Non-minimal bridge position of $2$-cable links

A knot in $S^{3}$ is said to be in bridge position with respect to a bridge sphere, the original notion introduced by Schubert [12], if the knot intersects each of the $3$-balls bounded by the bridge sphere in a collection of $\partial$-parallel arcs. It is generalized to knots (and links) in $3$-manifolds with the development of Heegaard splitting theory, and is related to many interesting problems concerning, e.g. bridge number, (Hempel) distance, and incompressible surfaces in $3$-manifolds.

From any $n$-bridge position, we can always get an $(n+1)$-bridge position by creating a new local minimum point and a nearby local maximum point of the knot. A bridge position isotopic to one obtained in this way is said to be perturbed. (It is said to be stabilized in some context.) A bridge position of the unknot is unique in the sense that any $n$-bridge ($n>1$) position of the unknot is perturbed [7]. The uniqueness also holds for $2$-bridge knots [8]. See also [11], where all bridge surfaces for $2$-bridge knots are considered. Ozawa [9] showed that non-minimal bridge positions of torus knots are perturbed. Zupan [14] showed such property for iterated torus knots and iterated cables of $2$-bridge knots. More generally, he showed that if $K$ is an mp-small knot and every non-minimal bridge position of $K$ is perturbed, then every non-minimal bridge position of a $(p,q)$-cable of $K$ is also perturbed [14]. (Here, a knot is mp-small if its exterior contains no essential meridional planar surface.) We remark that there exist examples of a knot with a non-minimal bridge position that is not perturbed [10] and furthermore, knots with arbitrarily high index bridge positions that are not perturbed [5].

In this paper, we consider non-minimal bridge position of $2$-cable links of a knot $K$ without the assumption of mp-smallness of $K$.

For the proof, we use the notion of t-incompressibility and t-$\partial$-incompressibility of [3]. We isotope an annuls $A$ whose boundary is $L$ to a good position so that it is t-incompressible and t-$\partial$-incompressible in one side, say $B_{2}$, of the bridge sphere. Then $A\cap B_{2}$ consists of bridge disks and (possibly) properly embedded disks. By using the idea of changing the order of t-$\partial$-compressions in [2] or [3], we show that in fact $A\cap B_{2}$ consists of bridge disks only. Then by a further argument, we find a cancelling pair of disks for the bridge position.

A trivial tangle $T$ is a union of properly embedded arcs $b_{1},\ldots,b_{n}$ in a $3$-ball $B$ such that each $b_{i}$ cobounds a disk $D_{i}$ with an arc $s_{i}$ in $\partial B$, and $D_{i}\cap(T-b_{i})=\emptyset$. By standard argument, $D_{i}$’s can be taken to be pairwise disjoint.

Let $F$ denote a surface in $B$ satisfying $F\cap(\partial B\cup T)=\partial F$. A t-compressing disk for $F$ is a disk $D$ in $B-T$ such that $D\cap F=\partial D$ and $\partial D$ is essential in $F$, i.e. $\partial D$ does not bound a disk in $F$. A surface $F$ is t-compressible if there is a t-compressing disk for $F$, and $F$ is t-incompressible if it is not t-compressible.

An arc $\alpha$ properly embedded in $F$ with its endpoints on $F\cap\partial B$, is t-essential if $\alpha$ does not cobound a disk in $F$ with a subarc of $F\cap\partial B$. In particular, an arc in $F$ parallel to a component of $T$ can be t-essential. See Figure 1. (Such an arc will be called bridge-parallel in Section 3.) A t-$\partial$-compressing disk for $F$ is a disk $\Delta$ in $B-T$ such that $\partial\Delta$ is an endpoint union of two arcs $\alpha$ and $\beta$, and $\alpha=\Delta\cap F$ is t-essential, and $\beta=\Delta\cap\partial B$. A surface $F$ is t-$\partial$-compressible if there is a t-$\partial$-compressing disk for $F$, and $F$ is t-$\partial$-incompressible if it is not t-$\partial$-compressible.

Hayashi and Shimokawa [3] classified t-incompressible and t-$\partial$-incompressible surfaces in a compression body in more general setting. Here we give a simplified version of the theorem.

