Exhausting pants graphs of punctured spheres by finite rigid sets

TL;DR

The paper constructs an increasing sequence of finite rigid sets in the pants graph of an n-punctured sphere that exhausts the entire graph, providing new insights into its structure.

Contribution

It introduces a method to build an increasing sequence of finite rigid sets that cover the entire pants graph of punctured spheres.

Findings

Sequence of finite rigid sets exhausts the pants graph

Provides a new approach to understanding the structure of pants graphs

Establishes a foundation for further topological and geometric analysis

Abstract

Let $S_{0,n}$ be an $n$-punctured sphere. For $n\geq 4$, we construct a sequence $(\mathcal{X}_i)_{i\in\mathbb{N}}$ of finite rigid sets in the pants graph $\mathcal{P}(S_{0,n})$ such that $\mathcal{X}_1 \subset \mathcal{X}_2 \subset ...\subset\mathcal{P}(S_{0,n})$ and $\bigcup_{i\geq1}\mathcal{X}_i=\mathcal{P}(S_{0,n})$.

Equations

$$f^Q:\mathcal{P}(S_{0,n})\to \mathcal{P}(S_{0,m}),$$

$$\varphi :\mathcal{X}\to \mathcal{P}(S_{0,m}),$$

$$f_1^{Q_1}{|}_{\mathcal{X}}=f_2^{Q_2}{|}_{\mathcal{X}},$$

$$Q_1=Q_2\text{ and }{f}_1=f_2,$$

$$\bigcup _{i\in \mathbb{N}}{\mathcal{X}}_i=\mathcal{P}(S_{0,n}).$$

$${\mathcal{X}}_{n+1}:={\mathcal{X}}_n\cup \bigcup _{\alpha \in C}{T}_{\alpha }^{\pm \frac{1}{2}}({\mathcal{X}}_n),$$

$$\bigcup _{i\in \mathbb{N}}{\mathcal{X}}_i=\mathcal{P}(S_{0,n}).$$

$$X_5=Z_5\cup \bigcup _{c\in {\Gamma }_5}{T}_c^{\pm \frac{1}{2}}(Z_5),$$

$$X_5^W=h^W(X_5)=\{h(u)\cup W\mid u\in X_5\},$$

$$X_n=Z_n\cup \bigcup _W{X}_5^W,$$

$$f_1^{Q_1}{|}_{X_n\cap T_{\alpha }^i(X_n)}=f_2^{Q_2}{|}_{X_n\cap T_{\alpha }^i(X_n)},$$

$$f_1=f_2\text{ or }{f}_1=f_2\circ e,$$

$$C=\{T^{\pm \frac{1}{2}}(\alpha )|\alpha \text{ a chain curve}\}$$

Full text

1 Introduction

Let $S=S_{g,n}$ be an orientable surface of genus $g$ with $n$ punctures and let $\operatorname{Mod}^{\pm}(S)=\pi_{0}(\operatorname{Homeo}(S))$ be the extended mapping class group. Ivanov [6], Korkmaz [7], and Luo [8] proved that, for most surfaces, the curve complexes $\mathcal{C}(S)$ is rigid, that is, $\operatorname{Aut}(\mathcal{C}(S))\cong\operatorname{Mod}^{\pm}(S)$. In [2], Aramayona and Leininger proved that curve complexes contain finite rigid sets; meaning a finite subgraph such that every simplicial embedding is a restriction of an element of $\operatorname{Mod}^{\pm}(S)$. Later in [3], they showed that there exists an exhaustion of the curve complexes by finite rigid sets.

For the pants graphs $\mathcal{P}(S)$, the rigidity property was proved by Margalit [9] using the result of Ivanov, Korkmaz, and Lou. Aramayona [1] extended Margalit’s result to prove a stronger form of rigidity, that is, if $S$ and $S^{\prime}$ are surfaces such that the complexity of $S$ is at least $2$, then every injective simplicial map $\phi:\mathcal{P}(S)\to\mathcal{P}(S^{\prime})$ is induced by a $\pi_{1}$-injective embedding $f:S\to S^{\prime}$. In [10], we refined Aramayona’s result by showing that the pants graphs of punctured spheres are finitely rigid.

