHom(X)f^{*}\circ g^{*}\in\text{Hom}^{*}(X). Composition of functions is associative. Moreover, by Definition 23.1.3, the identity map on XX induces the identity map on 𝒫(X)\mathcal{P}(X), which we denote by Id(Hom(X))\operatorname{Id}^{*}(\in\text{Hom}^{*}(X)).

By Definition 23.1.3, it is not hard to see that fHom(X)\forall f^{*}\in\text{Hom}^{*}(X), ff^{*} is bijective. fHom(X)\forall f\in\text{Hom}(X), let (f)1=(f1){{(f^{*})}^{-1}}=({{f}^{-1}})^{*}. Then fHom(X)\forall f\in\text{Hom}(X), by Lemma 23.1.12, (f)1f=(f1)f=(f1f)=Id{{(f^{*})}^{-1}}\circ f^{*}=({{f}^{-1}})^{*}\circ f^{*}=({{f}^{-1}}\circ f)^{*}=\operatorname{Id}^{*}, and analogously, f(f)1=Idf^{*}\circ{{(f^{*})}^{-1}}=\operatorname{Id}^{*}.

Therefore, Hom(X)\text{Hom}^{*}(X) constitutes a group with composition of functions as the binary operation.

Then from Proposition 23.1.4, we can tell that Hom(X)AutT(𝒫(X))\text{Hom}^{*}(X)\subseteq\operatorname{Aut}_{T}(\mathcal{P}(X)). By Proposition 2.4.3, AutT(𝒫(X))\operatorname{Aut}_{T}(\mathcal{P}(X)) constitutes a group with composition of functions as the binary operation. Thus Hom(X)\text{Hom}^{*}(X) constitutes a subgroup of AutT(𝒫(X))\operatorname{Aut}_{T}(\mathcal{P}(X)). ∎

Moreover, Hom(X)\text{Hom}^{*}(X) is isomorphic to Hom(X)\text{Hom}(X), as shown below.

Proposition 23.1.16.

IX|Hom(X){{I}_{X}}{{|}_{\emph{Hom}(X)}}, which is the restriction of IX{{I}_{X}} to Hom(X)\emph{Hom}(X), is a group isomorphism from Hom(X)\emph{Hom}(X) to Hom(X)\emph{Hom}^{*}(X).

Remark.

IXI_{X} was defined by Notation 23.1.8.

Proof.

By Proposition 23.1.11, IX|Hom(X){{I}_{X}}{{|}_{\text{Hom}(X)}} is an injective map from Hom(X)\text{Hom}(X) to Hom(X)(={f|fHom(X)})\text{Hom}^{*}(X)(=\{f^{*}\,|\,f\in\text{Hom}(X)\}). It follows that IX|Hom(X){{I}_{X}}{{|}_{\text{Hom}(X)}} is a bijective map between the two groups. And by Lemma 23.1.12,

IX(fg)=IX(f)IX(g),f,gHom(X).{{I}_{X}}(f\circ g)={{I}_{X}}(f)\circ{{I}_{X}}(g),\forall f,g\in\text{Hom}(X).

Therefore, IX|Hom(X){{I}_{X}}{{|}_{\text{Hom}(X)}} is a group isomorphism. ∎

23.2. Galois correspondences on topological spaces

Firstly, we need to recall some notations. By Definitions 2.5.1, 2.6.1, 3.2.1 and 3.2.3, we immediately have

Proposition 23.2.1.

Let B𝒫(X)B\subseteq\mathcal{P}(X) and let HEndT(𝒫(X))H\subseteq\operatorname{End}_{T}(\mathcal{P}(X)). Then

𝒫(X)H={A𝒫(X)|σ(A)=A,σH},\mathcal{P}{{(X)}^{H}}=\{A\in\mathcal{P}(X)\,|\,\sigma(A)=A,\forall\sigma\in H\},
GMnT(𝒫(X)/B)={σEndT(𝒫(X))|σ(A)=A,AB},\operatorname{GMn}_{T}(\mathcal{P}(X)/B)=\{\sigma\in\operatorname{End}_{T}(\mathcal{P}(X))\,|\,\sigma(A)=A,\forall A\in B\},
IntTEnd(𝒫(X)/B)={𝒫(X)LB|LEndT(𝒫(X))},\operatorname{Int}_{T}^{\operatorname{End}}(\mathcal{P}(X)/B)=\{\mathcal{P}{{(X)}^{L}}\supseteq B\,|\,L\subseteq\operatorname{End}_{T}(\mathcal{P}(X))\},

and

GSMnT(𝒫(X)/B)={GMnT(𝒫(X)/K)|BK𝒫(X)}.\operatorname{GSMn}_{T}(\mathcal{P}(X)/B)=\{\operatorname{GMn}_{T}(\mathcal{P}(X)/K)\,|\,B\subseteq K\subseteq\mathcal{P}(X)\}.
Remark.

EndT(𝒫(X))\operatorname{End}_{T}(\mathcal{P}(X)) was given in Proposition 23.1.2.