Let $S$ be a $2$-sphere decomposing $S^{3}$ into two $3$-balls $B_{1}$ and $B_{2}$. Let $K$ denote a knot (or link) in $S^{3}$. Then $K$ is said to be in bridge position with respect to $S$ if $K\cap B_{i}$ $(i=1,2)$ is a trivial tangle. Each arc of $K\cap B_{i}$ is called a bridge. If the number of bridges of $K\cap B_{i}$ is $n$, we say that $K$ is in $n$-bridge position. The minimum such number $n$ among all bridge positions of $K$ is called the bridge number $b(K)$ of $K$. A bridge cobounds a bridge disk with an arc in $S$, whose interior is disjoint from $K$. We can take a collection of $n$ pairwise disjoint bridge disks by standard argument, and it is called a complete bridge disk system. For a bridge disk $D$ in, say $B_{1}$, if there exists a bridge disk $E$ in $B_{2}$ such that $D\cap E$ is a single point of $K$, then $D$ is called a cancelling disk. We call $(D,E)$ a cancelling pair.

A perturbation is an operation on an $n$-bridge position of $K$ that creates a new local minimum and local maximum in a small neighborhood of a point of $K\cap S$, resulting in an $(n+1)$-bridge position of $K$. A bridge position obtained by a perturbation admits a cancelling pair by the construction. Conversely, it is known that a bridge position admitting a cancelling pair is isotopic to one obtained from a lower index bridge position by a perturbation. (See e.g. [11, Lemma 3.1]). Hence, as a definition, we say that a bridge position is perturbed if it admits a cancelling pair.

Let $V_{1}$ be a standard solid torus in $S^{3}$ with core $\alpha$, and $V_{2}$ be a solid torus in $S^{3}$ whose core is a knot $C$. A meridian $m_{1}$ of $V_{1}$ is uniquely determined up to isotopy. Let $l_{1}\subset\partial V_{1}$ be a longitude of $V_{1}$ such that the linking number $\textrm{lk}(l_{1},\alpha)=0$, called the preferred longitude. Similarly, let $m_{2}$ and $l_{2}$ be a meridian and a longitude of $V_{2}$ respectively such that $\textrm{lk}(l_{2},C)=0$. Take a $(p,q)$-torus knot (or link) $T_{p,q}$ in $\partial V_{1}$ that wraps $V_{1}$ longitudinally $p$ times; more precisely, $|T_{p,q}\cap m_{1}|=p$ and $|T_{p,q}\cap l_{1}|=q$. Let $h:V_{1}\to V_{2}$ be a homeomorphism sending $m_{1}$ to $m_{2}$ and $l_{1}$ to $l_{2}$. Then $K=h(T_{p,q})\subset S^{3}$ is called a $(p,q)$-cable of $C$. Concerning the bridge number, it is known that $b(K)=p\cdot b(C)$ [12], [13].

Let $K$ be a knot (or link) in $n$-bridge position with respect to a decomposition $S^{3}=B_{1}\cup_{S}B_{2}$, so $K\cap B_{1}$ is a union of bridges $b_{1},\ldots,b_{n}$. Let $\mathcal{R}=R_{1}\cup\cdots\cup R_{n}$ be a complete bridge disk system for $\bigcup b_{i}$, where $R_{i}$ is a bridge disk for $b_{i}$. Let $F$ be a surface bounded by $K$ and $F_{1}=F\cap B_{1}$. When we move $K$ to an isotopic bridge position, $F_{1}$ moves together. We consider $\mathcal{R}\cap F_{1}$. By isotopy we assume that in a small neighborhood of $b_{i}$, $\mathcal{R}\cap F_{1}=b_{i}$.

An arc $\gamma$ in $F_{1}$ is bridge-parallel (b-parallel briefly) if $\gamma$ is parallel, in $F_{1}$, to some $b_{i}$ and cuts off a rectangle $P$ from $F_{1}$ whose four edges are $\gamma$, $b_{i}$, and two arcs in $S$. Let $\alpha$($\neq b_{k}$) denote an arc of $\mathcal{R}\cap F_{1}$ which is outermost in some $R_{k}$ and cuts off the corresponding outermost disk $\Delta$ disjoint from $b_{k}$. The following lemma will be used in Section 6.

Now we consider a sufficient condition for a bridge position to be perturbed.