In this paper, we modify the tools Aramayona and Leininger built in [3], together with the finite rigid sets we constructed [10], to prove that we can exhaust the pants graphs of punctured spheres by finite rigid sets:

Theorem 1.1**.**

Let $S_{0,n}$ be an $n$-punctured sphere. For $n\geq 4$, there exists a sequence of finite rigid sets $\mathcal{X}_{1}\subset\mathcal{X}_{2}\subset...\subset\mathcal{P}(S_{0,n})$ such that $\bigcup_{i\geq 1}\mathcal{X}_{i}=\mathcal{P}(S_{0,n})$.

Outline of the paper. Section 2 contains the relevant background and definitions. In Section 3 we describe the adjustments to the tools Aramayona and Leininger [3] developed to enlarge their rigid sets in the curve complex so we can use them in our setting. We use these tools to prove the main theorem in Section 4.

Acknowledgments. The author would like to thank Christopher J. Leininger for his guidance, useful conversations, and suggestions.

2 Background and definitions

This section contains necessary definitions and background restricted to punctured spheres, for general settings see [1] and [9]. Let $S=S_{0,n}$ be an $n$-punctured sphere. A simple closed curve on $S$ is essential if it does not bound a disk or a once-punctured disk on $S$. Throughout this paper, a curve is a homotopy class of essential simple closed curves on $S$. Given two curves $\gamma$ and $\gamma^{\prime}$, we denote their geometric intersection number by $i(\gamma,\gamma^{\prime})$, which is the minimum number of transverse intersection points among the representatives of $\gamma$ and $\gamma^{\prime}$. The two curves are disjoint if $i(\gamma,\gamma^{\prime})=0$

A multicurve $Q$ is a set of pairwise distinct, disjoint curves on $S$. For a given multicurve $Q$, the nontrivial piece $(S-Q)_{0}$ of the complement of the curves in $Q$ is the union of the non thrice-punctured sphere components of the complement. We call a thrice-punctured sphere, a pair of pants.

A pants decomposition $P$ is a maximal multicurve: the complement in $S$ is a disjoint union of pairs of pants. A pants decomposition always contains $n-3$ curves and we call this number the complexity $\kappa(S)$ of $S$. The deficiency of a multicurve $Q$ is the number $\kappa(S)-|Q|$. If $Q$ is a deficiency-$1$ multicurve then $(S-Q)_{0}$ is homeomorphic to $S_{0,4}$.

Let $P$ and $P^{\prime}$ be pants decompositions of $S$. We say that $P$ and $P^{\prime}$ differ by an elementary move if there are curves $\alpha,\alpha^{\prime}$ on $S$ and a deficiency-$1$ multicurve $Q$ such that $P=\{\alpha\}\cup Q,P^{\prime}=\{\alpha^{\prime}\}\cup Q$ and $i(\alpha,\alpha^{\prime})=2$; see Figure 1 for an example of elementary moves.

The pants graph $\mathcal{P}(S)$ of $S$ is a graph with the set of vertices corresponding to pants decompositions. Two vertices are connected by an edge if the corresponding pants decompositions differ by an elementary move. The pants graph $\mathcal{P}(S)$ is connected and the pants graph $\mathcal{P}(S_{0,4})$ of a $4$-punctured sphere is isomorphic to a Farey graph, see [5].

A path in $\mathcal{P}(S)$ is an edge path determined by a sequence of distinct adjacent vertices of $\mathcal{P}(S)$. A cycle in $\mathcal{P}(S)$ is a subgraph homeomorphic to a circle. We call a cycle, a triangle, rectangle, or pentagon if it has $3$, $4$, or $5$ vertices, respectively.

Let $\mathcal{X}\subset\mathcal{P}(S_{0,n})$ and $\phi:\mathcal{X}\to\mathcal{P}(S_{0,m})$ be an injective simplicial map. We say that a $\pi_{1}$-injective embedding $f:S_{0,n}\to S_{0,m}$ induces $\phi$ if there is a deficiency-$(n-3)$ multicurve $Q$ on $S_{0,m}$ such that $f(S_{0,n})=(S_{0,m}-Q)_{0}$ and the simplicial map

[TABLE]

defined by $f^{Q}(u)=f(u)\cup Q$ satisfies $f^{Q}(u)=\phi(u)$ for any vertex $u\in\mathcal{X}$.

Definition 2.1**.**

For $n\geq 4$, we say that $\mathcal{X}\subset\mathcal{P}(S_{0,n})$ is rigid if for any punctured sphere $S_{0,m}$ and any injective simplicial map

[TABLE]

there exists a $\pi_{1}$-injective embedding $f:S_{0,n}\to S_{0,m}$ that induces $\phi$.