Then by applying Corollary 3.2.5 to a topological space XX, we immediately get the following, which, roughly speaking, shows the Galois correspondence between the Galois TT-monoids of 𝒫(X)\mathcal{P}(X) and the fixed subsets of 𝒫(X)\mathcal{P}(X) under the actions of the TT-endomorphisms of 𝒫(X)\mathcal{P}(X).

Corollary 23.2.2.

Let XX be a topological space, let 𝒫(X)\mathcal{P}(X) be the power set of XX, and let B𝒫(X)B\subseteq\mathcal{P}(X). Then the correspondences

γ:H𝒫(X)H\gamma:H\mapsto\mathcal{P}{{(X)}^{H}} and δ:KGMnT(𝒫(X)/K)\delta:K\mapsto\operatorname{GMn}_{T}(\mathcal{P}(X)/K)

define inclusion-inverting mutually inverse bijective maps between GSMnT(𝒫(X)/B)\operatorname{GSMn}_{T}(\mathcal{P}(X)/B) and IntTEnd(𝒫(X)/B)\operatorname{Int}_{T}^{\operatorname{End}}(\mathcal{P}(X)/B).

Moreover, we may employ Theorem 5.1.2 (resp. Theorem 5.3.2) to investigate whether we can endow 𝒫(X)\mathcal{P}(X) with a topology such that

IntTEnd(𝒫(X)/B)\{}=\operatorname{Int}_{T}^{\operatorname{End}}(\mathcal{P}(X)/B)\backslash\{\emptyset\}={BSq𝒫(X)|SB\subseteq{S}^{\prime}\leq_{q}\mathcal{P}(X)\,|\,{S}^{\prime} is closed and nonempty}

(resp. whether we can define a topology on EndT(𝒫(X)){}_{T}(\mathcal{P}(X)) such that GSMnT(𝒫(X)/B)\operatorname{GSMn}_{T}(\mathcal{P}(X)/B) is the set of all closed submonoids of GMnT(𝒫(X)/B)\operatorname{GMn}_{T}(\mathcal{P}(X)/B)).

We will not go further for this subject.

Analogously, by Proposition 23.1.2 and Definition 2.4.1, we have

AutT(𝒫(X))={{}_{T}(\mathcal{P}(X))=\{bijective σ:𝒫(X)𝒫(X)|σ(A¯)=σ(A)¯,A𝒫(X)}\sigma:\mathcal{P}(X)\to\mathcal{P}(X)\,|\,\sigma(\overline{A})=\overline{\sigma(A)},\forall A\in\mathcal{P}(X)\}.

And by Definitions 2.6.1, 3.2.1 and 3.2.3, we immediately have

Proposition 23.2.3.

Let B𝒫(X)B\subseteq\mathcal{P}(X). Then

GGrT(𝒫(X)/B)={σAutT(𝒫(X))|σ(A)=A,AB},\operatorname{GGr}_{T}(\mathcal{P}(X)/B)=\{\sigma\in\operatorname{Aut}_{T}(\mathcal{P}(X))\,|\,\sigma(A)=A,\forall A\in B\},
IntTAut(𝒫(X)/B)={𝒫(X)HB|HAutT(𝒫(X))},\operatorname{Int}_{T}^{\operatorname{Aut}}(\mathcal{P}(X)/B)=\{\mathcal{P}{{(X)}^{H}}\supseteq B\,|\,H\subseteq\operatorname{Aut}_{T}(\mathcal{P}(X))\},

and

GSGrT(𝒫(X)/B)={GGrT(𝒫(X)/K)|BK𝒫(X)}.\operatorname{GSGr}_{T}(\mathcal{P}(X)/B)=\{\operatorname{GGr}_{T}(\mathcal{P}(X)/K)\,|\,B\subseteq K\subseteq\mathcal{P}(X)\}.

Then by applying Corollary 3.2.6 to a topological space XX, we immediately have the following, which shows the Galois correspondence between the Galois TT-groups of 𝒫(X)\mathcal{P}(X) and the fixed subsets of 𝒫(X)\mathcal{P}(X) under the actions of the TT-automorphisms of 𝒫(X)\mathcal{P}(X).

Corollary 23.2.4.

Let XX be a topological space, let 𝒫(X)\mathcal{P}(X) be the power set of XX, and let B𝒫(X)B\subseteq\mathcal{P}(X). Then the correspondences

γ:H𝒫(X)H\gamma:H\mapsto\mathcal{P}{{(X)}^{H}} and δ:KGGrT(𝒫(X)/K)\delta:K\mapsto\operatorname{GGr}_{T}(\mathcal{P}(X)/K)

define inclusion-inverting mutually inverse bijective maps between GSGrT(𝒫(X)/B)\operatorname{GSGr}_{T}(\mathcal{P}(X)/B) and IntTAut(𝒫(X)/B)\operatorname{Int}_{T}^{\operatorname{Aut}}(\mathcal{P}(X)/B).

Moreover, we may employ Theorem 5.2.2 (resp. Theorem 5.4.2) to investigate whether we can endow 𝒫(X)\mathcal{P}(X) with a topology such t