Let $K$ be a knot such that every non-minimal bridge position of $K$ is perturbed. Let $L$ be a $(2,2q)$-cable link of $K$, with components $K_{1}$ and $K_{2}$. Suppose that $L$ is in non-minimal bridge position with respect to a bridge sphere $S$ bounding $3$-balls $B_{1}$ and $B_{2}$. Each $L\cap B_{i}$ $(i=1,2)$ is a trivial tangle. Since $L$ is a $2$-cable link, $L$ bounds an annulus, denoted by $A$. We take $A$ so that $|A\cap S|$ is minimal.

Since our goal is to show that the bridge position of $L$ is perturbed, from now on we assume that $L$ is not the unlink. By Claim 1, $A\cap B_{2}$ is t-incompressible in $B_{2}$. If $A\cap B_{2}$ is t-$\partial$-compressible in $B_{2}$, we do a t-$\partial$-compression.

A t-$\partial$-compression simplifies a surface because it cuts the surface along a t-essential arc. So if we maximally t-$\partial$-compress $A\cap B_{2}$, we obtain a t-$\partial$-incompressible $A\cap B_{2}$. Note that the effect on $A$ of a t-$\partial$-compression of $A\cap B_{2}$ is just pushing a neighborhood of an arc in $A$ into $B_{1}$, which is called an isotopy of Type $A$ in [4]. After a maximal sequence of t-$\partial$-compressions, $A\cap B_{2}$ is both t-incompressible and t-$\partial$-incompressible by Claim 2. Then by applying Lemma 2.1,

• ($\ast$)

  $A\cap B_{2}$ consists of bridge disks $D_{i}$’s and properly embedded disks $C_{j}$’s.

Take an annulus $A$ bounded by $L$ so that $(\ast)$ holds and the number $m$ of properly embedded disks $C_{j}$ is minimal. In this section, we will show that $m=0$, i.e. $A\cap B_{2}$ consists of bridge disks only.

Suppose that $m>0$. Then $A\cap B_{1}$ is homeomorphic to an $m$-punctured annulus. A similar argument as in the proof of Claim 1 leads to that $A\cap B_{1}$ is t-incompressible. By Lemma 2.1 again, $A\cap B_{1}$ is t-$\partial$-compressible. We can do a sequence of t-$\partial$-compressions on $A\cap B_{1}$ until it becomes t-$\partial$-incompressible. Note that the t-incompressibility of $A\cap B_{1}$ is preserved.

Now we are going to define a t-$\partial$-compressing disk $\Delta_{i}$ ($i=0,1,\ldots,s$ for some $s$) and its dual disk $U_{i+1}$ inductively. Let $A_{0}=A$. Let $\Delta_{0}$ be a t-$\partial$-compressing disk for $A_{0}\cap B_{1}$ and $\alpha_{0}=\Delta_{0}\cap A_{0}$ and $\beta_{0}=\Delta_{0}\cap S$. By a t-$\partial$-compression along $\Delta_{0}$, a neighborhood of $\alpha_{0}$ is pushed along $\Delta_{0}$ into $B_{2}$ and thus a band $b_{1}$ is created in $B_{2}$. Let $A_{1}$ denote the resulting annulus bounded by $L$. Let $U_{1}$ be a dual disk for $\Delta_{0}$, that is, a disk such that an isotopy of Type A along $U_{1}$ recovers a surface isotopic to $A_{0}$. For the next step, let $\mathcal{U}_{1}=U_{1}$.

Let $\Delta_{1}$ be a t-$\partial$-compressing disk for $A_{1}\cap B_{1}$ and $\alpha_{1}=\Delta_{1}\cap A_{1}$ and $\beta_{1}=\Delta_{1}\cap S$. After a t-$\partial$-compression along $\Delta_{1}$, a band $b_{2}$ is created in $B_{2}$. Let $A_{2}$ denote the resulting annulus bounded by $L$. There are three cases to consider.

Case $1$. $\beta_{1}$ intersects the arc $\mathcal{U}_{1}\cap S$ more than once.

The band $b_{2}$ cuts off small disks $U_{2,1},U_{2,2},\ldots,U_{2,k_{2}}$ from $\mathcal{U}_{1}$, which are mutually parallel along the band. We designate any one among the small disks, say $U_{2,1}$, as the dual disk $U_{2}$. Let $\mathcal{R}_{2}=\bigcup^{k_{2}}_{j=2}U_{2,j}$ be the union of others.

Case $2$. $\beta_{1}$ intersects $\mathcal{U}_{1}\cap S$ once.