For $n=4$, the isotopy class of $f$ is unique up to precomposing by an element $\sigma\in\operatorname{Mod}(S_{0,4})$ inducing the identity on $\mathcal{P}(S_{0,4})$.

For $n\geq 5$, the isotopy class of $f$ is unique.

The following theorem is a refinement of Aramayona’s result [1] that we proved in [10].

Theorem 2.2**.**

For $n\geq 4$, there exists a finite rigid subgraph $X_{n}\subset\mathcal{P}(S_{0,n})$.

3 Tools for enlarging rigid sets

This section contains the definitions and theorems Aramayona and Leininger [3] developed to enlarge their rigid sets in the curve complexes. We make some necessary adjustments to them in order to enlarge rigid sets in the pants graphs.

Definition 3.1**.**

Let $n\geq 5$. A set $\mathcal{X}\subset\mathcal{P}(S_{0,n})$ is said to be weakly rigid if whenever $f_{1},f_{2}:S_{0,n}\to S_{0,m}$ are $\pi_{1}$-injective embeddings satisfy

[TABLE]

for some deficiency-$(n-3)$ multicurves $Q_{1}$ and $Q_{2}$ on $S_{0,m}$, then

[TABLE]

up to isotopy.

It is easy to see from the definition that a superset of a weakly rigid set is also weakly rigid.

Lemma 3.2**.**

For $n\geq 5$, let $\mathcal{X}_{1},\mathcal{X}_{2}\subset\mathcal{P}(S_{0,n})$ be rigid sets. If $\mathcal{X}_{1}\cap\mathcal{X}_{2}$ is weakly rigid then $\mathcal{X}_{1}\cup\mathcal{X}_{2}$ is rigid.

Proof.

Let $\phi:\mathcal{X}_{1}\cup\mathcal{X}_{2}\to\mathcal{P}(S_{0,m})$ be an injective simplicial map. Since $\mathcal{X}_{i}$ is rigid, there exist a $\pi_{1}$-injective embedding $f_{i}:S_{0,n}\to S_{0,m}$ and a deficiency-$(n-3)$ multicurve $Q_{i}$ such that $f_{i}^{Q_{i}}|_{\mathcal{X}_{i}}=\phi|_{\mathcal{X}_{i}}$. Hence $f^{Q_{1}}_{1}|_{\mathcal{X}_{1}\cap\mathcal{X}_{2}}=\phi|_{\mathcal{X}_{1}\cap\mathcal{X}_{2}}=f^{Q_{2}}_{2}|_{\mathcal{X}_{1}\cap\mathcal{X}_{2}}$. The weakly rigidity of $\mathcal{X}_{1}\cap\mathcal{X}_{2}$ implies that $Q_{1}=Q_{2}=Q$ and $f_{1}=f_{2}=f$. Therefore $f$ is a $\pi_{1}$-injective embedding such that $f^{Q}|_{\mathcal{X}_{1}\cup\mathcal{X}_{2}}=\phi$ which implies the rigidity of $\mathcal{X}_{1}\cup\mathcal{X}_{2}$. ∎

The following proposition is the key to enlarge rigid sets.

Proposition 3.3**.**

For $n\geq 5$, let $\mathcal{X}\subset\mathcal{P}(S_{0,n})$ be a finite rigid set such that $\operatorname{Mod}(S_{0,n})\cdot\mathcal{X}=\mathcal{P}(S_{0,n})$. Suppose there exists a finite subset $C$ of curves on $S_{0,n}$ such that:

$(1)$* The set $\{T^{\pm\frac{1}{2}}_{\alpha}~{}|~{}\alpha\in C\}$ generates $\operatorname{Mod}(S_{0,n})$;*

$(2)$* $\mathcal{X}\cap T^{i}_{\alpha}(\mathcal{X})$ is weakly rigid, for all $\alpha\in C$, and $i\in\{-\frac{1}{2},\frac{1}{2}\}$.*

Then there exists a sequence $\mathcal{X}=\mathcal{X}_{1}\subset\mathcal{X}_{2}\subset...\subset\mathcal{X}_{n}\subset...$ such that each $\mathcal{X}_{i}$ is a finite rigid set, and

[TABLE]

Proof.

