Semi-coarse Spaces: Fundamental Groupoid and the van Kampen Theorem

Jonathan Treviño-Marroquín

Abstract.

In algebraic topology, the fundamental groupoid is a classical homotopy invariant which is defined using continuous maps from the closed interval to a topological space. In this paper, we construct a semi-coarse version of this invariant, using as paths a finite sequences of maps from $\mathbb{Z}_{1}$ to a semi-coarse space, connecting their tails through semi-coarse homotopy. In contrast to semi-coarse homotopy groups, this groupoid is not necessarily trivial for coarse spaces, and, unlike coarse homotopy, it is well-defined for general semi-coarse spaces. In addition, we show that the semi-coarse fundamental groupoid which we introduce admits a version of the Van Kampen Theorem.

This work was supported by the grant N62909-19-1-2134 from the US Office of Naval Research Global and the Southern Office of Aerospace Research and Development of the US Air Force Office of Scientific Research. The author was also supported by the CONACYT postgraduate studies scholarship number 839062, and this document is part of his PhD project.

1. Introduction

The category of semi-coarse spaces is one of several topological constructs that has been taken up to study algebraic-topological invariants on graphs and semipseudometric spaces, which together serve as a foundation for topological data analysis. Other categories which have been proposed for this purpose include Čech closure spaces [Bubenik_Milicevic_2021b, Rieser_2021] and Choquet Spaces [Rieser_arXiv_2022]. Semi-coarse spaces generalize Roe’s axiomatization of coarse spaces [Roe_2003] by removing the product of sets axiom. This is fundamental because it provides us with a collection of interval objects for semi-coarse spaces, and we can define homotopy maps in this category [rieser2023semicoarse]. Additionally, unlike in coarse geometry, compact (metric) spaces admit semi-coarse structures which are no longer necessarily equivalent to the point, allowing one to coarsen a compact metric space up to a pre-defined scale [Zava_2019] without making it trivial as a semi-coarse space. Canonical examples of semi-coarse spaces are the (undirected) graphs, such as the cyclic graphs and the Cayley graph of the integers with the unit as generator ( $\mathbb{Z}_{1}$ ), and semipseudometric spaces, in particular those constructed from metric spaces with a privileged scale. The loss of the product axiom for coarse spaces produces two possible definitions of bounded sets and fails to fulfill that the union of bounded sets is bounded in connected spaces. Hence, we use bornological maps as our morphisms, which avoids the use of bounded sets to define, but complicates the adaptation of coarse homotopy to semi-coarse spaces (for more on this, see Appendix A).

In previous work, joint with Antonio Rieser [rieser2023semicoarse], we defined homotopy groups for arbitrary semi-coarse spaces. There we observe that, when applied to coarse spaces, $\pi_{0}$ quantifies the number of (coarsely) connected components, and the rest of the homotopy groups vanishe. In this paper, we construct an invariant which may be non-trivial on both coarse and non-coarse semi-coarse spaces, the semi-coarse fundamental groupoid, inspired by the one in topology [Brown_Higgins_Sivera_2011] and by the visual boundary in semi-geodesic spaces [Kitzmiller_2009]. This construction generalizes the semi-coarse fundamental group, and reduces to it under some assumptions. While the definition of the new fundamental groupoid is slightly technical, it is indeed a semi-coarse invariant which is non-trivial on coarse spaces and is able to distinguish them. Moreover, under some specific cases, we have an analogue to van Kampen theorem for semi-coarse spaces. In related work, the fundamental groupoid has been also realized for graphs in $A$ -theory [kapulkin2023fundamental] and for $\times$ -homotopy [Chih_Scull_2022].

The article is organized as follows. We introduce the category of semi-coarse spaces in Section 2. There, we recall some results and definitions of [rieser2023semicoarse], and we introduce semi-coarse homotopy and the fundamental group. In semi-coarse spaces, we do not have a canonical collection of sets with which to construct useful covers, i.e. like open sets in topological spaces, and in Section 3 we partially resolve this problem by defining when a pair of subsets well-split a space (3.6) and we establish several relations between well-splitting, coarse spaces, and connectedness. The definition of well-splitting provides us with a sufficient condition for proving a van Kampen theorem for the semi-coarse fundamental groupoid. In Section 4, we define strings of paths and an equivalence relation on them, and we use these to construct the semi-coarse fundamental groupoid.

In Section 5, we begin with the main results of the paper, and define the relative fundamental groupoid and prove a semi-coarse version of the Van Kampen Theorem give some mild hypotheses. Finally, in Appendix A we discuss several issues which arise when trying to generalize coarse homotopy to both coarse and semi-coarse spaces. We do not discard the possbility that this notion can be extended, but it appears that such a generalization is difficult to construct.

2. Semi-coarse Spaces and its Homotopy

Our first first approximation was looking for a space where we can take some ideas of coarse geometry to study any (undirected) graphs and metric spaces with privileged scale. This category received the name of semi-coarse spaces [Zava_2019] and we developed the homotopy with some ideas of [Babson_etal_2006] using the cartesian product instead the inductive (or box) product. This section is a summary of [rieser2023semicoarse], we enunciate the definitions and results we use in the subsequent sections.

Definition 2.1 (Semi-coarse space; [rieser2023semicoarse]).

Let $X$ be a set, and let $\mathcal{V}\subset\mathcal{P}(X\times X)$ be a collection of subsets of $X\times X$ which satisfies

(sc1)

$\Delta_{X}\in\mathcal{V}$ ,
(sc2)

If $B\in\mathcal{V}$ and $A\subset B$ , then $A\in\mathcal{V}$ ,
(sc3)

If $A,B\in\mathcal{V}$ , then $A\cup B\in\mathcal{V}$ ,
(sc4)

If $A\in\mathcal{V}$ , then $A^{-1}\in\mathcal{V}$ .

We call $\mathcal{V}$ a semi-coarse structure on $X$ , and we say that the pair $(X,\mathcal{V})$ is a semi-coarse space. If, in addition, $\mathcal{V}$ satisfies

(sc5)

If $A,B\in\mathcal{V}$ , then $A\circ B\in\mathcal{V}$ .

then $\mathcal{V}$ will be called a coarse structure, and $(X,\mathcal{V})$ will be called a coarse space, as in [Roe_2003].

A map $f:X\rightarrow Y$ between the coarse spaces $(X,\mathcal{V})$ and $(Y,\mathcal{W})$ is called $(\mathcal{V},\mathcal{W})$ -bornologous if $f\times f(V)\coloneqq\{(f(a),f(b))\mid(a,b)\in V\}\in\mathcal{W}$ for every $V\in\mathcal{V}$ . We observe that semi-coarse spaces with bornologous maps is a category.

Semi-coarse spaces is a topological construct, the practical effect for this paper is that final and initial structures are defined for every collection of semi-coarse spaces [Zava_2019] (to see the formal definition of Topological Constructs and some properties see [Beattie_Butzmann_2002, Preuss_2002]). Some important structures are: subspace, product space, quotients and disjoint union space.

Definition 2.2 (Semi-coarse Subspace; 2.2.3 [rieser2023semicoarse]).

Let $(X,\mathcal{V})$ be a semi-coarse space and let $Y\subset X$ . $(Y,\mathcal{V}_{Y})$ is called a semi-coarse subdspace of $X$ with $\mathcal{V}_{Y}$ the collection

\displaystyle\{V\cap(Y\times Y)\mid V\in\mathcal{V}\}

In the following sense, bornologous maps might be seen locally: If all of the controlled sets might be as the union of the intersections with sets of the way $X_{i}\times X_{i}$ , then we can observe if the maps are bornologous there, and conclude they are bornologous in the whole.

Proposition 2.3 ([rieser2023semicoarse] 2.2.4).

Let $(X,\mathcal{V})$ and $(Y,\mathcal{W})$ be semi-coarse spaces, and suppose that $(X_{i},\mathcal{V}_{i})\subset(X,\mathcal{V})$ , $i\in\{1,\ldots,n\}$ , are subspaces of $(X,\mathcal{V})$ such that $\cup_{i=1}^{n}X_{i}=X$ and every set $V\in\mathcal{V}$ may be written in the form

(1)

\displaystyle V=\bigcup_{i=1}^{n}V_{i}

where each $V_{i}\in\mathcal{V}_{i}$ . Now suppose that $f:X\rightarrow Y$ is a map such that the restrictions $f\mid_{X_{i}}:(X_{i},\mathcal{V}_{i})\rightarrow(Y,\mathcal{W})$ are bornologous for all $i\in\{1,\ldots,n\}$ . Then $f:(X,\mathcal{V})\rightarrow(Y,\mathcal{W})$ .

Definition 2.4.

[Semi-coarse Product; 2.3.7 [rieser2023semicoarse]] Let $(X,\mathcal{V})$ and $(Y,\mathcal{W})$ be semi-coarse spaces. $(X\times Y,\mathcal{V}\times\mathcal{W})$ is called the product space, and $\mathcal{V}\times\mathcal{W}$ the product structure, where $U\in\mathcal{V}\times\mathcal{W}$ if there exists $V\in\mathcal{V}$ and $W\in\mathcal{W}$ such that $U\subset V\boxtimes W$ with

\displaystyle V\boxtimes W\coloneqq\{(v,w,v^{\prime},w^{\prime})\mid(v,v^{\prime})\in V,(w,w^{\prime})\in W\}

Since semi-coarse spaces are a topological construct, inductive product is also defined and it defines another kind of homotopy (see examples in other categories in [Babson_etal_2006, Bubenik_Milicevic_2021b].) For the objectives in this paper, it is enough to work with (categorical) product space; it could be interesting to explore the later definitions with that product.

Definition 2.5 (Quotient Space; 2.4.1 and 2.4.2 [rieser2023semicoarse]).

Let $(X,\mathcal{V})$ be a semi-coarse space, let $Y$ be a set, and let $g:X\rightarrow Y$ be a surjective function. We define

\displaystyle\mathcal{V}_{g}\coloneqq\{(g\times g)(V)\mid V\in\mathcal{V}\}.

Then $(Y,\mathcal{V}_{g})$ is called the quotient space (or semi-coarse space inductively generated by the function $g$ ) and $\mathcal{V}_{g}$ is called the quotient structure (or semi-coarse structure inductively generated by the function $g$ ).

Theorem 2.6 ([rieser2023semicoarse] 2.4.3).

Let $(Y,\mathcal{V}_{g})$ be a quotien semi-coarse space inductively generated by the function $g:X\rightarrow Y$ and let $(Z,\mathcal{Z})$ be a semi-coarse space. A function $f:(Y,\mathcal{V}_{g})\rightarrow(Z,\mathcal{Z})$ is bornologous if and only if $f\circ g:X\rightarrow Z$ is bornologous.

Definition 2.7 (Disjoint Union; 2.5.1 and 2.5.2 [rieser2023semicoarse]).

Let $\{(X_{\lambda},\mathcal{V}_{\lambda})\}_{\lambda\in\Lambda}$ be a collection of semi-coarse spaces indexed by the set $\Lambda$ , and let $\sqcup_{\lambda\in\Lambda}\mathcal{V}_{\lambda}$ be the collection of sets of the form

\displaystyle\bigsqcup_{\lambda\in\Lambda^{\prime}}A_{\lambda},

where $|\Lambda^{\prime}|<\infty$ and each $A_{\lambda}\in\mathcal{V}_{\lambda}$ . (Note that any given $A_{\lambda}$ may be the empty set.)

We call $\sqcup_{\lambda\in\Lambda}\mathcal{V}_{\lambda}$ the disjoint union semi-coarse structure, and $(\sqcup_{\lambda\in\Lambda}X_{\lambda},\sqcup_{\lambda\in\Lambda}\mathcal{V}_{\lambda})$ the disjoint union of the semi-coarse spaces $\{(X_{\lambda},\mathcal{V}_{\lambda})\}_{\lambda\in\Lambda}$ .

Proposition 2.8.

Let $\{(X_{\lambda},\mathcal{V}_{\lambda}\}$ be a collection of semi-coarse spaces indexed bu the set $\Lambda$ and $(Z,\mathcal{Z})$ be a semi-coarse space. Define $i_{\lambda}:X_{\lambda}\rightarrow X$ such that $i_{\lambda}(x)=(x,\lambda)$ . Then $g:(\sqcup X_{\lambda},\sqcup\mathcal{V}_{\lambda})\rightarrow(Z,\mathcal{Z})$ is a bornologous maps if and only if $g\circ i_{\lambda}:X_{\lambda}\rightarrow Z$ is bornologous for every $\lambda\in\Lambda$ .

Proof.

By defintion, $i_{\lambda}$ sends controlled sets in controlled set, then it is a bornologous map. Thus, if $g$ is bornologous, we have that $g\circ i_{\lambda}$ is bornologous for every $\lambda\in\Lambda$ .

On the other hand, consider that $g\circ i_{\lambda}$ is bornologous for every $\lambda$ . Take a controlled set $A$ of $\sqcup X$ , then there exists $\lambda_{1},\ldots,\lambda_{n}\in\Lambda$ and $A_{\lambda_{j}}\in\mathcal{V}_{i}$ such that $A=\sqcup A_{\lambda_{j}}$ . Since $g\circ i_{\lambda_{i}}$ is bornologous, then $g(A_{\lambda_{j}})\in\mathcal{Z}$ , then $\bigcup g(A_{\lambda_{j}})\in\mathcal{Z}$ . Therefore $g$ is bornologous. ∎

Through the disjoint union, we can generate the coarser coarse space which contains a semi-coarse space.

Definition 2.9 (Product Extension; 2.5.9 and 2.5.10 [rieser2023semicoarse]).

Let $(X,\mathcal{V})$ be a semi-coarse space, and define

\displaystyle\mathcal{V}^{PE}\coloneqq\{C\subset X\times X:\exists A,B\in\mathcal{V}\text{ with }C\subset A\circ B\}.

We call the structure $\mathcal{V}^{EP}$ the set product extension of $\mathcal{V}$ , and the ordered pair $(X,\mathcal{V}^{EP})$ is called the set product extension of $(X,\mathcal{V})$ . For any $k\in\mathbb{N}$ , we recursively define $\mathcal{V}^{kPE}$ to be the set product extension of $\mathcal{V}^{(k-1)PE}$ .

Observe that $\mathcal{V}\subset\mathcal{V}^{PE}$ .

Definition 2.10 (Product Extension; 2.5.14 and 2.5.15 [rieser2023semicoarse]).

Let $(X,\mathcal{V})$ be a semi-coarse space, and let $\{(X,\mathcal{V}^{kPE},\iota_{i}^{j},\mathbb{N}\}$ be the directed system of semi-coarse spaces such that all the $\iota:(X,\mathcal{V}^{iPE})\rightarrow(X,\mathcal{V}^{jPE})$ are the identity map. We call that coarse structure the coarse structure induced by $\mathcal{V}$ , and which we denote by $\mathcal{V}^{\infty}$ .

To close this section, we introduce briefly the semi-coarse homotopy and the fundamental semi-coarse group. The first concept is necessary to define what a string is (4.5); the second one is to illustrate that all of the fundamental semi-coarse groups of $X$ are contained in the fundamental groupoid.

From here on out, we denote by $\mathbb{Z}_{1}$ the semi-coarse space on $\mathbb{Z}$ such that the controlled sets are subsets of $\{(i,i+j)\mid i\in\mathbb{Z},j\in\{-1,0,1\}\}$ .

Definition 2.11 (Semi-coarse Homotopy; 3.1.3 [rieser2023semicoarse]).

Let $X$ and $Y$ semi-coarse spaces. $f,g:X\rightarrow Y$ bornologous maps are homotopic if there exists $H:X\times\mathbb{Z}_{1}\rightarrow Y$ bornologous and $M,N\in\mathbb{Z}$ such that $H(x,k)=f(x)$ for every $k\leq M$ and $H(x,k)=g(x)$ for every $k\geq N$ .

Definition 2.12 (Homotopy of Maps of Pairs and Triples; 3.1.6 [rieser2023semicoarse]).

Let $X$ and $Y$ be semi-coarse spaces, let $B\subset A\subset X$ and $D\subset C\subset Y$ be endowed with the subspace structure, and let $f,g:(X,A,B)\rightarrow(Y,C,D)$ be bornologous maps of a triple, i.e. such that $f\mid_{A}\subset C$ and $f\mid_{B}\subset D$ . We say that $f$ is relatively homotopic to $g$ and weite $f\simeq_{sc}g$ if and only if there is a homotopy $H:X\times\mathbb{Z}_{1}\rightarrow Y$ such that $H\mid_{A\times\mathbb{Z}}\subset C$ and $H\mid_{B\times\mathbb{Z}}\subset D$ .