We take the subdisk that $b_{2}$ cuts off from $\mathcal{U}_{1}$ as the dual disk $U_{2}$, and let $\mathcal{R}_{2}=\emptyset$ in this case.

Case $3$. $\beta_{1}$ does not intersect $\mathcal{U}_{1}\cap S$.

We take a dual disk $U_{2}$ freely, and let $\mathcal{R}_{2}=\emptyset$ in this case.

In any case, let $\mathcal{U}_{2}=\mathcal{U}_{1}\cup U_{2}-\mathrm{int}\,\mathcal{R}_{2}$.

In general, assume that $A_{i}$ and $\mathcal{U}_{i}$ are defined. Let $\Delta_{i}$ be a t-$\partial$-compressing disk for $A_{i}\cap B_{1}$ and $\alpha_{i}=\Delta_{i}\cap A_{i}$ and $\beta_{i}=\Delta_{i}\cap S$. After a t-$\partial$-compression along $\Delta_{i}$, a band $b_{i+1}$ is created in $B_{2}$. Let $A_{i+1}$ denote the resulting annulus bounded by $L$.

Case a. $\beta_{i}$ intersects the collection of arcs $\mathcal{U}_{i}\cap S$ more than once.

The band $b_{i+1}$ cuts off small disks $U_{i+1,1},U_{i+1,2},\ldots,U_{i+1,k_{i+1}}$ from $\mathcal{U}_{i}$, which are mutually parallel along the band. We designate any one among the small disks, say $U_{i+1,1}$, as the dual disk $U_{i+1}$. Let $\mathcal{R}_{i+1}=\bigcup^{k_{i+1}}_{j=2}U_{i+1,j}$ be the union of others.

Case b. $\beta_{i}$ intersects $\mathcal{U}_{i}\cap S$ once.

We take the subdisk that $b_{i+1}$ cuts off from $\mathcal{U}_{i}$ as the dual disk $U_{i+1}$, and let $\mathcal{R}_{i+1}=\emptyset$ in this case.

Case c. $\beta_{i}$ does not intersect $\mathcal{U}_{i}\cap S$.

We take a dual disk $U_{i+1}$ freely, and let $\mathcal{R}_{i+1}=\emptyset$ in this case.

In any case, let $\mathcal{U}_{i+1}=\mathcal{U}_{i}\cup U_{i+1}-\mathrm{int}\,\mathcal{R}_{i+1}$.

Later, we do isotopy of Type A, dual to the t-$\partial$-compression, in reverse order along $U_{i+1},U_{i},\ldots,U_{1}$. Let us call it dual operation for our convenience.. Let $b_{i+1}$ be the band mentioned above, cutting off $U_{i+1,1},U_{i+1,2},\ldots,U_{i+1,k_{i+1}}$ from $\mathcal{U}_{i}$ (in Case a). When the dual operation along $U_{i+1}$ is done, we modify every $U_{j}$ and $\mathcal{U}_{j}$ ($j\leq i$) containing any $U_{i+1,s}$ ($s>1)$ of $\mathcal{R}_{i+1}$, by replacing each $U_{i+1,s}$ ($s>1$) with the union of a subband of $b_{i+1}$ between $U_{i+1,s}$ and $U_{i+1}$($=U_{i+1,1}$) and a copy of $U_{i+1}$, and doing a slight isotopy. We remark that, although it is not illustrated in Figure 5, some $U_{j}$’s and $\mathcal{U}_{j}$’s temporarily become immersed when the subband passes through some removed region, say $U_{r,s}$ ($s>1$). But the $U_{r,s}$ ($s>1$) is also modified as we proceed the dual operations, and the $U_{j}$’s and $\mathcal{U}_{j}$’s again become embedded. (In Figure 5 and Figure 6, the dual operation along $U_{i+1}$ is done, and the dual operation along $U_{i}$ is not done yet.) Actually, before the sequence of dual operations, $U_{j}$’s and $\mathcal{U}_{j}$’s ($j\leq k$) are modified in advance so that $\mathcal{U}_{k}$ is disjoint from the union of certain band $b_{k+1}$ and a disk $C_{l}$ (which will be explained later). See Figure 6.