Since $\mathcal{X}$ is rigid and a half twist is a homeomorphism, $T^{i}_{\alpha}(\mathcal{X})$ is rigid for all $\alpha\in C$, and $i\in\{-\frac{1}{2},\frac{1}{2}\}$. Given $\alpha,\beta\in C$ and $i,j\in\{-\frac{1}{2},\frac{1}{2}\}$. By assumption $(2)$ and by applying Lemma 3.2, we see that $\mathcal{X}\cup T^{i}_{\alpha}(\mathcal{X})$ is rigid. Recall that a superset of a weakly rigid set is also weakly rigid. Hence $(\mathcal{X}\cup T^{i}_{\alpha}(\mathcal{X}))\cap T^{j}_{\beta}(\mathcal{X})$, which contains $\mathcal{X}\cap T^{j}_{\beta}(\mathcal{X})$, is weakly rigid. Apply Lemma 3.2, we see that $\mathcal{X}\cup T^{i}_{\alpha}(\mathcal{X})\cup T^{j}_{\beta}(\mathcal{X})$ is weakly rigid. By repeating above arguments, the set $\mathcal{X}_{2}:=\mathcal{X}\cup\bigcup_{\alpha\in C}T^{\pm\frac{1}{2}}_{\alpha}(\mathcal{X})$ is rigid. We define

[TABLE]

for $n\geq 2$. Since a weakly rigid set $\mathcal{X}\cap T^{i}_{\alpha}(\mathcal{X})$ is a subset of $\mathcal{X}_{n}\cap T^{i}_{\alpha}(\mathcal{X}_{n})$ , $\mathcal{X}_{n}\cap T^{i}_{\alpha}(\mathcal{X}_{n})$ is weakly rigid. Again, by applying 3.2 repeatedly and use induction, we conclude that $\mathcal{X}_{n}$ is rigid for all $n$. Then the first claim is proved.

Since $\{T^{\pm\frac{1}{2}}_{\alpha}~{}|~{}\alpha\in C\}$ generates $\operatorname{Mod}(S_{0,n})$ and $\operatorname{Mod}(S_{0,n})\cdot\mathcal{X}=\mathcal{P}(S_{0,n})$,

[TABLE]

∎

4 The proof of the main theorem

We note that for $n\leq 3$, the pants graphs $\mathcal{P}(S_{0,3})$ is empty. We give a separate proof for $n=4$ as follow.

Proof of Theorem 1.1 for $S=S_{0,4}$..

The pants graph of $S_{0,4}$ is isomorphic to the Farey graph. Any triangle in $S_{0,4}$ is rigid as proved in [10]. Then we let $\mathcal{X}_{1}$ to be a triangle. Each edge in a pants graph of any punctured sphere is contained in exactly two triangles which are both in the same Farey graph. Then we can define $\mathcal{X}_{n+1}$ inductively; let $\mathcal{X}_{n+1}$ be an enlargement of $\mathcal{X}_{n}$ obtained by attaching one more triangle to each edge of $\mathcal{X}_{n}$ contained in only one triangle. Hence $\mathcal{X}_{n+1}$ is rigid for all $n\geq 1$, and by the construction, $\bigcup_{i\in\mathbb{N}}(\mathcal{X}_{i})=\mathcal{P}(S_{0,4})$. We conclude that sequence $(\mathcal{X}_{n})_{n\in\mathbb{N}}$ is an exhaustion of $\mathcal{P}(S_{0,4})$. ∎

For $n\geq 5$, we begin by recalling the construction of finite rigid sets $X_{n}$ in [10]. First we construct $S_{0,n}$ with a set of curves, then define $X_{5}$, and finally, define $X_{n}$ for $n\geq 6$.

Consider a regular $n$-gon with the $n$ vertices removed and label the sides as $1,2,...,n$, cyclically. For each non-adjacent pair of sides of the $n$-gon, draw a straight line segment to connect the two sides. Then double the $n$-gon to obtain $S_{0,n}$ and a set of curves $\Gamma_{n}$, see Figure 2 for the case of $S_{0,8}$ and Figure 3 for the case of $S_{0,5}$. Let $a_{i,j}\in\Gamma_{n}$ be the curve connecting the $i^{th}$ side to the $j^{th}$ side of $S_{n}$. We call $a_{i,j}$ such that $i-j\equiv\pm 2\mod n$, a chain curve. Compare to [2, Section 3].