We define a homotopy between maps of pairs $f,g:(X,A)\rightarrow(Y,C)$ to be the homotopy between maps of a triple as above with $B=A$ and $C=D$ .

We summarize the building of the fundamental group in [rieser2023semicoarse] from section 3.2. Fix a semi-coarse space $X$ and consider all of the $\{0,\ldots,n\}$ as subspace of $\mathbb{Z}_{1}$ , and consider $f:\{0,\ldots,n\}\rightarrow X$ as bornologous maps such that $f(0)=f(n)=x$ for a fixed $x\in X$ . For every $n<m\in\mathbb{N}_{0}$ we might extend the bornologous maps $f:\{0,\ldots,n\}\rightarrow X$ to $f^{\prime}:\{0,\ldots,m\}\rightarrow X$ such that $f^{\prime}(i)=f(i)$ for every $0\leq i\leq n$ and $f^{\prime}(i)=f(n)$ for every $n<i\leq m$ .

Calling that extension $i_{n}^{m}$ . That construction produces the directed set

\displaystyle([f:\{0,\ldots,n\}\rightarrow X],i_{n}^{m},\mathbb{N})

where we apply the direct limit which we call $\pi_{1}^{sc}(X,x)$ . We name the classes there as $[f]$ and we define the homotopy as $[f]\simeq_{sc}[g]$ if there are $f$ .

Finally, we define a product among these function. Let $m,m^{\prime}\in\mathbb{N}$ and suppose that $f:\{0,\ldots,m\}\rightarrow X$ and $g:\{0,\ldots,m^{\prime}\}\rightarrow X$ . We define the $\star$ -product $f\star g:I_{m+m^{\prime}}\rightarrow X$ such that

\displaystyle f\star g(j)\coloneqq\left\{\begin{array}[]{ll}f(j)&\text{ if }0\leq j\leq m\\ f(j-m)&\text{ if }j>m.\end{array}\right.

This map induces a product in $\pi_{1}^{sc}(X,x)$ and for Theorem 3.2.18 in [rieser2023semicoarse] we have that that collection is a groupo with $\star$ -operation.

3. Connectedness and Splitting-well

Topological spaces have a considerable advantage over other categories: open sets. When we have an open cover $\{U,V\}$ , we already get that the pushout of $U\leftarrow U\cap V\rightarrow V$ is exactly $X$ up isomorphisms; furthermore, we would know that $X$ is disconnected if $U\cap V=\varnothing$ . In general, semi-coarse spaces don not have a family of sets with such characteristics. To remedy this deficiency, we study the property that two subsets $A,B$ well-split $X$ (3.6). This kind of division help is an important tool to prove Van Kampen theorem for specific subsets.

Definition 3.1.

Let $X$ be a semi-coarse space. We say that $X$ is connected if for every $x,y\in X$ there exists a path $\gamma:\{0,1,\ldots,n\}\rightarrow X$ , $\{0,1,\ldots,n\}$ subspace of $\mathbb{Z}_{1}$ , such that $\gamma(0)=x$ and $\gamma(n)=y$ .

Definition 3.2.

Let $A,B,C$ be semi-coarse spaces and $f:C\rightarrow A$ and $g:C\rightarrow B$ be bornologous maps. We define $P_{B}/f(c)\sim g(c)$ is called the pushout of $A\sqcup_{C}B\coloneqq A\sqcup B/f(c)\sim g(c)$ is called the pushout of .

Proposition 3.3 (Universal Property of Pushout).

Let $A,B,C$ be semi-coarse spaces and $f:C\rightarrow A$ and $g:C\rightarrow B$ be bornologous maps. The diagram

is commutative, and for every other commutative diagram

we have that there exists a unique bornologous map $h:A\sqcup_{C}B\rightarrow X$ such that $j_{1}=hi_{1}$ and $j_{2}=hi_{2}$ .

Proof.

Using the classical notation, $A\sqcup B=(A\times\{1\})\cup(B\times\{2\})$ . Moreover, we call $i_{1}(a)=[a,1]$ and $i_{2}(b)=[b,1]$ .

We define the maps

	$\displaystyle h:A\sqcup B\rightarrow X$	$\displaystyle\text{ such that }h(a,1)=j_{1}(a)\text{ and }h(b,2)=j_{2},$
	$\displaystyle\overline{h}:A\sqcup_{C}B\rightarrow X$	$\displaystyle\text{ such that }h[a,1]=j_{1}(a)\text{ and }h[b,2]=j_{2}.$

We want to prove that $\overline{h}$ is well-defined bornologous map, and it is the unique bornologous map such that $j_{2}=\overline{h}\circ i_{2}$ and $j_{1}=\overline{h}\circ i_{1}$ .

To prove that $\overline{h}$ is well-defined, we obtain the following cases:

•

If $(a,1)\in[b,2]$ , then there exists $c\in C$ such that $f(c)=a$ and $g(c)=b$ . Thus $\overline{h}[a,1]=\overline{h}i_{1}(a)=\overline{h}i_{1}f(c)=\overline{h}i_{2}g(c)=\overline{h}i_{2}(b)=\overline{h}[b,2]$ .
•

If $(a^{\prime},1)\in[a,1]$ , then there exists $c,c^{\prime}\in C$ such that $f(c)=a$ , $f(c^{\prime})=a^{\prime}$ and $g(c)=g(c^{\prime})$ . Thus $\overline{h}[a,1]=\overline{h}[g(c),2]=\overline{h}[g(c^{\prime}),3]=\overline{h}[a^{\prime},1]$ .

We also have analogous cases for $(b,2)\in[a,1]$ and $(b^{\prime},2)\in[b,2]$ , the procedure is completely analogous.

Now, we can prove that $\overline{h}$ is bornologous using the universal properties of the disjoint union and quotient, and observing that the following diagrams are commutative

Lastly, we observe that $\overline{h}$ is the unique map which satisfies that $j_{2}=\overline{h}\circ i_{2}$ and $j_{1}=\overline{h}\circ i_{1}$ . Suppose that there exists $h^{\prime}:A\sqcup_{C}B\rightarrow X$ such that $h^{\prime}[a,1]\neq\overline{h}[a,1]$ , for some $a\in A$ , then $h^{\prime}i_{1}(a)\neq\overline{h}i_{1}(a)=j_{2}(a)$ . Analogously if we take $[b,2]$ with $b\in B$ . ∎

Proposition 3.4.

Let $(X,\mathcal{V})$ a semi-coarse space. If $\{(x_{1},x_{2})\}\in\mathcal{V}_{A}\sqcup_{\mathcal{V}_{A\cap B}}\mathcal{V}_{B}$ and:

(1)

$x_{1}\in A$ , $x_{2}\notin A$ , then $x_{1}\in B$ .
(2)

$x_{1}\in B$ , $x_{2}\notin B$ , then $x_{1}\in A$ .

Proof.

Let $(X,\mathcal{V})$ a semi-coarse space, $A,B\subset X$ which well-split $X$ and $\{(x_{1},x_{2})\}\in\mathcal{V}_{A}\sqcup_{\mathcal{V}_{A\cap B}}\mathcal{V}_{B}$ . Both cases are completely analogous; thus we prove just the first case. Therefore, consider that $x_{1}\in A$ and $x_{2}\notin A$ .

We call $(Y,\mathcal{W})$ the subspace $\{1,2,3\}$ of $\mathbb{Z}_{1}$ , and define the maps

	$\displaystyle j_{1}:A\rightarrow Y$	$\displaystyle\text{ such that }j_{1}(A\cap B)=2\text{ and }j_{1}(A-B)=1$
	$\displaystyle j_{2}:B\rightarrow Y$	$\displaystyle\text{ such that }j_{2}(A\cap B)=2\text{ and }j_{2}(B-A)=3$

Since $\{1,2\}\times\{1,2\},\{2,3\}\times\{2,3\}\in\mathcal{W}$ , then $j_{1},j_{2}$ are bornologous maps. Moreover, the diagram

is commutative. Thus, since $A\sqcup_{A\cap B}B$ is a pushout, there exists a unique $h:A\sqcup_{A\cap B}B\rightarrow Y$ bornologous such that $j_{2}=hi_{2}$ and $j_{1}=hi_{1}$ . Since $h$ is bornologus, then $\{h(x_{1}),h(x_{2})\}\in\mathcal{W}$ . In addition, $x_{1}\in A$ and $x_{2}\in B-A$ , thus $h(x_{2})=3$ and $h(x_{1})\in\{1,2\}$ . $h(x_{1})\neq 1$ because $\{(1,3)\}\notin\mathcal{W}$ . Therefore $h(x_{1})=2$ and $x_{1}\in A\cap B$ . ∎

Corollary 3.5.

Let $X$ be a semi-coarse space and $A,B\subset X$ . Then every path $\gamma$ in $A\sqcup_{A\cap B}B$ satisfies that

•

if $\gamma(i)\in A$ and $\gamma(i+1)\notin A$ , then $\gamma(i)\in B$ , and
•

if $\gamma(i)\in B$ and $\gamma(i+1)\notin B$ , then $\gamma(i)\in A$ .

Definition 3.6.

Let $(X,\mathcal{V})$ be a semi-coarse space and $A,B\subset X$ non-empty. Then $A,B$ well-split $X=A\cup B$ if for every $x,y,y^{\prime}\in X$ such $\{(x,y),(x,y^{\prime})\}\in\mathcal{V}-A\sqcup_{A\cap B}B$ and $\{(y,y^{\prime})\}\in A\sqcup_{A\cap B}B$ :

(1)

There exists a path $\gamma,\gamma^{\prime}:\{0,1,2\}\rightarrow X$ in $A\sqcup_{A\cap B}B$ such that $\gamma(0)=\gamma^{\prime}(0)=x$ , $\gamma(2)=y$ , $\gamma^{\prime}(2)=y^{\prime}$ and $\gamma(1)=\gamma^{\prime}(1)$ .
(2)

The set $\{\gamma(1)\mid\gamma\text{ is a path in }A\sqcup_{A\cap B}B\text{ such that}\gamma(0)=x,\gamma(2)=y\}$ is connected as subspace of $X$ .

there exists a path $\gamma:\{0,1,2\}\rightarrow X$ in $A\sqcup_{A\cap B}B$ such that $\lambda\simeq_{sc}\gamma$ .

Figure 1. Well-splitting example: Red points are

A-B

, blue points are

B-A

and violet points are

A\cap B

X

with the semi-coarse space induce by the graph.

Condition two is equivalent to say that every $f:\{0,1,2\}\rightarrow X$ with $f(0)=x$ and $f(2)=y$ are homotopic. This implies that every $A,B$ with the first condition in a coarse space well-split the space. Indeed, the reason to include these both conditions is to include coarse spaces in our analyze.

Proposition 3.7.

Let $X$ be a semi-coarse space and $A,B\subset C$ . If $X$ is (semi-coarse) homeomorphic to $A\sqcup_{A\cap B}B$ , then $A,B\subset X$ well-split $X$ .

Example in Figure 1 is a clearly example that the other direction is not necessary true. It might be tempting to define take only $A,B\subset X$ such that $X$ and $X$ and $A\sqcup_{A\cap B}B$ are semi-coarse isomorphic; however, we might leave out the interesting coarse cases.

Consider for example $\mathbb{R}$ with its coarse structure induced by the metric, and $A\coloneqq[0,\infty)$ , $B\coloneqq(-\infty,0]$ . $A\cup B=X$ , but $A\sqcup_{A\cap B}B$ is not a coarse space and thus is not isomorphic to $X$ (see Figure 2 to compare both spaces).

Figure 2. Every controlled set in

A\sqcup_{A\cap B}B

and

X

, respectively, is contained in some red area. Observe that there are not controlled sets in the second and fourth quadrants in

A\sqcup_{A\cap B}B

, in contrast with

X

Proposition 3.8.

Let $(X,\mathcal{V})$ be a semi-coarse space and $A,B\subset X$ non-empty. $X$ is disconnected if and only if there exist $A,B$ which well-split $X$ such that $A\cap B=\varnothing$ .

Proof.

Let $(X,\mathcal{V})$ be a semi-coarse space and $A,B\subset X$ non-empty.

$(\Rightarrow)$ Suppose that $X$ is disconnected and $x\in X$ . Take $A$ as the component of $x$ and $B\coloneqq X-A$ . Observe that $A\sqcup_{A\cap B}B=A\sqcup B=X$ as semi-coarse spaces. Thus, $A,B$ well-split $X$ .

$(\Leftarrow)$ Suppose that $A,B$ well-split $X$ and $A\cap B=\varnothing$ . Let $a\in A$ and $b\in A$ , and consider that there is a path $\gamma:\{0,\ldots,n\}\rightarrow X$ such that $\gamma(0)=a$ and $\gamma(n)=b$ . Since $A\cup B=X$ , there there exists $i$ such that $\gamma(i)\in A$ and $\gamma(i+1)\in B$ . Hence, there is $\lambda:\{0,1,2\}\rightarrow X$ in $A\sqcup_{A\cap B}B$ such that $\lambda(0)=\gamma(i)$ and $\lambda(2)=\gamma(i+1)$ . By 3.4, we obtain that at least one of the three elements are in $A\cap B$ $(\rightarrow\leftarrow)$ . Therefore, there is not a path between $a,b\in X$ ; $X$ is disconnected. ∎

4. Semi-coarse Strings

Building a groupoid in semi-coarse spaces demands some tricks and considerations. On one side, we would like to preserve the fundamental groups as a part of this groupoid, restricting to just one object and some morphisms. On the other hand, our motivation is to find and invariant which doesn’t lose all of the structure when we work in coarse spaces.

Therefore, we are going to explore bornologous maps $f:\mathbb{Z}_{1}\rightarrow X$ for a semi-coarse space $X$ . From this point, we call tails to the restrictions $f\mid_{[n,\infty)}$ and $f\mid_{(-\infty,m]}$ for some $m,n,\in\mathbb{Z}$ . The tails are our main focus in the subsequent construction, and sooner we realize that it contains almost all of the information.

First, we give the main ingredients to create a groupoid: objects, morphism and a composition rule; for this construction we have symmetric maps, string and $\star$ -operation, respectively.

Definition 4.1.

Let $f:\mathbb{Z}_{1}\rightarrow X$ :

•

$f$ is called symmetric if $f(x)=f(-x)$ .
•

$\overline{f}$ is called the opposite direction of $f:\mathbb{Z}_{1}\rightarrow X$ with $\overline{f}(x)=f(-x)$

Through the tails of the symmetric maps, we can develop a equivalence relation.

Definition 4.2.

Let $f,g$ be symmetric maps. $f$ and $g$ are eventually equal if there exists $N,M\in\mathbb{N}_{0}$ such that $f(N+k)=g(M+k)$ for every $k\in\mathbb{N}_{0}$ .

Proposition 4.3.

The relation of two symmetric maps being eventually equal is a equivalence relation. We write $\langle f\rangle_{\infty}$ to refer to the class of maps eventually equal to $f$ .

Taking the coarse space $\mathbb{Z}_{1}$ as domain of our maps provides some structure in the image, that is, it compels the element $f(z)$ to be related with $f(z+1)$ and $f(z-1)$ . However, being labeled by the integers proves to be really restrictive. The definition below unlabel these maps. This can be seen as being capable to move the functions in the integers axis and say they are exactly the same function. Later, that definition allows to obtain well defined constructions of merging and splitting these maps.

Definition 4.4.

Let $f,g:\mathbb{Z}_{1}\rightarrow X$ . $f$ and $g$ are called affine if there exists $N\in\mathbb{N}_{0}$ such that $f(x)=g(x+k)$ .

Affine is also an equivalence relation. We denote by $\langle f\rangle_{A}\coloneqq\{g\mid f\text{ and }g\text{ are affine}\}$ .

In 4.5, we are considering any map of the strings as affine classes.

Definition 4.5.