The t-$\partial$-compressing disk $\Delta_{i}$ is taken to be disjoint from two copies of $\Delta_{j}$ ($j<i$). Moreover, for every $i$ we can draw $\alpha_{i}$ on $A$($=A_{0}$). Since the t-$\partial$-compression along $\Delta_{i}$ is equivalent to cutting along $\alpha_{i}$ and the sequence $\Delta_{0},\Delta_{1},\ldots,\Delta_{s}$ is maximal, every $c_{j}=\partial C_{j}$ is incident to some $\alpha_{i}$.

Let $k$ be the smallest index such that $\alpha_{k}$ connects some $c_{j}$, say $c_{l}$, to other component (other $c_{j}$ or $D_{i}\cap S$). If $k=0$, then by a t-$\partial$-compression along $\Delta_{0}$, either $C_{l}$ and other $C_{j}$ are merged into one properly embedded disk, or $C_{l}$ and a bridge disk are merged into a new bridge disk. This contradicts the minimality of $m$. So we assume that $k\geq 1$.

Suppose that we performed t-$\partial$-compressions along $\Delta_{0},\Delta_{1},\ldots,\Delta_{k}$. Consider the small disks that the band $b_{k+1}$ cuts off from $\mathcal{U}_{k}$. They are parallel along $b_{k+1}$. We replace the small disks one by one, the nearest one to $C_{l}$ first, so that $\mathcal{U}_{k}$ is disjoint from $b_{k+1}\cup C_{l}$. Let $\Delta$ be the small disk nearest to $C_{l}$. Let $\Delta^{\prime}$ be the union of a subband of $b_{k+1}$ and $C_{l}$ that $\Delta\cap b_{k+1}$ cuts off from $b_{k+1}\cup C_{l}$. For every $U_{j}$ and $\mathcal{U}_{j}$ ($j\leq k$) containing $\Delta$, we replace $\Delta$ with $\Delta^{\prime}$. Then again let $\Delta$ be the (next) small disk nearest to $C_{l}$ and we repeat the above operation until $\mathcal{U}_{k}$ is disjoint from $b_{k+1}\cup C_{l}$.

Now we do the dual operation on $A_{k}$ in reverse order along $U_{k},U_{k-1},\ldots,U_{1}$. Let $A^{\prime}_{i-1}$ ($i=1,\ldots,k$) be the resulting annulus after the dual operation along $U_{i}$. The shape of the dual disk $U_{k}$ is possibly changed but the number of circle components and arc components of $A^{\prime}_{k-1}\cap S$ is same with those of $A_{k-1}\cap S$. After the dual operation along $U_{k}$, it is necessary to modify some $U_{j}$’s and $\mathcal{U}_{j}$’s ($j\leq k-1$) further as in Figure 6 so that $U_{k-1}$ is disjoint from $b_{k+1}\cup C_{l}$. In this way, we do the sequence of dual operations, and the number of circle components of $A^{\prime}_{0}\cap S$ is also $m$. Then because $b_{k+1}$ is disjoint from $U_{1},\ldots,U_{k}$, we can do the t-$\partial$-compression of $A^{\prime}_{0}$ along $\Delta_{k}$ first and the number $m$ of properly embedded disks $C_{j}$ is reduced, contrary to our assumption.

We have shown the following claim.

By Claim 4, $A\cap B_{2}$ consists of bridge disks. Let $d_{0},\ldots,d_{k-1}$ and $e_{0},\ldots,e_{l-1}$ be bridges of $K_{1}\cap B_{2}$ and $K_{2}\cap B_{2}$ respectively that are indexed consecutively along each component. Let $D_{i}\subset A\cap B_{2}$ ($i=0,\ldots,k-1$) be the bridge disk for $d_{i}$ and $s_{i}=D_{i}\cap S$, and let $E_{j}\subset A\cap B_{2}$ ($j=0,\ldots,l-1$) be the bridge disk for $e_{j}$ and $t_{j}=E_{j}\cap S$. Let $\mathcal{R}=R_{1}\cup\cdots\cup R_{k+l}$ be a complete bridge disk system for $L\cap B_{1}$ and let $F=A\cap B_{1}$. We consider $\mathcal{R}\cap F$ except for the bridges $L\cap B_{1}$.