Let $Z_{n}$ be a subgraph of $\mathcal{P}(S_{0,n})$ induced by the set of vertices corresponding to pants decompositions consisting of curves from $\Gamma_{n}$.

For $\mathcal{P}(S_{0,5})$, we defined

[TABLE]

where $T^{\frac{1}{2}}_{c}$ is a simplicial map on $\mathcal{P}(S_{0,5})$ induced by the half-twist around the curve $c$.

See Figure 3 for a partial figure of $X_{5}$. The subgraph $X_{5}$ consists of the alternating pentagon $Z_{5}$ and $10$ of its images under the twists. Those $10$ images form $10$ triangles attached to $Z_{5}$. In [10], we proved that $X_{5}$ is rigid.

For $n\geq 6$, we construct $X_{n}$ as follows. Let $W\subset\Gamma_{n}$ be a deficiency-$2$ multicurve such that $(S_{0,n}-W)_{0}\cong S_{0,5}$. Let $\Gamma^{W}_{5}=\{\alpha\in\Gamma_{n}~{}|~{}\alpha$ is disjoint from all curves in $W\}$. There is a natural homeomorphism $h:S_{0,5}\to(S_{0,n}-W)_{0}$ such that $h(\Gamma_{5})=\Gamma_{5}^{W}$, see [10, Lemma 3.1]. Let

[TABLE]

where $h^{W}:P(S_{0,5})\to P(S_{0,n})$ is the induced map of $h$ defined by $h^{W}(u)=h(u)\cup W$. Then $X_{5}^{W}\cong X_{5}$. Finally we let

[TABLE]

where the union is taken over all deficiency-$2$ multicurves in $\Gamma_{n}$ with a $5$-punctured sphere component. In [10], we proved that $X_{n}$ is rigid.

We need the following lemmas to prove the main theorem for $n\geq 5$.

Lemma 4.1**.**

$\operatorname{Mod}(S_{0,n})\cdot X_{n}=\mathcal{P}(S_{0,n})$**

Proof.

In the first part of this proof, we will show that, for a given vertex $P$ in $\mathcal{P}(S_{0,n})$, there exist a vertex $P^{\prime}$ in $X_{n}$ and $f\in\operatorname{Mod}(S_{0,n})$ such that $f(P^{\prime})=P$. To do this, we obtain a pants decomposition $P^{\prime}$ from a dual graph of the pants decomposition $P$. For the second part, we will show that there is a homeomorphism that send a given edge in $\mathcal{P}(S_{0,n})$ to an edge in $Z_{n}\subset X_{n}$.

Given a vertex $P$ in $\mathcal{P}(S_{0,n})$. Recall that we consider $S_{0,n}$ as a double of a regular $n$-gon. Consider $P$ as a pants decomposition on $S_{0,n}$. The following construction of a dual graph of $P$ was given in [5]. For each pair of pants component of $(S_{0,n}-P)$, we mark a vertex on the interior of the component. We also mark the $n$ punctures as $n$ vertices. Two vertices are connected by an edge if (1) they are vertices on the interior of two pants components which have a common boundary, or (2) one of the vertices is on the interior of a pair of pants component and another vertex is a puncture of the same component. The result is a tree with $2n-2$ vertices; all puncture-vertices have degree $1$, while the rest of the vertices have degree $3$, see Figure 4.

Since a tree is planar, we can redraw this tree on the plane inside a regular $n$-gon so that all $n$ puncture-vertices are the $n$ vertices of the $n$-gon. We reconstruct a pants decomposition consisting of curves in $\Gamma_{n}$ by drawing a curve connecting two sides of the regular $n$-gon whenever this curve can cross exactly one edge of the tree and both endpoints of this edge are not puncture-vertices. Double the regular $n$-gon. We now have a pants decomposition $P^{\prime}$ consisting of curves in $\Gamma_{n}$, i.e., $P^{\prime}$ is a vertex in $Z_{n}\subset X_{n}$.

The above construction of $P^{\prime}$ from $P$ gives a one-to-one correspondence between the pants components $S_{0,n}-P$ and the pants components $S_{0,n}-P^{\prime}$. This correspondence describes a homeomorphism $f$ such that $f(P^{\prime})=P$, as desired.

Next we show that if $P_{1}$ and $P_{2}$ are adjacent vertices in $\mathcal{P}(S_{0,n})$, then after applying some homeomorphisms on $S_{0,n}$ to $P_{1}$ and $P_{2}$, we get two vertices that are adjacent in $Z_{n}$.