Let $n\in\mathbb{N}$ . We say that a $F=(f_{1},\ldots,f_{n})$ is an $n$ -string from $\langle f\rangle_{\infty}$ to $\langle g\rangle_{\infty}$ if we have the following:

(1)

$\overleftarrow{f_{1}}\in\langle f\rangle_{\infty}$ such that

$\displaystyle\overleftarrow{f_{1}}(x)\coloneqq\left\{\begin{array}[]{ll}f_{1}(x)&x\leq 0\\ f_{1}(-x)&x>0.\end{array}\right.$
(2)

$\overrightarrow{f_{n}}\in\langle g\rangle_{\infty}$ such that

$\displaystyle\overrightarrow{f_{n}}(x)\coloneqq\left\{\begin{array}[]{ll}f_{n}(x)&x\leq 0\\ f_{n}(-x)&x>0.\end{array}\right.$
(3)

For every $i\in\{1,\ldots n-1\}$ there exists $f_{i,R}\in\langle f_{i}\rangle_{A}$ and $f_{i+1,L}\in\langle f_{i+1}\rangle_{A}$ such that $f_{i,R}\mid_{[0,\infty)}\simeq_{sc}\overline{f_{i+1,L}}\mid_{[0,\infty)}$ .

Strings are the elements of the collection of all of the $n$ -strings for every $n\in\mathbb{N}$ .

We use the following notation in the subsequent sections, specially when we talk about the Van Kampen theorems:

•

$\mathcal{F}_{X,A,n,f,g}$ is the collection of all of the $n$ -string in $X$ from $\langle f\rangle_{\infty}$ to $\langle g\rangle_{\infty}$ such that their tails land in $A\subset X$ .
•

If we omit the subset $A$ , we consider that $A=X$ .
•

If we omit the number $n$ we consider is all of the $n$ -strings with $n\in\mathbb{N}$ .
•

If we omit $f$ and $g$ , we consider that the strings can start and end in any element.

When the context is clear, we omit $X$ .

Definition 4.6.

Let $F=(f_{1},\ldots,f_{n})$ be a path from $\langle f\rangle_{\infty}$ to $\langle g\rangle_{\infty}$ and $G=(g_{1},\ldots,g_{m})$ be a path from $\langle g\rangle_{\infty}$ to $\langle h\rangle_{\infty}$ . We define $(f_{1},\ldots,f_{n})\star(g_{1},\ldots g_{m})\coloneqq(f_{1},\ldots,f_{n},g_{1},\ldots,g_{m})$ .

$\overline{F}\coloneqq(\overline{f_{n}},\ldots,\overline{f_{1}})$ is called the opposite direction of $F$ .

Although we define an operation between pair of strings, we are not ready to obtain a groupoid from this set. By this moment, it is not clear which element is the identity and how the product of an element with its inverse element produces such identity. To achieve that goal, we are interested in splitting and merging maps from $\mathbb{Z}_{1}$ and deleting consecutive opposite maps.

The technical definition of this is divided in two relations are written in 4.7 and 4.13. The consecutive results and definitions are included to look for the correct way to merge maps and to see that this relation eventually build an equivalence relation.

Definition 4.7.

Let $F=(f_{1},\ldots,f_{n})\in\mathcal{F}$ .

•

For $i\in\{2,\ldots,n-2\}$ . If $f_{i}\in\langle\overline{f_{i+1}}\rangle_{A}$ , then we say that the result of deleting two consecutive opposite maps in $i$ is $F^{\prime}=(f_{1},\ldots,f_{i-1},f_{i+2},\ldots,f_{n})$ . We express this relation as $F\xrightarrow{d_{op}(i)}F^{\prime}$
•

For $i\in\{2,\ldots,n\}$ . Let $g$ we say that the result of adding the opposite maps $g$ and $\overline{g}$ in $i$ is $F^{\prime}=(f_{1},\ldots,f_{i-1},g,\overline{g},f_{i},\ldots,f_{n})$ if $F^{\prime}\in\mathcal{F}$ . We express this relation as $F\xrightarrow{a_{op}(i,g)}F^{\prime}$ .

Lemma 4.8.

Let $F\in\mathcal{F}$ . Then

(1)

If $F\xrightarrow{d_{op}(i)}F^{\prime}$ , then $F^{\prime}\in\mathcal{F}$ and $F^{\prime}\xrightarrow{a_{op}(i,f_{i})}F$ .
(2)

If $F\xrightarrow{a_{op}(i,g)}F^{\prime}$ , then $F^{\prime}\xrightarrow{d_{op}(i)}F$ .

Proof.

Let $F\in\mathcal{F}$ .

(1) Suppose that $F\xrightarrow{d_{op}(i)}F^{\prime}$ . By definition, $f_{i}\in\langle\overline{f_{i+1}}\rangle_{A}$ . Without loss of generality, suppose that precisely $f_{i}=\overline{f_{i+1}}$ . In addition, there exists $r_{i-1},l_{i},r_{i+1},l_{i+2},D,U\in\mathbb{Z}$ and $H_{1},H_{2}:[0,\infty)\times\mathbb{Z}_{1}\rightarrow X$ such that

	$\displaystyle H_{1}(z,k)\ =\$	$\displaystyle f_{i-1}(r_{i-1}+z)\text{ if }k\leq D$
	$\displaystyle H_{1}(z,k)\ =\$	$\displaystyle f_{i}(l_{i}-z)\text{ if }k\geq 0$
	$\displaystyle H_{2}(z,k)\ =\$	$\displaystyle f_{i+1}(r_{i}+z)\text{ if }k\leq 0$
	$\displaystyle H_{2}(z,k)\ =\$	$\displaystyle f_{i+2}(l_{i+1}-z)\text{ if }k\geq U.$

Since $f_{i}=\overline{f_{i+1}}$ , $f_{i+1}(z-l_{i})=f_{i}(l_{i}-z)$ . Take $a\coloneqq\max\{r_{i},-l_{i}\}$ . Displace $H_{1}$ and $H_{2}$ such that

	$\displaystyle H^{\prime}_{1}(z,k)\ =\$	$\displaystyle f_{i-1}(r_{i-1}+\|a+l_{i}\|+z)\text{ if }k\leq D$
	$\displaystyle H^{\prime}_{1}(z,k)\ =\$	$\displaystyle f_{i}(l_{i}-\|a+l_{i}\|-z)\text{ if }k\geq 0$
	$\displaystyle H^{\prime}_{2}(z,k)\ =\$	$\displaystyle f_{i+1}(r_{i}+\|r_{i}-a\|+z)\text{ if }k\leq 0$
	$\displaystyle H^{\prime}_{2}(z,k)\ =\$	$\displaystyle f_{i+1}(l_{i+1}-\|r_{i}-a\|-z)\text{ if }k\geq U.$

Therefore, we built a homotopy from some $f^{\prime}_{i-1,R}\in\langle f_{i-1}\rangle_{A}$ to some $f^{\prime}_{i+2,L}\in\langle f_{i+1}\rangle_{A}$ . Thus $F^{\prime}\in\mathcal{F}$ .

On the other hand, since $F\xrightarrow{d_{op}(i)}F^{\prime}$ , then $F\in\mathcal{F}$ and $f_{i}\in\langle\overline{f_{i+1}}\rangle_{A}$ , thus $F^{\prime}\xrightarrow{a_{op}(i,g)}F$ .

(2) Let $F\xrightarrow{a_{s}(i,g)}F^{\prime}$ . By definition $F^{\prime}\xrightarrow{d_{s}(i)}F$ because the element after $g$ in $F^{\prime}$ is precisely $\overline{g}$ . ∎

Definition 4.9.

Let $f:\mathbb{Z}\rightarrow X$ be a map.

•

$f$ is called periodic if there exists $T\in\mathbb{N}$ such that $f(x)=f(x+T)$ for every $x\in X$ .
•

$f$ is called eventually right periodic if there exist $T\in\mathbb{N}$ and $z\in\mathbb{Z}$ such that $f(x)=f(x+T)$ for every $x\geq z$ .
•

$f$ is called eventually left periodic if there exists $T\in\mathbb{N}$ and $z\in\mathbb{Z}$ such that $f(x)=f(x-T)$ for evry $x\leq z$ .

Lemma 4.10.

Let $f:\mathbb{Z}\rightarrow X$ be a map.

(1)

If $f$ is periodic, then $f$ is both eventually right and left periodic.
(2)

If $f$ is not periodic and is eventually right periodic, then

$\displaystyle\min\{z\in\mathbb{Z}\mid f(x)=f(x+T)\text{ for every }x\geq z\}\in\mathbb{Z}.$
(3)

If $f$ is not periodic and is eventually left periodic, then

$\displaystyle\max\{z\in\mathbb{Z}\mid f(x)=f(x-T)\text{ for every }x\leq z\}\in\mathbb{Z}.$

Proof.

Let $f:\mathbb{Z}\rightarrow X$ be a map.

(1) Let $f$ be periodic, then there exists $T\in\mathbb{N}$ such that $f(x)=f(x+T)$ for every $x\in X$ . Thus $f(x)=f(x-T)$ for every $x\in X$ . In particular, $f(x)=f(x+T)$ for every $x\geq 0$ and $f(y)=f(y-T)$ for every $y\leq 0$ .

(2) Let $f$ be eventually right periodic and not periodic. Then

	$\displaystyle A\coloneqq$	$\displaystyle\{z\in\mathbb{Z}\mid f(x)=f(x+T)\text{ for every }x\geq z\}\in\mathbb{Z}\neq\varnothing\text{ and }$
	$\displaystyle z\in$	$\displaystyle A\Rightarrow z+1\in A.$

If $\min\ A\notin\mathbb{Z}$ , then $f(x)=f(x+T)$ for every $x\in\mathbb{Z}$ $(\rightarrow\leftarrow)$ . Thus $\min\ A\in\mathbb{Z}$ .

(3) Let $f$ be eventually left periodic and not periodic. Then

	$\displaystyle A\coloneqq$	$\displaystyle\{z\in\mathbb{Z}\mid f(x)=f(x-T)\text{ for every }x\leq z\}\in\mathbb{Z}\neq\varnothing\text{ and }$
	$\displaystyle z\in$	$\displaystyle A\Rightarrow z-1\in A.$

If $\max\ A\notin\mathbb{Z}$ , then $f(x)=f(x-T)$ for every $x\in\mathbb{Z}$ $(\rightarrow\leftarrow)$ . Thus $\max\ A\in\mathbb{Z}$ . ∎

Lemma 4.11.

Let $f,g:\mathbb{Z}_{1}\rightarrow X$ be not periodic bornologous maps such that there exists $a,b,a^{\prime},b^{\prime}\in\mathbb{Z}$ such that $f(a+z)=g(b-z)$ and $f(a^{\prime}+z)=g(b^{\prime}-z)$ for every $z\geq 0$ . Then $a-a^{\prime}=b^{\prime}-b$ or $g$ is eventually left periodic.

Proof.

Without loss of generality, suppose that $a\leq a^{\prime}$ , and consider $d\coloneqq|b^{\prime}-b|$ . Then, $a^{\prime}-a+d\geq 0$ .

By hypothesis , we obtain that

	$\displaystyle f(a+(a^{\prime}-a+d+z))=$	$\displaystyle g(b-(a^{\prime}-a+d+z))\text{ and }$
	$\displaystyle f(a+(a^{\prime}-a+d+z))=$	$\displaystyle f(a^{\prime}+(d+z))=g(b^{\prime}-(d+z)).$

for every $z\geq 0$ .

On one hand, suppose that $b^{\prime}-b\geq 0$ . Then

\displaystyle g(b^{\prime}-(d+z))=g(b^{\prime}-(b^{\prime}-b+z))=g(b-z).

Thus, $g(b-T-z)=g(b-z)$ for every $z\geq 0$ with $T=(a^{\prime}-a)+d$ .

On the other hand, suppose that $b-b^{\prime}\geq 0$ . Then

	$\displaystyle g(b-(a^{\prime}-a+d+z))=$	$\displaystyle g(b-(a^{\prime}-a+z)-b+b^{\prime})=g(b^{\prime}-(a^{\prime}-a+z))$
	$\displaystyle g(b^{\prime}-(d+z))=$	$\displaystyle g(b^{\prime}-(b-b^{\prime}+z))$

If $b-b^{\prime}=a^{\prime}-a$ , we cannot ensure any about the periodicity. If they are different, we can define $m=\max\{b-b^{\prime},a^{\prime}-a\}$ and see that $g(b^{\prime}-m-T-z)=g(b^{\prime}-m-z)$ for every $z\geq 0$ and with $T=|a^{\prime}-a+b^{\prime}-b|$ . In both cases $g$ is eventually left cyclic or $b-b^{\prime}=a^{\prime}-a$ . ∎

Proposition 4.12.

Let $f,g:\mathbb{Z}_{1}\rightarrow X$ be not periodic bornologous maps such that $f\notin\langle\overline{g}\rangle_{A}$ and there exist $a,b\in\mathbb{Z}$ such that $f(a+z)=g(b-z)$ for every $z\geq 0$ . There exists unique $a^{\prime},b^{\prime}\in\mathbb{Z}$ such that $f(a^{\prime}+z)=g(b^{\prime}-z)$ for every $z\geq 0$ and $f(a^{\prime}-1)\neq g(b^{\prime}+1)$ .

Proof.

We divide this proof in two case, whether $f$ is eventually right periodic. We can observe that $f$ is eventually right periodic if and only if $g$ is eventually left periodic, so it’s enough to consider these two cases.

Suppose that $f$ is not eventually right periodic. If $a^{\prime},b^{\prime}\in\mathbb{Z}$ satisfy that $f(a^{\prime}+z)=g(b^{\prime}-z)$ for every $z\geq 0$ , then, by 4.11, we observe that $a^{\prime}-a=b-b^{\prime}$ ; therefore we can write it as $f(a+z)=g(b-z)$ for every $z\geq a^{\prime}-a$ . Since $f\notin\langle\overline{g}\rangle_{A}$ we have that there exists $m<0$ such that $f(a+m)=g(b+m)$ . Then,

\displaystyle M\coloneqq\{z^{\prime}\in\mathbb{Z}\mid f(a+z)=g(b-z)\text{ for every }z\geq z^{\prime}\}\in\mathbb{Z}

Thus, $f(a+M+z)=g(b-M-z)$ for every $z\geq 0$ and $f(a+M-1)\neq g(b-M+1)$ .

Suppose that $f$ is eventually right periodic with period $T$ , then $g$ is eventually left periodic with period $T$ . Applying 4.10, there exists $A,B\in\mathbb{Z}$ such that

	$\displaystyle A=$	$\displaystyle\min\{z\in\mathbb{Z}\mid f(x)=f(x+T)\text{ for every }x\geq z\}$
	$\displaystyle B=$	$\displaystyle\max\{z\in\mathbb{Z}\mid g(x)=g(x-T)\text{ for every }x\leq z\}$

First we work with the case $a\leq A$ . Observe that $f(A-1)\neq g(A+T-1)$ . If $f(A-1)=g(b-A+a+1)$ , then $g(b-A+a+1)\neq g(b-A+a+1-T)$ , and $g(b-A+a-z)=g(b-A+a-T-z)$ for every $z\geq 0$ ; thus $b-A+a+1=B+1$ . Hence $b-B=A-a\geq 0$ . We obtain that

\displaystyle f(A)=f(a+(A-a))=g(b-(A-a))=g(b-(b-B))=g(B).

The procedure is completely analogous for not eventually right periodic, obtaining analogous values $a^{\prime}$ and $b^{\prime}$ .

On the other hand, if $f(A-1)\neq g(b-A+a+1)$ , then $a\geq A$ . Hence $a=A$ and $B\geq b$ . We consider

\displaystyle b^{\prime}\coloneqq\max\{b^{\prime}\in\mathbb{Z}\mid f(A+z)=g(b^{\prime}-z)\text{ for every }z\geq 0\},

observing that $b^{\prime}\geq b$ . Hence, for this case, $a^{\prime}\coloneqq a=A$ and $b^{\prime}$ is the element we have found.

We now consider that $a>A$ , which implies that $B\geq b$ , because at least $g(b)=g(b-T)$ . Define $d\coloneqq\{a-A,B-b\}$ , then $f(a-d+z)=g(b+d-z)$ by periodicity.

Quickly we observe that:

•

If $a-A=B-b$ , then $f(A+z)=g(B-z)$ and we proceed in similar way as not eventually right periodic.

•

If $a-A<B-b$ , we compute

\displaystyle a^{\prime}\coloneqq\min\{a^{\prime}\in\mathbb{Z}\mid f(a^{\prime}-d+z)=g(b+d-z)\text{ for every }z\geq 0\}

and the values we want to get was $a^{\prime}-d$ and $b+d$ .