If there is an inessential circle component of $\mathcal{R}\cap F$ in $F$, it can be removed by standard innermost disk argument. If there is an essential circle component of $\mathcal{R}\cap F$ in $F$, then $L$ would be the unlink as in Case $1$ of the proof of Claim 1. So we assume that there is no circle component of $\mathcal{R}\cap F$. If there is an inessential arc component of $\mathcal{R}\cap F$ in $F$ with both endpoints on the same $s_{i}$ (or $t_{j}$), then the arc can be removed by standard outermost disk argument. So we assume that there is no arc component of $\mathcal{R}\cap F$ with both endpoints on the same $s_{i}$ (or $t_{j}$).

If $\mathcal{R}\cap F=\emptyset$, we easily get a cancelling pair, say $(R_{m},D_{i})$ or $(R_{m},E_{j})$, so we assume that $\mathcal{R}\cap F\neq\emptyset$. Let $\alpha$ denote an arc of $\mathcal{R}\cap F$ which is outermost in some $R_{m}$ and let $\Delta$ denote the outermost disk that $\alpha$ cuts off from $R_{m}$. Applying Lemma 3.1, $\alpha$ is not b-parallel. Suppose that one endpoint of $\alpha$ is in, say $s_{i_{1}}$, and the other is in $s_{i_{2}}$ with the cyclic distance $d(i_{1},i_{2})=\textrm{min}\{|i_{1}-i_{2}|,|k-(i_{1}-i_{2})|\}$ greater than $1$. Then after the t-$\partial$-compression along $\Delta$, we get a subdisk of $A$ satisfying the assumption of Lemma 3.2, hence $L$ is perturbed. So without loss of generality, we assume that one endpoint of $\alpha$ is in $s_{0}$ and the other is in $t_{0}$.

Let $A_{1}$ be the annulus obtained from $A$ by the t-$\partial$-compression along $\Delta$ and $F_{1}=A_{1}\cap B_{1}$. The bridge disks $D_{0}$ and $E_{0}$ are connected by a band, and let $P_{1}$ be the resulting rectangle with four edges $d_{0},e_{0}$, and two arcs in $S$, say $p_{1},p_{2}$. Let $\alpha_{1}$ denote an arc of $\mathcal{R}\cap F_{1}$ which is outermost in some $R_{m}$ and let $\Delta_{1}$ denote the outermost disk cut off by $\alpha_{1}$. If at least one endpoint of $\alpha_{1}$ is contained in $p_{1}$ or $p_{2}$, or one endpoint of $\alpha_{1}$ is in $s_{i_{1}}$($t_{j_{1}}$ respectively) and the other is in $s_{i_{2}}$($t_{j_{2}}$ respectively), then similarly as above,

• •

  either $\alpha_{1}$ is inessential with both endpoints on the same component of $F_{1}\cap S$, or

• •

  $\alpha_{1}$ is b-parallel, or

• •

  Lemma 3.2 can be applied.

Hence we may assume that one endpoint of $\alpha_{1}$ is in $s_{i}$ ($i\neq 0$) and the other is in $t_{j}$ ($j\neq 0$). After the t-$\partial$-compression along $\Delta_{1}$, $D_{i}$ and $E_{j}$ are merged into a rectangle. Arguing in this way, each $D_{i}$ ($i=0,\ldots,k-1$) is merged with some $E_{j}$ because of the fact that $\mathcal{R}\cap s_{i}=\emptyset$ gives us a cancelling pair. Moreover, we see that $k=l$. After $k$ successive t-$\partial$-compressions on $A$, the new annulus $A^{\prime}$ intersects $B_{1}$ and $B_{2}$ alternately, in rectangles.

Note that $b(L)=2b(K)$ and $L$ is in $2k$-bridge position. Since $L$ is in non-minimal bridge position, $k>b(K)$. So by the assumption of the theorem, the bridge position of $K_{i}$ ($i=1,2$) is perturbed. Let $(D,E)$ be a cancelling pair for $K_{1}$ with $D\subset B_{1}$ and $E\subset B_{2}$. However, $D$ and $E$ may intersect $K_{2}$. Let $P_{i}$ and $P_{i+1}$ be any adjacent rectangles of $A^{\prime}$ in $B_{1}$ and $B_{2}$ respectively. We remove any unnecessary intersection of $D\cap P_{i}$ and $E\cap P_{i+1}$, the nearest one to $K_{2}$ first, by isotopies along subdisks of $P_{i}$ and $P_{i+1}$ respectively. See Figure $7$ for an example. Then $(D,E)$ becomes a cancelling pair for the bridge position of $L$ as desired.

The author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2018R1D1A1A09081849).