Given adjacent vertices $P_{1}$ and $P_{2}$ in $\mathcal{P}(S_{0,n})$, then there exist curves $u_{1},u_{2}$ on $S_{0,n}$ and a deficiency-$1$ multicurve $Q$ such that $P_{1}=\{u_{1}\}\cup Q$ and $P_{2}=\{u_{2}\}\cup Q$. By the first part of the proof, there is $f\in\operatorname{Mod}(S_{0,n})$ such that $f(P_{1})$ is a vertex in $Z_{n}$. If $f(P_{2})$ is also in $Z_{n}$, then we are done.

Suppose $f(P_{2})$ is not in $Z_{n}$. Use Figure 5 as a reference for the rest of the proof. We note that $f(Q)\subset\Gamma_{n}$ and it has deficiency-$1$. The nontrivial component $(S_{0,n}-f(Q))_{0}\cong S_{0,4}$ contains exactly two curves in $\Gamma_{n}$; one curve is $f(u_{1})$ and we call the other curve $\alpha$. Then $i(f(u_{2}),\alpha)=2n$ for some $n\in\mathbb{N}$. Applying one full twist around $f(u_{1})$ in an appropriate direction reduces the intersection number by $4$. $f(P_{1})$ is invariant under this full twist. So we can choose a new $f$ (by composing the old one with some power of full twists) and assume that $i(f(u_{2}),\alpha)=0$ or $i(f(u_{2}),\alpha)=2$. If $i(f(u_{2}),\alpha)=0$, then $f(u_{2})=\alpha$ and we are done.

Suppose $i(f(u_{2}),\alpha)=2$. We compose $f$ by an appropriate half twist $T$ around $f(u_{1})$: here a half twist in $f(u_{1})$ is a homeomorphism on $S_{0,n}$, whose square is the Dehn twist in $f(u_{1})$, although we note that it does not necessary restrict to a homeomorphism of $(S_{0,n}-f(Q))_{0}\cong S_{0,4}$. We choose the half twist that essentially “flips over” half of the $n$-gon, cut along $f(u_{1})$; see Figure 6 and also Figure 5. Then $T\circ f(u_{2})=\alpha$ and the edge $\{T\circ f(P_{1}),T\circ f(P_{2})\}$ is in $Z_{n}$ as desired. ∎

Let $\alpha$ be a curve on $S_{0,n}$. We defined $\mathcal{P}_{\alpha}(S_{0,n})$ to be a subgraph of $\mathcal{P}(S_{0,n})$ induced by vertices corresponding to pants decompositions containing $\alpha$.

The following lemma is proved in [10] and we use this lemma to prove Lemma 4.3.

Lemma 4.2**.**

For $n\geq 6$, let $\alpha$ be a chain curve on $S_{0,n}$ and let $X^{\alpha}_{n-1}=X_{n}\cap\mathcal{P}_{\alpha}(S_{0,n})$.

Then $X^{\alpha}_{n-1}\cong X_{n-1}$. Moreover, this isomorphism is induced by $h:S_{0,n-1}\to(S_{0,n}-\alpha)_{0}$ as $h^{\alpha}(v)=h(v)\cup\{\alpha\}\in X_{n-1}^{\alpha}$.

Lemma 4.3**.**

$X_{n}\cap T^{i}_{\alpha}(X_{n})$* is weakly rigid, for $i\in\{-\frac{1}{2},\frac{1}{2}\}$ and for all chain curves $\alpha$ in $S_{0,n}$.*

Proof.

Let $\alpha$ be a chain curve and $i\in\{-\frac{1}{2},\frac{1}{2}\}$. Suppose $f_{1},f_{2}:S_{0,n}\to S_{0,m}$ are $\pi_{1}$-injective embeddings such that

[TABLE]

for some deficiency-$(n-3)$ multicurves $Q_{1}$ and $Q_{2}$ on $S_{0,m}$.