•

If $a-A>B-b$ , we compute.

\displaystyle b^{\prime}\coloneqq\min\{b^{\prime}\in\mathbb{Z}\mid f(a-d+z)=g(b^{\prime}+d-z)\text{ for every }z\geq 0\}

and the values we want to get was $a-d$ and $b^{\prime}+d$ .∎

For a technical reason, we are just going to merge maps which are neither eventually right periodic nor eventually left periodic, with the exception of period 1, 2 or 3.

Definition 4.13.

Let $F=(f_{1},\ldots,f_{n})\in\mathcal{F}$ . For $i\in\{1,\ldots,n-1\}$ , $f_{i}\notin\langle\overline{f_{i+1}}\rangle_{A}$ and there exists $a,b\in\mathbb{Z}$ such that $f_{i}(a+z)=f_{i+1}(b-z)$ for every $z\geq 0$ , we say that the result of merging $f_{i}$ and $f_{i+1}$ is $F^{\prime}=(f_{1},\ldots,f_{i-1},g,f_{i+2},\ldots f_{n})$ such that

•

If $f_{i}$ is periodic with $T\in\{1,2,3\}$ and $f_{i+1}$ is not periodic, then by 4.10 there exists $b^{\prime}$ such $f_{i+1}(x)=f_{i+1}(x-T)$ for every $x\leq b^{\prime}$ and $f_{i+1}(b+1)\neq f_{i+1}(b+1-T)$ , and

\displaystyle g(z)\coloneqq\left\{\begin{array}[]{ll}f_{i}(a+z)&\text{ if }z\leq 0\\ f_{i+1}(b^{\prime}+z)&\text{ if }z>0.\end{array}\right.

•

If $f_{i+1}$ is periodic with $T\in\{1,2,3\}$ and $f_{i}$ is not periodic, then by 4.10 there exists $a^{\prime}$ such $f(x)=f(x+T)$ for every $x\geq b^{\prime}$ and $f(a-1)=f(a-1+T)$ , and

\displaystyle g(z)\coloneqq\left\{\begin{array}[]{ll}f_{i}(a^{\prime}+z)&\text{ if }z\leq 0\\ f_{i+1}(b+z)&\text{ if }z>0.\end{array}\right.

•

If $f_{i},f_{i+1}$ are not periodic, and $f_{i}$ is eventually right periodic with $T\in\{1,2,3\}$ or is not eventually right periodic. Applying 4.12, we find $a^{\prime},b^{\prime}\in\mathbb{Z}$ such that $f_{i}(a^{\prime}+z)=f_{i+1}(b^{\prime}-z)$ for every $z\in\mathbb{Z}$ and $f_{i}(a^{\prime}-1)\neq f_{i+1}(b^{\prime}+1)$ . We define

\displaystyle g(z)\coloneqq\left\{\begin{array}[]{ll}f_{i}(a^{\prime}+z)&\text{ if }z\leq 0\\ f_{i+1}(b^{\prime}+z)&\text{ if }z>0.\end{array}\right.

On the other hand, if $f_{i}\in\langle\overline{f_{i+1}}\rangle_{A}$ and $f_{i}$ is eventually left periodic with $T\in\{1,2\}$ we define $g(2\mathbb{Z})=f_{i}(a)$ and $g(2\mathbb{Z}-1)=f_{i}(a-1)$ where $a$ satisfies that $f_{i}(x)=f_{i}(x-T)$ for every $x\leq a$ .

We express this relation as $F\xrightarrow{m(i)}F^{\prime}$ (see an exemplification in Figure 3).

Figure 3. Merging

f_{i}

and

f_{i+1}

which are maps from

\mathbb{Z}_{2}

\mathbb{R}^{2}

. The darker the color in the point, the higher its value in the integers.

For $i\in\{1,\ldots,n\}$ , we sat that the result of split $f_{i}$ in $g$ and $h$ at $z_{i}$ is $F^{\prime}=(f_{1},\ldots,f_{i-1},g,h,f_{i+1},\ldots,f_{n})$ is a string with:

•

If $f_{i}$ is periodic with $T\in\{1,2\}$ and there exist $m,n\in\mathbb{Z}$ such that

	$\displaystyle g(m+k)=\$	$\displaystyle h(n-k)\ \forall k\in\mathbb{Z},$
	$\displaystyle g(m+k)=\$	$\displaystyle f_{i}(k)\ \forall k\leq 0.$

•

If there exists $z\in\mathbb{Z}$ such that $f_{i}(z-1)\neq f_{i}(z+1)$ , take $z_{i}\in\mathbb{Z}$ with that condition and there exist $m,n\in\mathbb{Z}$ such that

	$\displaystyle g(m+k)=\$	$\displaystyle h(n-k)\ \forall k\geq 0,$
	$\displaystyle g(m+k)=\$	$\displaystyle f_{i}(z_{i}+k)\ \forall k\leq 0,$
	$\displaystyle h(n+k)=\$	$\displaystyle f_{i}(z_{i}+k)\ \forall k\geq 0.$

where $g$ is eventually right periodic with $T\in\{1,2,3\}$ or is not eventually right periodic. We express this relation as $F\xrightarrow{s(i,g,h)}F^{\prime}$ (see an exemplification in Figure 3).

Figure 4. Splitting

f_{i}

, a map from

\mathbb{Z}_{2}

\mathbb{R}^{2}

, in

g

and

h

Lemma 4.14.

Let $F=(f_{1},\ldots,f_{n})\in\mathcal{F}$ . Then

(1)

$m(i)$ is well-defined.
(2)

If $F\xrightarrow{m(i)}F^{\prime}$ , then $F^{\prime}\xrightarrow{s(i,f_{i},f_{i+1})}F$ .
(3)

If $F\xrightarrow{s(i,g,h)}F^{\prime}$ , then $F^{\prime}\xrightarrow{m(i)}F$ .

Proof.

Let $F\in\mathcal{F}$ .

(1) Suppose that for $i\in\{1,\ldots,n-1\}$ we have that $F\xrightarrow{m(i)}F^{\prime}$ . Then $f_{i}\notin\langle\overline{f_{i+1}}\rangle_{A}$ and there exists $a,b\in\mathbb{Z}$ such that $f_{i}(a+z)=f_{i+1}(b-z)$ for every $z\geq 0$ . Hence $F^{\prime}$ is a string because $g(z)=f_{i}(a+z)$ for every $z\leq 0$ and $g(z)=f_{i+1}(b+z)$ for every $z>0$ .

•

Suppose that $f_{i}$ is periodic with $T\in\{1,2,3\}$ and $f_{i+1}$ is not periodic. Consider $a^{\prime\prime}\in\mathbb{Z}$ such that $f_{i}(a^{\prime\prime}+z)=f_{i+1}(b^{\prime}-z)$ for every $z\in\mathbb{Z}$ , then $T=|a-a^{\prime\prime}|$ we define

\displaystyle g^{\prime\prime}(z)\coloneqq\left\{\begin{array}[]{ll}f_{i}(a^{\prime\prime}+z)&\text{ if }z\leq 0\\ f_{i+1}(b^{\prime}+z)&\text{ if }z>0.\end{array}\right.

Since $f_{i}(a+z)=f_{i}(a^{\prime\prime}-(a^{\prime\prime}-a)+z)=f_{i}(a^{\prime\prime}+z)$ . Thus $g=g^{\prime}$ .

•

Suppose that $f_{i+1}$ is periodic with $T\in\{1,2,3\}$ and $f_{i}$ is not periodic. In analogous way of the case above, we can select any $a,b\in\mathbb{Z}$ such that $f_{i}(a+z)=f_{i+1}(b-z)$ for every $z\in\mathbb{Z}$ .
•

Suppose that $f_{i},f_{i+1}$ are not periodic, and $f_{i}$ is eventually right periodic with $T\in\{1,2,3\}$ or is not eventually right periodic. Since $a^{\prime},b^{\prime}$ are unique, then $m(i)$ is well defined.

(2) Suppose that $F\xrightarrow{m(i)}F^{\prime}$ . We explore the whole four cases, just in the first three cases we consider that $f_{i}\notin\langle f_{i+1}\rangle_{A}$ :

•

Suppose that $f_{i}$ is periodic with $T\in\{1,2,3\}$ and $f_{i+1}$ is not periodic. By 4.10, there exists $b^{\prime}$ such that $f_{i+1}(x)=f_{i+1}(x-T)$ for every $x\leq b^{\prime}$ and $f(b+1)\neq f(b+1-T)$ . Hence, building $g$ as in 4.13, $g(-1)\neq g(1)$ and by definition

	$\displaystyle f_{i}(a+k)=$	$\displaystyle f_{i+1}(b^{\prime}-k)\text{ for every }k\geq 0,$
	$\displaystyle f_{i}(a+k)=$	$\displaystyle g(k)\text{ for every }k\leq 0,$
	$\displaystyle f_{i+1}(b^{\prime}+k)=$	$\displaystyle g(k)\text{ for every }k\geq 0.$

Thus $F^{\prime}\xrightarrow{s(0,f_{i},f_{i+1})}F$ .

•

Suppose that $f_{i+1}$ is periodic with $T\in\{1,2,3\}$ and $f_{i}$ is not periodic. The case analogous but we apply 4.10 in $f_{i}$ . Thus $F^{\prime}\xrightarrow{s(0,f_{i},f_{i+1})}F$ .

•

Suppose that $f_{i},f_{i+1}$ are not periodic, and $f_{i}$ is eventually right periodic with $T\in\{1,2,3\}$ or is not eventually right periodic. Then there exist $a^{\prime},b^{\prime}\in\mathbb{Z}$ such that $f_{i}(a^{\prime}+z)=f_{i+1}(b^{\prime}-z)$ for every $z\geq 0$ and $f_{i}(a^{\prime}-1)\neq f_{i+1}(b^{\prime}+1)$ . Hence, building $g$ as in 4.13, $g(-1)\neq g(1)$ and by definition

	$\displaystyle f_{i}(a^{\prime}+k)=$	$\displaystyle f_{i+1}(b^{\prime}-k)\text{ for every }k\geq 0,$
	$\displaystyle f_{i}(a^{\prime}+k)=$	$\displaystyle g(k)\text{ for every }k\leq 0,$
	$\displaystyle f_{i+1}(b^{\prime}+k)=$	$\displaystyle g(k)\text{ for every }k\geq 0.$

Thus $F^{\prime}\xrightarrow{s(i,f_{i},f_{i+1})}F$ .

•

Suppose that $f_{i}\in\langle\overline{f_{i+1}}\rangle_{A}$ and $f_{i}$ is eventually left periodic with $T\in\{1,2\}$ . Then there exist $a,b\in\mathbb{Z}$ such that $f_{i}(a+z)=f_{i+1}(b-z)$ for every $z\in\mathbb{Z}$ and there exists $a^{\prime}\in\mathbb{Z}$ such that $f_{i}(z)=f_{i}(z-T)$ for every $z\leq a^{\prime}$ and $f_{i}(a^{\prime}+1)\neq f_{i}(a^{\prime}+1-T)$ , and $g(2\mathbb{Z})=f_{i}(a^{\prime})$ and $g(2\mathbb{Z}+1)=f_{i}(a^{\prime}-1)$ . Hence,

	$\displaystyle f_{i}(a^{\prime}+k)=$	$\displaystyle f_{i+1}(b^{\prime}-(a^{\prime}-a)-k)\text{ for every }k\in\mathbb{Z},$
	$\displaystyle f_{i}(a^{\prime}+k)=$	$\displaystyle g(k)\text{ for every }k\leq 0.$

Thus $F^{\prime}\xrightarrow{s(i,f_{i},f_{i+1})}F$ .

(3) Suppose that $F\xrightarrow{s(i,g,h)}F^{\prime}$ . We consider both of the possible splittings.

•

Suppose that $f_{i}$ is periodic with $T\in\{1,2\}$ and there exist $m,n\in\mathbb{Z}$ such that $g(m+k)=h(n-k)$ for every $k\in\mathbb{Z}$ and $g(m+k)=f_{i}(k)$ for every $k\leq 0$ . Then we have that $g\in\langle\overline{h}\rangle_{A}$ and $g$ is eventually left periodic. Therefore we can merge them an create $g^{\prime}\in\langle f_{i}\rangle_{A}$ . Thus, $F^{\prime}\xrightarrow{m(i)}F$ .
•

Suppose that there exists $z_{0}\in\mathbb{Z}$ such that $f_{i}(z_{0}-1)\neq f_{i}(z_{0}+1)$ , and there exist $m,n\in\mathbb{Z}$ such that $g(m+k)=h(n-k)$ for every $k\geq 0$ , $g(m+k)=f_{i}(z_{0}+k)$ for every $k\leq 0$ , and $h(n+k)=f_{i}(z_{0}+k)$ for every $k\geq 0$ . Observe that $h(n+1)\neq g(m-1)$ . Then, we might apply any of the three possibilities to merge. Thus $F^{\prime}\xrightarrow{m(i)}F$ .∎

Definition 4.15.

Let $F,G\in\mathcal{F}$ . We say that we can convert $F$ in $G$ if there exists a finite sequence of arrows

\displaystyle F\coloneqq F_{0}\xrightarrow{\alpha_{0}}F_{1}\xrightarrow{\alpha_{1}}\cdots\xrightarrow{\alpha_{n-1}}F_{n}\eqqcolon G

Proposition 4.16.

$\simeq_{S}$ is an equivalence relation.

Proof.

Let $F,G,H\in\mathcal{F}$ with $F=(f_{1},\ldots,f_{n})$ .

Reflexive: In a first case, consider that there exist $z_{1}\in\mathbb{Z}$ such that $f_{1}(z_{1}-1)\neq f_{1}(z_{1}+1)$ . Define

\displaystyle g\coloneqq\left\{\begin{array}[]{ll}f_{1}(z_{1}+z)&\text{ if }z\leq 0\\ f_{1}(z_{1})&\text{ if }z>0\end{array}\right.,\ h\coloneqq\left\{\begin{array}[]{ll}f_{1}(z_{1}+z)&\text{ if }z\geq 0\\ f_{1}(z_{1})&\text{ if }z<0.\end{array}\right.

On the other hand, for every $z\in\mathbb{Z}$ , $f_{1}(z-1)=f_{1}(z)$ and $f_{1}$ is periodic with $T\in\{1,2\}$ . Define

\displaystyle g\coloneqq\left\{\begin{array}[]{ll}f_{1}(z)&\text{ if }z\leq 0\\ f_{1}(0)&\text{ if }z>0\end{array}\right.,\ h\coloneqq\left\{\begin{array}[]{ll}f_{1}(z)&\text{ if }z\geq 0\\ f_{1}(0)&\text{ if }z<0.\end{array}\right.

In both case we have that $F\xrightarrow{s(1,g,h)}(g,h,f_{2},\ldots)\xrightarrow{m(1)}F$ .

Symmetry: Suppose that $F\simeq_{S}G$ . Then, there exist the sequence

\displaystyle F\coloneqq F_{0}\xrightarrow{\alpha_{0}}F_{1}\xrightarrow{\alpha_{1}}\cdots\xrightarrow{\alpha_{n-1}}F_{n}\eqqcolon G

with $\alpha_{i}\in\{d_{op}(i)\mid i\in\mathbb{Z}\}\cup\{a_{op}(i,g)\mid i\in\mathbb{Z},\ g:\mathbb{Z}_{1}\rightarrow X\}\cup\{m(i)\mid i\in\mathbb{Z}\}\cup\{s(i,g,h)\mid i\in\mathbb{Z},\ g,h:\mathbb{Z}_{1}\rightarrow X\}$ . By 4.8 and 4.14, we know that every map $\alpha_{i}$ have his opposite arrow $\beta_{i}\in\{d_{op}(i)\mid i\in\mathbb{Z}\}\cup\{a_{op}(i,g)\mid i\in\mathbb{Z},\ g:\mathbb{Z}_{1}\rightarrow X\}\cup\{m(i)\mid i\in\mathbb{Z}\}\cup\{s(i,g,h)\mid i\in\mathbb{Z},\ g,h:\mathbb{Z}_{1}\rightarrow X\}$ such that

\displaystyle F_{n}\xrightarrow{\beta_{n-1}}F_{n-1}\xrightarrow{\alpha_{n-2}}\cdots\xrightarrow{\alpha_{0}}F_{0}

Hence $G\simeq_{S}F$ .

Transitive: Suppose that $F\simeq_{S}G$ and $G\simeq_{S}H$ . Then, there exist the sequence

	$\displaystyle F\coloneqq F_{0}\xrightarrow{\alpha_{0}}F_{1}\xrightarrow{\alpha_{1}}\cdots\xrightarrow{\alpha_{n-1}}F_{n}\eqqcolon G$
	$\displaystyle G\coloneqq G_{0}\xrightarrow{\beta_{0}}G_{1}\xrightarrow{\beta_{1}}\cdots\xrightarrow{\beta_{n-1}}G_{n}\eqqcolon H$

\displaystyle F\coloneqq F_{0}\xrightarrow{\alpha_{0}}F_{1}\xrightarrow{\alpha_{1}}\cdots\xrightarrow{\alpha_{n-1}}F_{n}=G=G_{0}\xrightarrow{\beta_{0}}G_{1}\xrightarrow{\beta_{1}}\cdots\xrightarrow{\beta_{n-1}}G_{n}\eqqcolon H

and thus $F\simeq_{S}H$ . ∎

Let $F=(f_{1},\ldots,f_{n})\in\mathcal{F}$ , we denote the class of $F$ as $[F]_{S}$ or $[f_{1},\ldots,f_{n}]_{S}$ . Finally, we develop the operation between classes.

Proposition 4.17.

Let $F,G\in\mathcal{F}$ . Then $[F]_{S}\star[G]_{S}=[F\star G]_{S}$ is well-defined.

Proof.

Let $F,G\in\mathcal{F}$ , $F^{\prime}\in[F]_{S}$ and $G^{\prime}\in[G]_{S}$ , then there exist the sequences

	$\displaystyle F\coloneqq F_{0}\xrightarrow{\alpha_{0}}F_{1}\xrightarrow{\alpha_{1}}\cdots\xrightarrow{\alpha_{n-1}}F_{n}\eqqcolon F^{\prime},$
	$\displaystyle G\coloneqq G_{0}\xrightarrow{\beta_{0}}G_{1}\xrightarrow{\beta_{1}}\cdots\xrightarrow{\beta_{n-1}}G_{m}\eqqcolon G^{\prime}.$

Then we have that

	$\displaystyle F\star G=$	$\displaystyle F_{0}\star G\xrightarrow{\alpha_{0}\star 1_{G}}F_{1}\star G\xrightarrow{\alpha_{1}\star 1_{G}}\cdots\xrightarrow{\alpha_{n-1}\star 1_{G}}F_{n}\star G=F^{\prime}\star G,$
	$\displaystyle F^{\prime}\star G=$	$\displaystyle F^{\prime}\star G_{0}\xrightarrow{1_{F^{\prime}}\star\beta_{0}}F^{\prime}\star G_{1}\xrightarrow{1_{F^{\prime}}\star\beta_{1}}\cdots\xrightarrow{1_{F^{\prime}}\star\beta_{m-1}}F^{\prime}\star G_{m}=F^{\prime}\star G^{\prime}.\qed$

Considering this class, we observe in the following lemma that with this operation every element has inverses and left and right identities. Hence we are able to define the groupoid.

Lemma 4.18.

Let $F=(f_{1},\ldots,f_{n})\in\mathcal{F}$ , then $[F]_{S}\star[\overline{F}]_{S}=[f_{1},\overline{f_{1}}]_{S}$ and $[\overline{F}]_{S}\star[F]_{S}=[\overline{f_{n}},f_{n}]_{S}$ . Moreover, $[f_{1},\overline{f_{1}}]_{S}\star[F]_{S}=[F]_{S}=[F]_{S}\star[\overline{f_{n}},f_{n}]_{S}$ .

Proof.

Let $F=(f_{1},\ldots,f_{n})\in\mathcal{F}$ . Suppose that $n>1$ , then

	$\displaystyle[F]_{S}\star[\overline{F}]_{S}=$	$\displaystyle[f_{1},\ldots,f_{n}]_{S}\star[\overline{f_{n}},\ldots,\overline{f_{1}}]_{S}=[f_{1},\ldots,f_{n},\overline{f_{n}},\ldots,\overline{f_{1}}]_{S}$
	$\displaystyle=$	$\displaystyle[f_{1},\dots,f_{n-1},\overline{f_{n-1}},\ldots,\overline{f_{1}}]_{S}=[f_{1},\overline{f_{1}}]_{S},$
	$\displaystyle[\overline{F}]_{S}\star[F]_{S}=$	$\displaystyle[\overline{f_{n}},\ldots,\overline{f_{1}}]_{S}\star[f_{1},\ldots,f_{n}]_{S}=[\overline{f_{n}},\ldots,\overline{f_{1}},f_{1},\ldots,f_{n}]_{S}$
	$\displaystyle=$	$\displaystyle[\overline{f_{n}},\ldots,\overline{f_{2}},f_{2},\ldots,f_{n}]_{S}=[\overline{f_{n}},f_{n}]_{S},$
	$\displaystyle[f_{1},\overline{f_{1}}]_{S}\star[F]_{S}=$	$\displaystyle[f_{1},\overline{f_{1}}]_{S}\star[f_{1},\ldots,f_{n}]_{S}=[f_{1},\overline{f_{1}},f_{1},\ldots f_{n}]_{S}$
	$\displaystyle=$	$\displaystyle[f_{1},\ldots,f_{n}]_{S}=[F]_{S}=[f_{1},\ldots,f_{n},\overline{f_{n}},f_{n}]_{S}$
	$\displaystyle=$	$\displaystyle[f_{1},\ldots,f_{n}]_{S}\star[\overline{f_{n}},f_{n}]_{S}=[F]_{S}\star[\overline{f_{n}},f_{n}]_{S}$

If $n=1$ , we can find $g,h:\mathbb{Z}_{1}\rightarrow X$ such that their tails are eventually equal and $F=(f_{1})=(g,h)$ . Then we have that $\overline{F}=(\overline{f_{1}})=\overline{(g,h)}=(\overline{h},\overline{g})$ . Therefore

	$\displaystyle[F]_{S}\star[\overline{F}]_{S}=$	$\displaystyle[g,h]_{S}\star[\overline{h},\overline{g}]_{S}=[g,h,\overline{h},\overline{g}]_{S}=[f_{1},\overline{f_{1}}]_{S}$
	$\displaystyle[\overline{F}]_{S}\star[F]_{S}=$	$\displaystyle[\overline{h},\overline{g}]_{S}\star[g,h]_{S}=[\overline{h},\overline{g},g,h]_{S}=[\overline{f_{1}},f_{1}]_{S}$
	$\displaystyle[f_{1},\overline{f_{1}}]_{S}\star[F]_{S}=$	$\displaystyle[g,h,\overline{h},\overline{g}]_{S}\star[g,h]_{S}=[g,h,\overline{h},\overline{g},g,h]_{S}=[g,h]_{S}=[F]_{S}$
	$\displaystyle=$	$\displaystyle[g,h,\overline{h},\overline{g},g,h]_{S}=[g,h]_{S}\star[\overline{h},\overline{g},g,h]_{S}=[F]_{S}\star[\overline{f_{1}},f_{1}]_{S}.\qed$

Lemma 4.19.

Let $f,g:\mathbb{Z}_{1}\rightarrow X$ bornologous maps such that there exist $m,n\in\mathbb{Z}$ such that $f(n-z)=g(m-z)$ for every $z\in\mathbb{Z}$ . Then, $[f,\overline{f}]_{S}=[g,\overline{g}]_{S}$ .

Proof.

Let $f,g:\mathbb{Z}_{1}\rightarrow X$ bornologous maps such that there exist $m,n\in\mathbb{Z}$ such that $f(n-z)=g(m-z)$ for every $z\in\mathbb{Z}$ .

Suppose that $f$ is eventually left periodic with $T\in\{1,2\}$ , then $g$ satisfies that condition too. By 4.13, we obtain that $[f,\overline{f}]_{S}=[h]_{S}=[g,\overline{g}]$ defining

\displaystyle h(z)\coloneqq\left\{\begin{array}[]{ll}f(z_{0})&\text{ if }z\in 2\mathbb{Z}\\ f(z_{0}-1)&\text{ if }z\in 2\mathbb{Z}+1\end{array}\right.

for $z_{0}$ big enough.

Suppose that $f$ is not eventually left periodic with $T\in\{1,2\}$ , then there exists $n^{\prime}<n$ such that $f(n^{\prime}-1)\neq f(n^{\prime}+1)$ . Observe that

\displaystyle g(m-(n-n^{\prime}+1))=f(n^{\prime}-1)\neq f(n^{\prime}+1)=g(m-(n-n^{\prime}-1)).

By 4.13, we have that $[f]_{S}=[h,f^{\prime}]_{S}$ and $[g]_{S}=[h,g^{\prime}]_{S}$ defining

	$\displaystyle h(z)\coloneqq\left\{\begin{array}[]{ll}f(n^{\prime}+z)&\text{ if }z\leq 0\\ f(n^{\prime})&\text{ if }z>0\end{array}\right.,$	$\displaystyle\ f^{\prime}(z)\coloneqq\left\{\begin{array}[]{ll}f(n^{\prime})&\text{ if }z\leq 0\\ f(n^{\prime}+z)&\text{ if }z>0\end{array}\right.,$
		$\displaystyle g^{\prime}(z)\coloneqq\left\{\begin{array}[]{ll}f(n^{\prime})&\text{ if }z\leq 0\\ g(m-n+n^{\prime}+z)&\text{ if }z>0.\end{array}\right.$

Hence, $[f,\overline{f}]_{S}=[h,f^{\prime},\overline{f^{\prime}},\overline{h}]_{S}=[h,\overline{h}]_{S}=[h,g^{\prime},\overline{g^{\prime}},\overline{h}]_{S}=[g,\overline{g}]_{S}$ . ∎

Definition 4.20.

The extended groupoid of $X$ , $\overline{\pi}_{\leq 1}(X)$ , is the groupoid with objects the symmetric maps to $X$ and morphisisms the strings.

Suppose that we have $X$ and $Y$ semi-coarse spaces and a bornologous map $h:X\rightarrow Y$ . If we have a string $(f_{1},\ldots,f_{n})$ on $X$ , then $h(f_{1},\ldots,f_{n})\coloneqq(hf_{1},\ldots,hf_{n})$ is a string on $Y$ . From this definition, we directly observe that the string are a coarse invariant.

Proposition 4.21.

Let $X,Y$ be isomorphic semi-coarse spaces. Then $\overline{\pi}_{\leq 1}(X)\cong\overline{\pi}_{\leq 1}(X)$ as groupoids.

Along the following construction, we are going to define classes between affine functions. The behavior in the tails are much more important than the behavior in the middle of the functions, then we allow to delete some middle points whenever the results is still a bornologous map.

Definition 4.22.

Let $f:\mathbb{Z}_{1}\rightarrow X$ be a bornologous map. Let $z_{0}\in\mathbb{Z}$ and $x_{0}\in X$ .

•

We said that $g$ is the result of deleting $z_{0}$ in $f$ , or we can delete $z_{0}$ in $f$ , if

$\displaystyle g(z)\coloneqq\left\{\begin{array}[]{ll}f(z)&\text{ if }z<z_{0}\\ f(z+1)&\text{ if }z\geq z_{0}.\end{array}\right.$

is bornologous. This is denoted as $f\xrightarrow{d(z_{0})}g$ .

•

We said that $g$ is the result of adding $x_{0}$ at $z_{0}$ in $f$ , or we can add $x_{0}$ at $z_{0}$ , if

\displaystyle g(z)\coloneqq\left\{\begin{array}[]{ll}f(z)&\text{ if }z<z_{0}\\ x_{0}&\text{ if }z=z_{0}\\ f(z-1)&\text{ if }z>z_{0}.\end{array}\right.

is bornologous. This is denoted as $f\xrightarrow{a(z_{0},x_{0})}g$ .

Lemma 4.23.

Let $X$ be a semi-coarse space and $f,g:\mathbb{Z}_{1}\rightarrow X$ be bornologous maps. Then, $f\xrightarrow{d(z_{0})}g$ if and only if $g\xrightarrow{a(z_{0},f(z_{0}))}f$ .

Proof.

Suppose that $f\xrightarrow{d(z_{0})}g$ . By definition, we have that $g(z)=f(z)$ for every $z<z_{0}$ and $g(z)=f(z+1)$ for every $z\geq z_{0}$ . Moreover, the result of adding $f(z_{0})$ at $z_{0}$ in $g$ is

\displaystyle h(z)=\left\{\begin{array}[]{ll}g(z)&\text{ if }z<z_{0}\\ f(z_{0})&\text{ if }z=z_{0}\\ g(z-1)&\text{ if }z>z_{0}.\end{array}\right.

Replacing the values of $g$ , we have that

\displaystyle h(z)=\left\{\begin{array}[]{ll}f(z)&\text{ if }z<z_{0}\\ f(z_{0})&\text{ if }z=z_{0}\\ f(z-1+1)=f(z)&\text{ if }z>z_{0}.\end{array}\right.

Thus $g\xrightarrow{a(z_{0},f(z_{0}))}f$ .

Suppose that $g\xrightarrow{a(z_{0},f(z_{0}))}f$ . By definition we have that $f(z)=g(z)$ for every $z<z_{0}$ and $f(z)=g(z-1)$ for every $z>z_{0}$ . Moreover, the result of deleting $z_{0}$ in $f$ is

\displaystyle h(z)=\left\{\begin{array}[]{ll}f(z)&\text{ if }z<z_{0}\\ f(z+1)&\text{ if }z\geq z_{0}.\end{array}\right.

Replacing the values of $f$ , we have that

\displaystyle h(z)=\left\{\begin{array}[]{ll}g(z)&\text{ if }z<z_{0}\\ g(z+1-1)=g(z)&\text{ if }z\geq z_{0}.\end{array}\right.

Thus $f\xrightarrow{a(z_{0})}f$ . ∎

Lemma 4.24.

Let $X$ be a semi-coarse space and $f:\mathbb{Z}_{1}\rightarrow X$ be a bornologous map, then

\displaystyle f_{a(z_{0})}\coloneqq\left\{\begin{array}[]{ll}f(z)&\text{ if }z<z_{0}\\ f(z_{0})&\text{ if }z=z_{0}\\ f(z-1)&\text{ if }z>z_{0}\end{array}\right.

is a bornologous map, and therefore $f\xrightarrow{a(z_{0},f(z_{0}))}f_{a(z_{0})}$ .

Proof.

Let $f:\mathbb{Z}_{1}\rightarrow X$ be a bornologous maps. We take the subsets $X_{1}\coloneqq(-\infty,z_{0}]$ , $X_{2}\coloneqq[z_{0},z_{0}+1]$ and $X_{3}\coloneqq[z_{0}+1,\infty)$ . Observe that every controlled set in $\mathbb{Z}_{1}$ is an union of controlled sets in $\mathbb{Z}_{1}\mid_{X_{1}}$ , $\mathbb{Z}_{1}\mid_{X_{2}}$ and $\mathbb{Z}_{1}\mid_{X_{3}}$ , and $X=X_{1}\cup X_{2}\cup X_{3}$ . Thus, we can use 2.3 to prove that $f_{a(z_{0})}$ is a bornologous map. Observe that $f_{a(z_{0})}\mid_{X_{1}}=f\mid_{X_{1}}$ and $f_{a(z_{0})}\mid_{X_{3}}=f\mid_{[z_{0},\infty)}$ , which are bornologous maps because they are restrictions of a bornologous map. Further, $f_{a(z_{0})}\mid_{X_{2}}=f\mid_{\{z_{0}\}}$ which goes from just one point, then it is a bornologous map. Thus, $f_{a(z_{0})}$ is a bornologous map. ∎

Definition 4.25.

Let $X$ be a semi-coarse space and $f,g:\mathbb{Z}_{1}\rightarrow X$ be bornologous maps. We say that $f\simeq_{d}g$ if there exist a finite sequence such that

\displaystyle f=f_{0}\xrightarrow{\alpha_{0}}f_{1}\xrightarrow{\alpha_{1}}\cdots\xrightarrow{\alpha_{n-1}}f_{n}=g

with $\alpha_{i}\in\{d(z)\mid z\in\mathbb{Z}\}\cup\{a(z,x)\mid z\in\mathbb{Z},x\in X\}$ .

Theorem 4.26.

$\simeq_{d}$ is an equivalence relation.

Proof.