We first prove the case of $n=5$. Recall the definition of $X_{5}$. By a direct calculation, we see that $X_{5}\cap T^{i}_{\alpha}(X_{5})$ consists of two alternating pentagons which are $Z_{5}=T^{i}_{\alpha}(T^{-i}_{\alpha}(Z_{5}))$ and $T^{i}_{\alpha}(Z_{5})$. They share an edge together with four triangles as shown in Figure 3. Since $Z_{5}$ is an alternating pentagon and $f^{Q_{1}}_{1}|_{Z_{5}}=f^{Q_{2}}_{2}|_{Z_{5}}$, [8, Lemma 4.2] implies that $Q_{1}=Q_{2}$ and

[TABLE]

where $e:S_{0,5}\to S_{0,5}$ is the involution exchanging the two pentagons (as we consider $S_{5}$ as a double of a pentagon). The map $e$ induces a simplicial map on $\mathcal{P}(S_{0,5})$ that fixes $Z_{5}$ and exchanges two triangles on each side of $Z_{5}$. But $f_{1}$ and $f_{2}$ also agree on the four triangles attached to $Z_{5}$ so $f_{1}=f_{2}$. Hence the case of $n=5$ is proved.

Let $n\geq 6$ and let $\alpha$ be any chain curve. By Lemma 4.2, a subgraph $X^{\alpha}_{n-1}=X_{n}\cap\mathcal{P}_{\alpha}(S_{0,n})\cong X_{n-1}$. Since each vertex of $X^{\alpha}_{n-1}$ contains $\alpha$, $T^{i}_{\alpha}(X^{\alpha}_{n-1})=X^{\alpha}_{n-1}$. Hence $X_{n}\cap T^{i}_{\alpha}(X_{n})$ contains $X^{\alpha}_{n-1}\cong X_{n-1}$. Consider the restrictions of $f_{1}$ and $f_{2}$ on the subsurface $(S_{0,n}-\{\alpha\})_{0}$. Since $X_{n-1}$ is rigid, so is $X^{\alpha}_{n-1}$, and the uniqueness part of Definition 2.1 implies that $f_{1}$ agrees with $f_{2}$ on $(S_{0,n}-\{\alpha\})_{0}$ and $Q_{1}\cup\{f_{1}(\alpha)\}=Q_{2}\cup\{f_{1}(\alpha)\}$.

We can see that $X^{\alpha}_{n-1}$ is a proper subgraph of $X_{n}\cap T^{i}_{\alpha}(X_{n})$. For example, choose a vertex $P$ in $Z_{n}\cap\mathcal{P}_{\alpha}(S_{0,n})\subset X^{\alpha}_{n-1}$. Then change $P$ to $P^{\prime}$ by the elementary move which replaces $\alpha$ by the other curve $\alpha^{\prime}$ in $\Gamma_{n}$. The vertex $T^{i}_{\alpha}(P^{\prime})$ is adjacent to $P$ and it is a vertex in $X_{n}\cap T^{i}_{\alpha}(X_{n})$. Hence $f_{1}$ and $f_{2}$ agree on $T^{i}_{\alpha}(P^{\prime})$. Since $Q_{1}$ and $Q_{2}$ are the intersections of all vertices in $f_{1}(X_{n}\cap T^{i}_{\alpha}(X_{n}))$ and $f_{2}(X_{n}\cap T^{i}_{\alpha}(X_{n}))$, respectively, and $\alpha\notin T^{i}_{\alpha}(P^{\prime})$, it follows $f_{1}(\alpha)=f_{2}(\alpha)$ is not in the intersection. Therefore $Q_{1}=Q_{2}$ and $f_{1}=f_{2}$. ∎

Proof of Theorem 1.1 for $S_{0,n},n\geq 5$.

We are ready to prove the main theorem for $n\geq 5$. We check that all conditions in Proposition 3.3 are satisfied. Let $\mathcal{X}=X_{n}$. Lemma 4.1 states that $\operatorname{Mod}(S_{0,n})\cdot\mathcal{X}=\mathcal{P}(S_{0,n})$. The set

[TABLE]

generates $\operatorname{Mod}(S_{0,n})$, see [4, Corollary 4.15]. And by Lemma 4.3, $X_{n}\cap T^{i}_{\alpha}(X_{n})$ is weakly rigid, for $i\in\{-\frac{1}{2},\frac{1}{2}\}$ and for all chain curves $\alpha$ in $S_{0,n}$. Therefore Proposition 3.3 gives us a sequence of finite rigid set $\mathcal{X}=\mathcal{X}_{1}\subset\mathcal{X}_{2}\subset...\subset\mathcal{X}_{m}\subset...$ such that $\bigcup_{i\in\mathbb{N}}\mathcal{X}_{i}=\mathcal{P}(S_{0,n})$, as desired.

∎