Let $X$ be a semi-coarse space and $f,g,h:\mathbb{Z}_{1}\rightarrow X$ be bornologous maps.

Reflexive: Since $f$ is bornologous, then $f_{a(z_{0})}$ is bornologous by 4.24. Moreover, by 4.23, we have that $f_{a(z_{0})}\xrightarrow{d(z_{0})}f$ . Then

\displaystyle f\xrightarrow{a(z_{0},f(z_{0}))}f_{a(z_{0})}\xrightarrow{d(z_{0})}f

Thus, $f\simeq_{d}g$ .

Symmetry: Suppose that $f\simeq_{d}g$ . Then there exist $f_{1},\ldots,f_{n-1}:\mathbb{Z}_{1}\rightarrow X$ bornologous maps and $\alpha_{0},\ldots,\alpha_{n-1}\in\{a(z,x)\mid z\in\mathbb{Z},x\in X\}\cup\{d(z)\mid z\in\mathbb{Z}\}$ such that

\displaystyle f=f_{0}\xrightarrow{\alpha_{0}}f_{1}\xrightarrow{\alpha_{1}}\cdots\xrightarrow{\alpha_{n-1}}f_{n}=g

For every $i$ :

•

If $\alpha_{i}=a(z_{i},x_{i})$ , then define $\beta_{i}\coloneqq d(z_{i})$ .
•

If $\alpha_{i}=d(z_{i})$ , then define $\beta_{i}\coloneqq a(z_{i},f_{i}(z_{i}))$ .

Thus, applying 4.23, we have the sequence

\displaystyle g=f_{n}\xrightarrow{\beta_{n-1}}f_{n-1}\xrightarrow{\beta_{n-2}}\cdots\xrightarrow{\beta_{0}}f_{0}=f,

and $g\simeq_{d}f$ .

Transitive: Suppose that $f\simeq_{d}g$ and $g\simeq_{d}h$ .Then there exist $f_{1},\ldots,f_{n-1},g_{1},\ldots,g_{m-1}:\mathbb{Z}_{1}\rightarrow X$ bornologous maps and $\alpha_{0},\ldots,\alpha_{n-1},\beta_{1},\ldots,\beta_{m-1}\in\{a(z,x)\mid z\in\mathbb{Z},x\in X\}\cup\{d(z)\mid z\in\mathbb{Z}\}$ such that

\displaystyle f=f_{0}\xrightarrow{\alpha_{0}}f_{1}\xrightarrow{\alpha_{1}}\cdots\xrightarrow{\alpha_{n-1}}f_{n}=g,\ g=g_{0}\xrightarrow{\beta_{0}}g_{1}\xrightarrow{\beta_{1}}\cdots\xrightarrow{\beta_{m-1}}f_{m}=h.

We can observe that

\displaystyle f=f_{0}\xrightarrow{\alpha_{0}}f_{1}\xrightarrow{\alpha_{1}}\cdots\xrightarrow{\alpha_{n-1}}f_{n}=g=g_{0}\xrightarrow{\beta_{0}}g_{1}\xrightarrow{\beta_{1}}\cdots\xrightarrow{\beta_{m-1}}f_{m}=h,

thus $f\simeq_{d}h$ . ∎

To work in the strings, we would like to see that this equivalence relation is well-defined in the affine classes. Suppose that $f\simeq_{d}g$ and take $\hat{f}\in\langle f\rangle_{A}$ . By definition, there exists a sequence

\displaystyle f\coloneqq f_{0}\xrightarrow{\alpha_{0}}f_{1}\xrightarrow{\alpha_{1}}\cdots\xrightarrow{\alpha_{n-1}}f_{n}\eqqcolon g,

and there exists $m\in\mathbb{Z}$ such that $\hat{f}(z)=f(z-m)$ for every $z\in\mathbb{Z}$ .

Define $\hat{f}_{i}:\mathbb{Z}_{1}\rightarrow X$ with the relation $\hat{f}_{i}(z)=f_{i}(z-m)$ for every $0<i\leq n$ .

First suppose that $\alpha_{i}=d(z_{i})$ , then

	$\displaystyle f_{i+1}(z)=\left\{\begin{array}[]{ll}f_{i}(z)&\text{ if }z<z_{i}\\ f_{i}(z+1)&\text{ if }z\geq z_{i}\end{array}\right.\ =\$	$\displaystyle\hat{f}_{i+1}(z+m)=\left\{\begin{array}[]{ll}\hat{f}_{i}(z+m)&\text{ if }z<z_{i}\\ \hat{f}_{i}(z+m+1)&\text{ if }z\geq z_{i}\end{array}\right.$
	$\displaystyle=\$	$\displaystyle\hat{f}_{i+1}(z)=\left\{\begin{array}[]{ll}\hat{f}_{i}(z)&\text{ if }z<z_{i}+m\\ \hat{f}_{i}(z+1)&\text{ if }z\geq z_{i}+m\end{array}\right.$

Then $\hat{f}_{i}\xrightarrow{d(z_{i}+m)}\hat{f}_{i+1}$ .

On the other hand, suppose that $\alpha_{i}=a(z_{i},x_{i})$ , then

	$\displaystyle f_{i+1}(z)=\left\{\begin{array}[]{ll}f_{i}(z)&\text{ if }z<z_{i}\\ x_{i}&\text{ if }z=z_{i}\\ f_{i}(z+1)&\text{ if }z>z_{i}\end{array}\right.\ =\$	$\displaystyle\hat{f}_{i+1}(z+m)=\left\{\begin{array}[]{ll}\hat{f}_{i}(z+m)&\text{ if }z<z_{i}\\ x_{i}&\text{ if }z=z_{i}\\ \hat{f}_{i}(z+m+1)&\text{ if }z\geq z_{i}\end{array}\right.$
	$\displaystyle=\$	$\displaystyle\hat{f}_{i+1}(z)=\left\{\begin{array}[]{ll}\hat{f}_{i}(z)&\text{ if }z<z_{i}+m\\ x_{i}&\text{ if }z=z_{i}+m\\ \hat{f}_{i}(z+1)&\text{ if }z\geq z_{i}+m\end{array}\right.$

Then $\hat{f}_{i}\xrightarrow{a(z_{i}+m,x_{i})}\hat{f}_{i+1}$ . Therefore $\hat{f}\simeq\hat{f}_{n}\eqqcolon\hat{g}$ , and $\hat{g}\in\langle g\rangle_{A}$ , what gives us the following definition.

Definition 4.27.

Let $X$ be a semi-coarse space and $f,g:\mathbb{Z}_{1}\rightarrow X$ be bornologous maps. We say that $\langle f\rangle_{A}\simeq_{d}\langle g\rangle_{A}$ if there exists $\hat{g}\in\langle g\rangle_{A}$ such that $f\simeq_{d}\hat{g}$ .

Observe that this relation is well defined for the procedure above; if we take $\hat{f}\in\langle f\rangle$ , there exists $h\in\langle\hat{g}\rangle_{A}=\langle g\rangle_{A}$ such that $\hat{f}\simeq_{d}h$ .

Proposition 4.28.

Let $X$ be a semi-coarse space and $f,g:\mathbb{Z}_{1}\rightarrow X$ be eventually equal symmetric maps. Then $\langle f\rangle_{A}\simeq_{d}\langle g\rangle_{A}$ .

Proof.

Let $f,g:\mathbb{Z}_{1}\rightarrow X$ be eventually equal symmetric maps. Then there exists $m,n\in\mathbb{N}$ such that $f(z+n)=g(z+m)$ for every $z\in\mathbb{N}_{0}$

We first reduce $f$ and apply an analogous algorithm with $g$ . Since $f$ is symmetric, then $f(n)=f(-n)$ for every $n\in\mathbb{N}_{0}$ ; thus, applying 2.3, we obtain that the result of deleting $0$ in $f$ , call it $f_{1}$ , is a bornologous map. Moreover, the result of deleting $0$ in $f_{1}$ , call it $f^{\prime}_{1}$ is bornologous because $f_{1}=(f_{1}^{\prime})_{a(-1)}$ . Observe that, if we define $\hat{f}_{1}:\mathbb{Z}_{1}\rightarrow X$ such that $\hat{f}_{1}(z)=f_{1}^{\prime}(z-1)$ , we obtain the following

\displaystyle\hat{f}_{1}(z)=\left\{\begin{array}[]{ll}f(z+1)&z\geq 0\\ f(z-1)&z<0,\end{array}\right.\text{ and }f\xrightarrow{d(0)}f_{1}\xrightarrow{d(0)}f^{\prime}_{1}\in\langle\hat{f}_{1}\rangle_{A}.

We can repeat this process from $1$ to $n-1$ ; it means, we call $f_{i+1}$ the result of deleting $0$ in $\hat{f}_{i}$ , $f_{i+1}^{\prime}$ the result of deleting $0$ in $f_{i+1}$ , and we define $\hat{f}_{i+1}:\mathbb{Z}_{1}\rightarrow X$ such that $\hat{f}_{i+1}(z)=f^{\prime}_{i+1}(z-1)$ , obtaining that

\displaystyle\hat{f}_{i+1}(z)=\left\{\begin{array}[]{ll}\hat{f}_{i}(z+1)&z\geq 0\\ \hat{f}_{i}(z-1)&z<0,\end{array}\right.\text{ and }\hat{f}_{i}\xrightarrow{d(0)}f_{i+1}\xrightarrow{d(0)}f^{\prime}_{i+1}\in\langle\hat{f}_{i+1}\rangle_{A}.

Therefore, $\langle f\rangle_{A}\simeq_{d}\langle\hat{f}_{n}\rangle_{A}$ . Using the same algorithm with $g$ until $m-1$ we obtain that $\langle g\rangle_{A}\simeq_{d}\langle\hat{g}_{m}\rangle_{A}$ . It only leaves to verify that, by definition

	$\displaystyle\hat{f}_{n}=\left\{\begin{array}[]{ll}f(z+n)&z\geq 0\\ f(z-n)&z<0\end{array}\right.\ =\$	$\displaystyle\hat{f}_{n}=\left\{\begin{array}[]{ll}g(z+m)&z\geq 0\\ g(z-m)&z<0\end{array}\right.$
	$\displaystyle\ =\$	$\displaystyle\hat{g}_{m}$

Concluding that $\langle f\rangle_{A}\simeq_{d}\langle g\rangle_{A}$ . ∎

In the result above we prove that 4.25 allows to make homotopy deformations in the path which joins two tails. This proposition is the reason for the fundamental groupoid (4.32) contains the semi-coarse fundamental group for every point in the space.

Proposition 4.29.

Let $X$ be a semi-coarse space and $f,g:\mathbb{Z}_{1}\rightarrow X$ such that there exists $N\in\mathbb{N}$ satisfying

•

$f\mid_{(-\infty,-N]\cup[N,\infty)}=g\mid_{(-\infty,-N]\cup[N,\infty)}$ .
•

$f\mid_{[-N,N]}\simeq_{sc}g\mid_{[-N,N]}$ through $H:[-N,N]\times\mathbb{Z}_{1}\rightarrow X$ such that $H(-N,-)=f(-N)$ and $H(N,-)=f(N)$ .

Then $\langle f\rangle_{A}\simeq_{d}\langle g\rangle_{A}$ .

Proof.

Let $X$ be a semi-coarse space and $f,g:\mathbb{Z}_{1}\rightarrow X$ and $\{0,1\}$ with the subspace structure. First we consider a homotopy with just one step. Suppose that there exists $N\in\mathbb{N}$ satisfying

•

$f\mid_{(-\infty,-N]\cup[N,\infty)}=g\mid_{(-\infty,-N]\cup[N,\infty)}$ .
•

There exists $H:[-N,N]\times\{0,1\}\rightarrow X$ bornologous map such that $H(-N,-)=f(-N)$ , $H(N,-)=f(N)$ , $H(z,0)=f(z)$ and $H(z,1)=g(z)$ .

By the condition two, observe that $\{(f(i),g(i+1)),(g(i+1),f(i+2))\}$ is controlled in $X$ for every $i\in\mathbb{Z}$ , in particular for $-N\leq i\leq N$ . Thus, we obtain the followin

\displaystyle f\xrightarrow{a(-N+1)}f^{\prime}_{1}\xrightarrow{a(-N+1,g(-N+1)}f^{\prime\prime}_{1}\xrightarrow{d(-N+2)}f^{\prime\prime\prime}_{1}.

Observe that $f(z)=f^{\prime\prime\prime}_{1}(z)$ for every $z\neq-N+1$ and $f^{\prime\prime\prime}_{1}(-N+1)=g(-N+1)$ . We can repeat this interaction as

\displaystyle f^{\prime\prime\prime}_{i}\xrightarrow{a(-N+i+1)}f^{\prime}_{i+1}\xrightarrow{a(-N+i+1,g(-N+i+1)}f^{\prime\prime}_{i+1}\xrightarrow{d(-N+i+2)}f^{\prime\prime\prime}_{i+1},

obtaining that $f_{i}^{\prime\prime\prime}(z)=f^{\prime\prime\prime}_{i+1}(z)$ for every $z\neq-N+i+1$ and $f^{\prime\prime\prime}_{i+1}(-N+i+1)=g(-N+i+1)$ . Making this algorithm until $i=2N-2$ we obtain $f^{\prime\prime\prime}_{2N-1}=g$ . Thus $f\simeq_{d}g$ .

In the general case, we have that there exists $N\in\mathbb{N}$ satisfying

•

$f\mid_{(-\infty,-N]\cup[N,\infty)}=g\mid_{(-\infty,-N]\cup[N,\infty)}$ .
•

$f\mid_{[-N,N]}\simeq_{sc}g\mid_{[-N,N]}$ through $H:[-N,N]\times\mathbb{Z}_{1}\rightarrow X$ such that $H(-N,-)=f(-N)$ and $H(N,-)=f(N)$ .

Since $H$ is a homotopy, there exists $a<b\in\mathbb{Z}$ such that $H(z,x)=f(x)$ for every $x\leq a$ and $H(z,x)=g(z)$ for every $x\geq b$ . Define $f_{i}:[-N,N]\rightarrow X$ such that $f_{i}(z)=H(z,i)$ for every $a\leq i\leq b$ Hence, taking $H_{i}\coloneqq H\mid_{[-N,N]\times\{i,i+1\}}$ we obtain that

\displaystyle f=f_{a}\simeq_{d}f_{a+1}\simeq_{d}\cdots\simeq_{d}f_{b-1}\simeq_{d}f_{b}=g

Concluding that $f\simeq_{d}g$ . ∎

Definition 4.30.

Let $(f_{1},\ldots,f_{n})$ and $(g_{1},\ldots,g_{n})$ be $n$ strings. We say that $(f_{1},\ldots,f_{n})\simeq_{d}(g_{1},\ldots,g_{n})$ if $f_{i}\simeq_{d}g_{i}$ .

Definition 4.31.

Let $F,G\in\mathcal{F}$ . We say that we can convert and deform $F$ in $G$ if there exists a finite sequence of arrows

\displaystyle F\coloneqq F_{0}\simeq_{\alpha_{0}}F_{1}\simeq_{\alpha_{1}}\cdots\simeq_{\alpha_{n-1}}F_{n}\eqqcolon G

with $\alpha_{i}\in\{S,d\}$ . We express this relation as $F\simeq G$ .

Definition 4.32.

The fundamental groupoid of $X$ , $\pi_{\leq 1}(X)$ , is the groupoid with objects the symmetric maps to $X$ and morphisisms the classes of strings under the delete and deform relation.

5. Relative Fundamental Groupoid and Van Kampen Theorem

In the semi-coarse fundamental groupoid, Van Kampen Theorem is not true in general. We can restrict the tails of the maps from $\mathbb{Z}_{1}$ to get a relative version of the fundamental groupoid; there we obtain the theorem with the correct partition of the space.

Definition 5.1.

Let $X$ be a semi-coarse space and $\mathcal{U}\subset 2^{X}$ . The $\mathcal{U}$ -relative fundamental groupoid, $\pi_{\leq 1}(X,\mathcal{U})$ , is the category with the following ingredients:

•

$Ob(\pi_{\leq 1}(X,\mathcal{U}))$ are the symmetric maps which are eventually equal to a symmetric map with image contained in some $U\in\mathcal{U}$ .
•
$Hom_{\pi_{\leq 1}(X,\mathcal{U})}(f,g)$ are strings $F=[f_{1},\ldots,f_{n}]$ such that for every $i\in\{0,\ldots,n\}$ there exists $U_{i}\in\mathcal{U}$ such that
- –
  
  The left tail of $f_{i}$ and the right tail if $f_{i+1}$ land in $U_{i}$ for $i\in{1,\ldots,n-1}$ .
- –
  
  The left tail of $f_{1}$ lands in $U_{0}$ .
- –
  
  The right tail of $f_{n}$ lans in $U_{n}$ .

First, we introduce a lemma which mimics the implementation of the lebesgue covering lemma [Brown_2006] in the van Kampen Theorem for the semi-coarse fundamental groupoid.

Lemma 5.2.

Let $X$ be a semi-coarse space, $A,B\subset X$ well-split $X$ and $f:\mathbb{Z}_{1}\rightarrow X$ . Additionally, suppose that each tail is within $A$ or $B$ . Then, there exists $f_{1},\ldots,f_{n}:\mathbb{Z}_{1}\rightarrow X$ such that $[f]=[f_{1}]\star[f_{2}]\star\ldots\star[f_{n}]$ , $f_{i}(\mathbb{Z})\subset A$ or $f_{i}(\mathbb{Z})\subset B$ for every $i$ and the right tail of $f_{i}$ is the same as the left tail of $f_{i+1}$

Proof.

Let $X$ be a semi-coarse space, $A,B\subset X$ well-split $X$ and $f:\mathbb{Z}_{1}\rightarrow X$ . Trying to simplify the redaction, we say that the left tail is eventually in a set $U$ and the right tail is eventually in a set $V$ .

If $f(\mathbb{Z})\subset A$ or $f(\mathbb{Z})\subset B$ , we already have what we wanted to. Suppose that $f(\mathbb{Z})\not\subset A$ and $f(\mathbb{Z})\not\subset B$ . Without loss of generality, suppose that $U\subset A$ . There exists $z\in\mathbb{Z}$ such that $f(z)\in B-A$ . Then we take the set $z_{1}<z_{2}<\ldots<z_{k}$ such that $f(z_{i})\in B-A$ and $f(z_{i+1})\in A-B$ for every $i\in\{1,3,\ldots,2\lfloor\frac{k}{2}\rfloor+1\}$ .

Figure 5. Dividing

f

If $f(z_{i})$ belongs to $A\cap B$ , we don’t do anything. Otherwise, since $A,B$ split well $X$ , there exists $\lambda_{i}:\{0,1,2\}\rightarrow X$ in $A\sqcup_{A\cap B}B$ such that $\lambda_{i}(0)=f(z_{i}-1)$ , $\lambda_{i}(2)=f(z_{i})$ and $\lambda_{i}(1)\in A\cap B$ . Then $f$ belongs to the same class of $f^{\prime}$ , with the addition of $\lambda_{i}(1)$ among $z_{i}-1$ and $z_{i}$ . Hence we obtain a new sequence $z^{\prime}_{1}<z^{\prime}_{2}<\ldots<z^{\prime}_{k}$ such that $f(z_{i}^{\prime})\in A\cap B$ , and $f(z^{\prime}_{i})\in B-A$ and $f(z^{\prime}_{i+1})\in A-B$ for every $i\in\{1,3,\ldots,2\lfloor\frac{k}{2}\rfloor+1\}$ .

Figure 6. Dividing

f^{\prime}

We define the following maps for $i\in\{1,\ldots k-1\}$

	$\displaystyle f_{L}(z)\coloneqq\left\{\begin{array}[]{ll}f^{\prime}(z)&\text{ if }z\leq z^{\prime}_{1}\\ f^{\prime}(z^{\prime}_{1})&\text{ if }z>z^{\prime}_{1},\end{array}\right.$	$\displaystyle f_{R}(z)\coloneqq\left\{\begin{array}[]{ll}f^{\prime}(z^{\prime}_{k})&\text{ if }z<z^{\prime}_{k}\\ f^{\prime}(z)&\text{ if }z\geq z^{\prime}_{k},\end{array}\right.$
		$\displaystyle f_{i}(z)\coloneqq\left\{\begin{array}[]{ll}f^{\prime}(z^{\prime}_{i})&\text{ if }z<z^{\prime}_{i}\\ f^{\prime}(z)&\text{ if }z^{\prime}_{i}\leq z\leq z^{\prime}_{i+1}\\ f^{\prime}(z^{\prime}_{i+1})&\text{ if }z>z^{\prime}_{i+1}.\end{array}\right.$

Hence, by construction $f_{L}(\mathbb{Z})\subset A$ , $f_{i}(\mathbb{Z})\subset B$ and $f_{i+1}(\mathbb{Z})\subset A$ for $i\in\{1,3,\ldots\}$ , and $f_{R}$ is subset of $A$ or $B$ . In addition, the right tail of $f_{i}$ and the left tail of $f_{i+1}$ are the same and constant. ∎

Previous to the main theorem’s proof, we are going to consider a connected semi-coarse space; remembering that this implies that $A\cap B\neq\varnothing$ if $A,B$ well-split $X$ , for 3.8.

Theorem 5.3.

Let $X$ be a connected semi-coarse space and $\mathcal{U}\subset 2^{X}$ such that $U\in\mathcal{U}$ is connected. Consider $A,B\subset X$ such that

(1)

$A,B$ well-split $X$ ,
(2)

for every component of $A$ , $B$ or $A\cap B$ there exists $U$ such that meets the component, and
(3)

for every $U\in\mathcal{U}$ , $U\subset A$ and $U\subset B$ .

Then the diagram

is a pushout.

Proof.

Let $X$ be a connected semi-coarse space and $\mathcal{U}\subset 2^{X}$ such that $U\in\mathcal{U}$ is connected.

Let $G$ be a groupoid and the following diagram be commutative

We proceed to construct a map $h:\pi_{\leq 1}(X,\mathcal{U})\rightarrow G$ such that is the unique such that $j_{2}=hi_{2}$ and $j_{1}=hi_{1}$ .

First we work with the objects; we define $h:Ob(\pi_{\leq 1}(X,\mathcal{U}))\rightarrow Ob(G)$ such that $h(\langle f\rangle_{A})=j_{1}(\langle f\rangle_{A})$ if $\langle f\rangle_{A}\in Ob(\pi_{\leq 1}(A,\mathcal{U}\cap A))$ and $h(\langle f\rangle_{A})=j_{2}(\langle f\rangle_{A})$ if $\langle f\rangle_{A}\in Ob(\pi_{\leq 1}(B,\mathcal{U}\cap B))$ . Since every $U\in\mathcal{U}$ satisfies that $U\subset A$ or $U\subset B$ , then for every $\langle f\rangle_{A}\in Ob(\pi_{\leq 1}(X,\mathcal{U}))$ the image of $g$ is completely contained in some $U\in\mathcal{U}$ , then it is totally contained in $A$ or $B$ . Hence $\langle f\rangle_{A}\in Ob(\pi_{\leq 1}(A,\mathcal{U}\cap A))$ or $\langle f\rangle_{A}\in Ob(\pi_{\leq 1}(B,\mathcal{U}\cap B))$ . If $\langle g\rangle_{A}$ is object of both categories, then is well-defined because $j_{2}i_{B}=j_{1}i_{A}$ .

By 5.2, we can write

\displaystyle[f]=[f_{1,L}]\star[f^{\prime}_{1,1}]\star\ldots[f^{\prime}_{1,k_{1}}]\star[f_{1,R},f_{2,L}]\star[f^{\prime}_{2,1}]\star\ldots\star[f_{n,R}]

such that every factor is a morphism in $\pi_{\leq 1}(A,\mathcal{U}\cap A)$ or in $\pi_{\leq 1}(B,\mathcal{U}\cap B)$ . Observe that every factor have constant tails, except the left tail of $f_{1,L}$ and the right tail of $f_{n,R}$ . In addition, this constant tails are in a component $A\cap B$ , then, there exists $U_{i,j}\in\mathcal{U}$ which meets such component. Then, there exists a bornologous map $\gamma^{\prime}_{i,j}:\{0,m_{i,j}\}\rightarrow X$ totally contained in the component such that $\gamma^{\prime}_{i,j}(0)$ is the right tail of $f_{i,j}$ and $f_{i,j}(m_{i,j})\in U_{i,j}$ . Through $\gamma^{\prime}_{i,j}$ , we define

\displaystyle\gamma_{i,j}(z)\coloneqq\left\{\begin{array}[]{ll}\gamma^{\prime}_{i,j}(0)&\text{ if }z\leq 0\\ \gamma^{\prime}_{i,j}(z)&\text{ if }0<z<m_{i,j}\\ \gamma^{\prime}_{i,j}(m_{i,j})&\text{ if }z\leq m_{i,j}.\end{array}\right.

with $f_{i,0}=f_{i,L}$ and $f_{i,k_{n}+1}=f_{i,R}$ . Therefore we take

\displaystyle[g_{i,j}]=[\gamma_{i,j-1}]\star[f_{i,j}]\star[\overline{\gamma_{i,j}}]\text{ and }[g_{i,R},g_{i+1,L}]=[\gamma_{i,k_{n}}]\star[f_{i,j}]\star[\overline{\gamma_{i+1,0}}]

with the exceptions $[g_{1,L}]=[f_{1,L}]\star[\gamma_{1,0}]$ and $[g_{n,R}]=[\overline{\gamma_{n,k_{n}}}]\star[f_{n,R}]$ . Every $g_{i,j}$ is completly contained in $A$ or $B$ , then $[g_{i,j}]$ is a string in $A$ or in $B$ . Thus

\displaystyle[f]=\iota_{1,0}[g_{1,L}]\star\iota_{1,1}[g_{1,1}]\star\ldots\iota_{n,k_{n}+1}[g_{n,R}]

where $\iota_{i,j}\in\{i_{1},i_{2}\}$ . In this order of ideas, we can suspect that, in case such $h$ homomorphism exists, then

\displaystyle h[f]=\kappa_{1,0}[g_{1,L}]\star\kappa_{1,1}[g_{1,1}]\star\ldots\kappa_{n,k_{n}+1}[g_{n,R}]

such that $\kappa_{i,j}=j_{1}$ if $\iota_{i,j}=i_{1}$ and $\kappa_{i,j}=j_{2}$ if $\iota_{i,j}=i_{2}$ .

Hence this assure the uniqueness of $h$ , and proves that $\pi_{\leq 1}(X,\mathcal{U})$ is generates as a grupoid by the images of $\pi_{\leq 1}(A,\mathcal{U}\cap A)$ and $\pi_{\leq 1}(B,\mathcal{U}\cap U)$ .

The laborious part of this proof is to show that this construction does not depend of the election of the element of $[f]$ or the election of $\gamma^{\prime}_{i,j}$ . To prove that it doesn’t depend of the election of the $\gamma^{\prime}_{i,j}$ , we take $U^{\prime}_{i,j}$ which meets the component and $\gamma^{\prime}_{i,j,2}:\{0,\ldots,m_{i,j,2}\}\rightarrow X$ totally contained in the component such that $\gamma^{\prime}_{i,j,2}(0)$ is the right tail of $f_{i,j}$ and $f_{i,j}(m_{i,j,2})\in U^{\prime}_{i,j}$ . We define $\gamma_{i,j,2}$ as the same way we define $\gamma_{i,j}$ and observe that:

	$\displaystyle\kappa_{i,j}[g_{i,j}]\kappa_{i^{>},j^{>}}[g_{i^{>},j^{>}}]=$	$\displaystyle\kappa_{i,j}[\overline{\gamma_{i^{<},j^{<}}},f_{i,j},\gamma_{i,j},\overline{\gamma_{i,j}},\gamma_{i,j,2},\overline{\gamma_{i,j,2}},\gamma_{i,j}]\kappa_{i^{>},j^{>}}[\overline{\gamma_{i,j}},f_{i^{>},j^{>}}\gamma_{i^{>},j^{>}}]$
	$\displaystyle=$	$\displaystyle\kappa_{i,j}[\overline{\gamma_{i^{<},j^{<}}},f_{i,j},\gamma_{i,j,2},\overline{\gamma_{i,j,2}},\gamma_{i,j}]\kappa_{i^{>},j^{>}}[\overline{\gamma_{i,j}},f_{i^{>},j^{>}}\gamma_{i^{>},j^{>}}]$
	$\displaystyle=$	$\displaystyle\kappa_{i,j}[\overline{\gamma_{i^{<},j^{<}}},f_{i,j},\gamma_{i,j,2}]\kappa_{i^{>},j^{>}}[\overline{\gamma_{i,j,2}},\gamma_{i,j},\overline{\gamma_{i,j}},f_{i^{>},j^{>}}\gamma_{i^{>},j^{>}}]$
	$\displaystyle=$	$\displaystyle\kappa_{i,j}[\overline{\gamma_{i^{<},j^{<}}},f_{i,j},\gamma_{i,j,2}]\kappa_{i^{>},j^{>}}[\overline{\gamma_{i,j,2}},f_{i^{>},j^{>}}\gamma_{i^{>},j^{>}}].$

With a similar argument, we observe that the image neither depends of the election of the value of $f^{\prime}(z^{\prime}_{k})$ in the construction 5.2. Consider that other $\lambda^{\prime}_{k}:\{0,1,2\}\rightarrow X$ in $A\sqcup_{A\cap B}B$ such that $\lambda^{\prime}_{k}(0)=f(z_{k}-1)$ and $\lambda^{\prime}_{k}(2)=f_{z_{k}}$ . Since $A,B$ well-split $X$ , there exists a path $\alpha:\{0,\ldots,m^{\prime}\}\rightarrow A\cap B$ such that $\alpha(0)=f^{\prime}(z^{\prime}_{k})$ and $\alpha(m^{\prime})=\lambda^{\prime}_{k}(1)$ . We define $\gamma_{i,j,2}=\alpha\star\gamma_{i,j}$ , $f^{\prime\prime}_{i,j}$ equal to $f_{i,j}$ but replacing its right tail by $\lambda^{\prime}_{k}(1)$ and $f^{\prime\prime}_{i^{>},j^{>}}$ is equal to $f_{i^{>},j^{>}}$ but replacing its left tail by $\lambda^{\prime}_{k}(1)$ . Thus

	$\displaystyle\kappa_{i,j}[g_{i,j}]\kappa_{i^{>},j^{>}}[g_{i^{>},j^{>}}]=$	$\displaystyle\kappa_{i,j}[\overline{\gamma_{i^{<},j^{<}}},f_{i,j},\gamma_{i,j}]\kappa_{i^{>},j^{>}}[\overline{\gamma_{i,j}},f_{i^{>},j^{>}},\gamma_{i^{>},j^{>}}]$
	$\displaystyle=$	$\displaystyle\kappa_{i,j}[\overline{\gamma_{i^{<},j^{<}}},f^{\prime\prime}_{i,j},\gamma_{i,j,2}]\kappa_{i^{>},j^{>}}[\overline{\gamma_{i,j,2}},f^{\prime\prime}_{i^{>},j^{>}},\gamma_{i^{>},j^{>}}]$
	$\displaystyle=$	$\displaystyle\kappa_{i,j}[\overline{\gamma_{i^{<},j^{<}}},f^{\prime\prime}_{i,j},\gamma_{i,j,2}]\kappa_{i^{>},j^{>}}[\overline{\gamma_{i,j,2}},f^{\prime\prime}_{i^{>},j^{>}},\gamma_{i^{>},j^{>}}]$

The last equality follow from we always form the graph in Figure 7.

Figure 7. Deleting

\alpha

’s

Now, to see that it does not depend of the representative, it is enough to work with the arrows defined in 4.7, 4.13 and 4.22. Then, we define $F\coloneqq(f_{1},\ldots,f_{n})$ and check case by case:

$F\xrightarrow{d_{op}(i)}G)$ We can apply the same algorithm as above to $f_{i}$ , getting that

\displaystyle[f_{i}]=[g_{i,L},g_{i,1},\ldots,g_{i,k_{i}},g_{i,R}],

and then we have obtain that

\displaystyle[\overline{f_{i}}]=[\overline{g_{i,R}},\overline{g_{i,k_{i}}},\ldots\overline{g_{i,1}},\overline{g_{i,L}}],

Thus,

\displaystyle[F]=[f_{1},\ldots,f_{i-1},g_{i,L},\overline{g_{i,L}},\ldots,f_{n}]=[G].

$F\xrightarrow{a_{op}(i,g)}G)$ The same as the case above, but applying the procedure to $g$ .

$F\xrightarrow{m(i)}G)$ We merge maps in four different ways. We work in the whole case up the second one, because is totally analogous to the first one

•

$f_{i}$ is periodic with $T\in\{1,2,3\}$ and $f_{i+1}$ is not periodic. Suppose that the right tail of $f_{i}$ is in $U\in\mathcal{U}$ and, without loss of generality, $U\subset A$ . Then $f_{i}(\mathbb{Z})\subset U\subset A$ . Thus $j_{1}([f_{i}])$ is defined. Divide $f_{i+1}$ as we construct in 5.2

$\displaystyle[f_{i+1}]=[g_{i+1,L},g_{i+1,1},\ldots,g_{i+1,n_{i}},g_{i+1,R}]$

Hence $g_{i+1,L}(\mathbb{Z})\subset A$ and $[f_{i},g_{i+1,L}]=[g^{\prime\prime}_{L}]\in\pi_{\leq 1}(A,\mathcal{U}\cap A)$ , with $g^{\prime\prime}_{L}$ the merging of $f_{i},g_{i+1,L}$ (4.13). Thus

$\displaystyle j_{1}[g_{i-1,R},f_{i},g_{i+1,L}]=$ $\displaystyle j_{1}[g_{i-1,R},g^{\prime\prime}_{i+1,L}]$

It only rest to observe that the merging of $f_{i},f_{i+1}$ , $g^{\prime\prime}$ is divided as

$\displaystyle[g^{\prime\prime}]=[g^{\prime\prime}_{L},f_{i+1,1},\ldots,f_{i+1,R}].$

Getting that $h$ is well defined for this way to merge.

•

$f_{i},f_{i+1}$ are not periodic, and $f_{i}$ is eventually right periodic with $T\in\{1,2,3\}$ or is not eventually right periodic. We already know that there exists $a^{\prime},b^{\prime}\in\mathbb{Z}$ such that $f_{i}(a^{\prime}+z)=f_{i+1}(b^{\prime}-z)$ for every $z\geq 0$ and $f(a^{\prime}-1)\neq f_{i+1}(b^{\prime}+1)$ . Let’s call $g$ the merging of $f_{i}$ and $f_{i+1}$ (4.13). By 5.2, we have

	$\displaystyle[f_{i}]=$	$\displaystyle[f_{i,L},f_{i,1},\ldots,f_{i,n_{i}},f_{i,R}],$
	$\displaystyle[f_{i+1}]=$	$\displaystyle[f_{i+1,L},f_{i+1,1},\ldots,f_{i+1,n_{i}},f_{i+1,R}],$
	$\displaystyle[g^{\prime\prime}]=$	$\displaystyle[g^{\prime\prime}_{i,L},g^{\prime\prime}_{i,1},\ldots,g^{\prime\prime}_{i,n^{\prime}},g^{\prime\prime}_{i,R}].$

Observe that $a^{\prime}$ lands in one $f_{i,j}$ with $j\in\{0,\ldots,n_{i}+1\}$ and $f_{i+1,j^{\prime}}$ with $j^{\prime}\in\{0,\ldots,n_{i+1}+1\}$ . Then we have that:

\displaystyle[f_{i},f_{i+1}]=[f_{i,L},f_{i,1},\ldots,f_{i,j},f_{i+1,j^{\prime}},f_{i+1,j^{\prime}+1},f_{i+1,R}]=[g^{\prime\prime}]

•

$f_{i}\in\langle\overline{f_{i+1}}\rangle_{A}$ and $f_{i}$ is eventually left periodic with $T\in\{1,2\}$ . The procedure is the same as the previous case.

$F\xrightarrow{a(i,f,g)}G)$ The sketchs of the proofs are the same as merging. You separate the maps through 5.2 and see how they work together.

In the cases of $\simeq_{d}$ , we can only work with specific maps; thus, we take $f_{i}$ and $g$

$f_{i}\xrightarrow{d(j)}g)$ Consider the the $z_{k}$ in $f_{i}$ as in 5.2. If $z_{k}<j<z_{k+1}$ , there is no problem in the construction when we remove the element in $j$ . Consider that $j=z_{k}$ for some $k$ .

Without loss of generality, suppose that $f_{i}(z_{k})\in B$ , and observe that $\{f_{i}(j-1),f_{i}(j+1))\}\in\mathcal{V}$ , since $f_{i}\xrightarrow{d(j)}g$ . If $f_{i}(j-1)\in A\cap B$ , there is nothing to do, because we select $z_{k}$ as $z^{\prime}_{k}$ . Assume that $f_{i}(j-1)\notin A\cap B$ . Then we look for the first element between $z_{k}$ and $z_{k+1}$ such that is in $B-A$ , let’s call this elements $\hat{z}_{k}$ .

Suppose that $\hat{z}_{k}=z_{k}+1$ , then $\{(f_{i}(z_{k}-1),f_{i}(z_{k}+1))\}\notin A\sqcup_{A\cap B}B$ . Hence, there exists $\lambda_{i,k},\lambda^{\prime}_{i,k}:\{0,1,2\}\rightarrow X$ in $A\sqcup_{A\cap B}B$ such that $\lambda_{i,k}(0)=\lambda^{\prime}_{i,k}(0)=f_{i}(z_{k}-1)$ , $\lambda_{i,k}(1)=\lambda^{\prime}_{i,k}(1)$ , $\lambda_{i,k}(2)=f_{i}(z_{k})$ , and $\lambda^{\prime}_{i,k}(2)=f_{i}(z_{k}+1)$ .

Figure 8. Deleting a point

Since $A,B$ well-split $X$ , then we can select $f^{\prime}(z^{\prime}_{k})$ of the original partition as $\lambda_{i,k}(1)$ because there is a path between them. Thus, the sequence $\lambda_{i,k}(1),f(z_{k}+1),\ldots,f(z_{k+1}-1)$ is equivalent in $\pi_{\leq 1}(B,\mathcal{U}\cap B)$ to $\lambda_{i,k}(1),f(z_{k}),f(z_{k}+1),\ldots f(z_{k+1}-1)$ . Getting what we wanted.

Figure 9. Representation of the process;

f_{i}(z_{k+1}+1)\notin A\cap B

and

z_{k}=0

and

\hat{z}_{k}=1

$f_{i}\xrightarrow{a(j,x_{0})}g)$ The case is analogous to the previous one. We only observe that happens if we add a new point which interrupts the sequence of element on $A$ or $B$ and we add one adequate point to recover the last case. ∎

Theorem 5.4.

Let $X$ be a semi-coarse space and $\mathcal{U}\subset 2^{X}$ , and write $\mathcal{U}^{\prime}$ the collection of all of the components of the elements of $\mathcal{U}$ . Consider $A,B\subset X$ such that

(1)

$A,B$ well-split $X$ ,
(2)

for every component of $A$ , $B$ $A\cap B$ within a component of $X$ which meets $\cup_{U\\ in\mathcal{U}}U$ , there exists $U^{\prime}\in\mathcal{U}^{\prime}$ such that meets the component, and
(3)

for every $U^{\prime}\in\mathcal{U}^{\prime}$ , $U^{\prime}\subset A$ and $U^{\prime}\subset B$ .

Then the diagram

is a pushout.

Proof.

We can divide $X$ in all its components and apply $X_{\alpha}$ and apply 5.3 to every $X_{\alpha}$ and $X_{\alpha}\cap\mathcal{U}^{\prime}$ . Obtaining the result. ∎

We conclude the section with two consequences of this theorem.

Corollary 5.5.

Let $X$ be a semi-coarse space and $\mathcal{U}=\{U\}$ for some $U\subset X$ , and write $\mathcal{U}^{\prime}$ the collection of all of the components of $U$ . Consider $A,B\subset X$ such that

(1)

$A,B$ well-split $X$ ,
(2)

for every component of $A$ , $B$ $A\cap B$ within a component of $X$ which meets $U$ , $U$ meets the component, and
(3)

for every $U^{\prime}\in\mathcal{U}^{\prime}$ , $U^{\prime}\subset A$ and $U^{\prime}\subset B$ .

Then the diagram

is a pushout.

Corollary 5.6.

Let $X$ be a semi-coarse space and $\mathcal{U}=\{\{x\}\mid x\in X\}$ . Consider $A,B\subset X$ such that $A,B$ well-split $X$ . Then the diagram

is a pushout. In addition, the groupoid $\pi_{\leq 1}(X,\mathcal{U})$ contains all of the $\pi_{1}^{sc}(X,x)$ .

6. Future Work

Most of the work in this document was to develop the fundamental groupoid and introduce the van Kampen theorem in this category. In the future, we would like to explore what other semi-coarse invariants we can study in an almost locally way with well-split notion. We did not find this concept in the literature and we hope this gives us some light about other constructions in this category.

Other authors have found relations between the coarse homotopy groups and the topological homotopy groups in compact manifolds (see [Weighill_LiftingCoarseHom_2021, Mitchener_2020].) We would like to study topological coarse spaces or topological semi-coarse spaces through this new approximation. Although we need deeper work, we expect that we will be capable to study open manifolds together its coarse structure (or maybe the semi coarse one.) In the same sense, we would like to observe the behavior of semi-coarse spaces adding other structure coming from “topological constructs”, for example closure spaces or limit spaces.

Appendix A Coarse Homotopy Fails in Semi-Coarse

Many coarse invariants has been studied before, and a natural question is why don’t we adapt some of them and study semi-coarse spaces. The problems with this kind of approximations are diverse, but generally: in some cases we do not recover the same information for coarse spaces, in others we lose some properties. In this brief appendix, we mention an example, the coarse homotopy [Mitchener_2020], and look at some of the problems with using semi-coarse spaces. We would like to remind that the semi-coarse homotopy groups are trivial in coarse spaces (3.2.19, [rieser2023semicoarse]).

In this section, $\mathbb{R}_{+}$ is endowed with the coarse structure

\displaystyle A\subset\{\mathbb{R}_{+}\times\mathbb{R}_{+}\mid\exists r>0\text{ such that }d(x,y)\leq r\ \forall(x,y)\in A\}.

In addition, we remind that in a coarse spaces:

Definition A.1.

Let $(X,\mathcal{V})$ and $(Y,\mathcal{W})$ be coarse spaces

•

$B\subset X$ is bounded in $(X,\mathcal{V})$ if there exists $x\in X$ and $A\in\mathcal{V}$ such that

$\displaystyle A[x]\coloneqq\{y\in X\mid(x,y)\in A\}=B.$
•

A map $f:X\rightarrow Y$ is proper if $f^{-1}(B)$ is bounded in $(X,\mathcal{V})$ for every $B$ bounded in $(Y,\mathcal{W})$ .
•

A map $f:X\rightarrow Y$ bornologous and proper is called coarse map.

Definition A.2 ( $p$ -Cylinder; [Mitchener_2020], 2.1).

Let $X$ be a coarse space, and let $p:X\rightarrow\mathbb{R}_{+}$ be a coarse map. Then we define the $p$ -cylinder

\displaystyle I_{p}X\coloneqq\{(x,t)\in X\times\mathbb{R}_{+}\mid t\leq p(x)+1\}

We have inclusions $i_{0}:X\rightarrow I_{p}X$ and $i_{1}:X\rightarrow I_{p}X$ defined by the formulas $i_{0}(x)=(x,0)$ and $i(x)=(x,p(x)+1)$ , respectively. We also have the canonical projection $q:I_{p}X\rightarrow X$ defined by the formula $q(x,t)=x$ .

Definition A.3 (Coarse Homotopy; [Mitchener_2020], 2.2).

Let $X$ and $Y$ be coarse spaces. A coarse homotopy is a coarse map $H:I_{p}X\rightarrow Y$ for some coarse map $p:X\rightarrow\mathbb{R}_{+}$ .

We call coarse maps $f_{0}:X\rightarrow Y$ and $f_{1}:X\rightarrow Y$ coarsely homotopic if there is a coarse homotopy $H:I_{p}X\rightarrow Y$ such that $f_{0}=H\circ i_{0}$ and $f_{1}=H_{1}\circ i_{1}$ .

This map $H$ is termed a coarse homotopy between the maps $f_{0}$ and $f_{1}$ .

We explore the following three modifications:

(1)

Let’s consider the definition of a bounded set $A$ as there exists a controlled set $B$ and $x$ such that $A=B[x]$ . Then we might lose several coarse maps.
(2)

Consider just bornologous maps, then we might obtain a coarser equivalence relation which makes trivial some maps.
(3)

In the same case, for semi-coarse spaces, we lose transitive.

In every case, we provide an examples which illustrates what we claim in the last list.

Example A.3.1.

Let $X$ be a coarse space and consider $f:\mathbb{Z}_{1}\rightarrow X$ be a coarse map (a bornologous proper map.) Since $(k,k+1)$ is a controlled set in $\mathbb{Z}_{1}$ , then we have that $\{f(0),f(5)\}$ is a bounded set in $X$ , however $\{0,5\}$ is not in $\mathbb{Z}_{1}$ . Thus, we don’t have any proper map from $\mathbb{Z}_{1}$ to $X$ . (We can mimic this procedure replacing $\mathbb{Z}_{1}$ for many semi-coarse spaces which are not coarse spaces.)

Example A.3.2.

Consider $\mathbb{Z}$ with the coarse structure induced by the metric. We can observe that the map $p:\mathbb{Z}\rightarrow\mathbb{R}_{+}$ such that $p(z)=|z|$ is bornologous (actually it is a coarse map.) Now we define $H:I_{p}(X)\rightarrow X$ such that $H(x,t)=t$ ; $H$ is a bornologous map and we can observe that it is not proper. Additionally, we observe that $H\circ i_{0}(x,t)=0$ and $H\circ i_{1}(x,t)=|t|+1$ , obtaining that a map which goes to infinity is homotpic to a constant map.

Example A.3.3.

Consider $X$ as the semi-coarse space induced by the graph

and the bornologous map (which is not proper) $p:\mathbb{Z}_{1}\rightarrow X$ such that $p(z)=0$ . Then we obtain that $I_{p}(\mathbb{Z}_{1})=\mathbb{Z}_{1}\times[0,1]$ . Let’s define the following bornologous functions:

	$\displaystyle u:\mathbb{Z}_{1}\rightarrow X$	$\displaystyle\text{ such that }u(3n)=0,\ u(3n+1)=2,\ u(3n+2)=3\text{ for every }n\in\mathbb{Z}$
	$\displaystyle m:\mathbb{Z}_{1}\rightarrow X$	$\displaystyle\text{ such that }u(3n)=0,\ u(3n+1)=u(3n+2)=2\text{ for every }n\in\mathbb{Z}$
	$\displaystyle d:\mathbb{Z}_{1}\rightarrow X$	$\displaystyle\text{ such that }u(3n)=0,\ u(3n+1)=2,\ u(3n+2)=1\text{ for every }n\in\mathbb{Z}$

Observe that we have the following homotopies:

	$\displaystyle H_{1}(z,t)=u(z)\text{ if }0\leq t<1,\ H_{1}(z,t)=m(z)\text{ if }t=1,$
	$\displaystyle H_{2}(z,t)=d(z)\text{ if }0\leq t<1,\ H_{2}(z,t)=m(z)\text{ if }t=1$

However, there is no $H$ such that $H\circ i_{0}=u$ and $H\circ i_{1}=d$ because ${1,3}$ is not controlled in $X$ .

As a last observation, we could try to replace $\mathbb{R}_{+}$ for a semi-coarse space in the definition of $I_{p}(X)$ or in the codomain of $H$ . Observe that if $(X,\mathcal{V})$ is a connected coarse space, $(Y,\mathcal{W})$ is a semi-coarse space and $f:X\rightarrow Y$ is a bornologous map, then $f(X)$ is a connected (in the coarse sense) space. This is a problem because it might limit our bornologous maps. To mention a brief example, if you replace $\mathbb{R}_{+}$ by $\mathbb{R}_{+,r}$ you will observe that every bornologous function $p$ maps a coarse space $X$ to $p(X)$ with coarse structure $2^{p(X)\times p(X)}$ . Then we can find a semi-coarse space where the transitive fails.

Acknowledgments

The author would like to thanks to Antonio Rieser, who is the supervisor of my PhD project, for the several discussions about this paper and the possible directions this work could take, as well as his advice to include a brief discussion about the coarse homotopy groups. The author would also like to thanks Noe Barcenas for his feedback on an event at the Casa Mexicana de Matemáticas in Oaxaca and another event in the CCM in Morelia.