# Graph morphisms and exhaustion of curve graphs of low-genus surfaces

Jesús Hernández Hernández

###### Abstract

This work is the extension of the results by the author in [7] and [6] for
low-genus surfaces. Let $S$ be an orientable, connected surface of finite
topological type, with genus $g\leq 2$, empty boundary, and complexity at
least $2$; as a complement of the results of [6], we prove that any graph
endomorphism of the curve graph of $S$ is actually an automorphism. Also, as a
complement of the results in [6] we prove that under mild conditions on the
complexity of the underlying surfaces any graph morphism between curve graphs
is induced by a homeomorphism of the surfaces.

To prove these results, we construct a finite subgraph whose union of iterated
rigid expansions is the curve graph $\mathcal{C}(S)$. The sets constructed,
and the method of rigid expansion, are closely related to Aramayona and
Leiniger’s finite rigid sets in [2]. Similarly to [7], a consequence of our
proof is that Aramayona and Leininger’s rigid set also exhausts the curve
graph via rigid expansions, and the combinatorial rigidity results follow as
an immediate consequence, based on the results in [6].
2020 AMS Mathematics Subject Classification: 57K20.
Keywords: Curve graph, low-genus surface, rigid expansions, graph morphisms.

## Introduction

In this work we suppose $S_{g,n}$ is an orientable surface of finte
topological type, with genus $g\geq 0$, $n\geq 0$ punctures, and empty
boundary. The (extended) mapping class group of $S_{g,n}$, denoted by
$\mathrm{Mod}^{\pm}(S_{g,n})$ is the group of isotopy classes of self-
homeomorphisms of $S_{g,n}$.

In 1979 (see [5]) Harvey introduced what is now one of the main tools for the
study of $\mathrm{Mod}^{\pm}(S_{g,n})$: the curve graph of $S_{g,n}$, denoted
by $\mathcal{C}(S_{g,n})$, is the simplicial graph whose vertices are isotopy
classes of (essential simple closed) curves on $S_{g,n}$, where two vertices
span an edge if the corresponding isotopy classes are different and have
disjoint representatives.

There is a natural action of $\mathrm{Mod}^{\pm}(S_{g,n})$ on
$\mathcal{C}(S_{g,n})$ by simplicial automorphisms. It is a well-known result
by Ivanov (see [13]), Korkmaz (see [14]), and Luo (see [15]), that if
$(g,n)\neq(1,2)$, then all the simplicial automorphisms of
$\mathcal{C}(S_{g,n})$ are induced by homeomorphisms of $S_{g,n}$.

This result was later extended by Schackleton (see [16]) to simplicial maps
that are locally injective (recall that a simplicial map is locally injective
if its restriction to the star of any vertex, is injective).

Afterwards, Aramayona and Leininger proved in [1] that there exist finite
subgraphs of $\mathcal{C}(S_{g,n})$ with this same property, i.e. there exists
a finite subgraph $\mathfrak{X}(S_{g,n})<\mathcal{C}(S_{g,n})$, such that all
locally injective maps $\mathfrak{X}(S_{g,n})\to\mathcal{C}(S_{g,n})$ are
restrictions to $\mathfrak{X}(S_{g,n})$ of an automorphism of
$\mathcal{C}(S_{g,n})$. Additionally, in [2] they call subgraphs with this
property rigid, and prove (using geometric methods) that there exists an (set-
theoretically) increasing sequence of rigid subgraphs of
$\mathcal{C}(S_{g,n})$ whose union is $\mathcal{C}(S_{g,n})$.

Note that this last result is not trivial since not any supergraph of a rigid
subgraph is rigid.

Later on, we reproved this result (see [7]) using combinatorial methods (first
introduced in [2]). The main technique used in [7] is called rigid expansions:

In the curve graph, a curve $\alpha$ is uniquely determined by a set of curves
$A$ if $\alpha\notin A$ and $\alpha$ is the unique curve on $S_{g,n}$ that is
disjoint from every element in $A$. Given a subgraph $Y<\mathcal{C}(S_{g,n})$,
the first rigid expansion of $Y$, denoted by $Y^{1}$, is the subgraph induced
by all the curves in $Y$ along with all the curves uniquely determined by
subsets of the vertex set of $Y$. The $n$-th rigid expansion of $Y$ (denoted
by $Y^{n}$) is defined recursively, and we define $Y^{\omega}$ as the union of
$Y^{n}$ for all $n\geq 1$. See Section 1 for more details.

In [2] it is proved that if $X$ is rigid, then $X^{n}$ is rigid for all $n\geq
1$, and in [7] we proved that if $g\geq 3$, then
$\mathfrak{X}(S_{g,n})^{\omega}=\mathcal{C}(S_{g,n})$. The majority of this
work is about extending this result to all surfaces with
$\kappa(S_{g,n})\displaystyle\mathrel{\mathop{:}}=3g-3+n\geq 2$. To do this,
we first prove there exists a convenient finite subgraph
$\mathscr{Y}(S_{g,n})$ that exhausts the curve graph via rigid expansions.

######  Theorem A.

Let $S_{g,n}$ be an orientable, connected surface of finite topological type,
with genus $g\leq 2$, $n\geq 0$ punctures, and empty boundary. If
$\kappa(S_{g,n})\geq 2$, then there exists a finite subgraph
$\mathscr{Y}(S_{g,n})$ of $\mathcal{C}(S_{g,n})$ whose union of iterated rigid
expansions is equal to $\mathcal{C}(S_{g,n})$.

Afterwards, we use this result to prove that $\mathfrak{X}(S_{g,n})$ also
exhausts the curve graph, with the methods varying depending on the genus of
the surface.

######  Theorem B.

Let $S_{g,n}$ be an orientable, connected surface of finite topological type
and empty boundary. If $\kappa(S_{g,n})\geq 2$, then
$\mathfrak{X}(S_{g,n})^{\omega}=\mathcal{C}(S_{g,n})$.

These results for the case $g\geq 3$, were used by the author in [6] to prove
that if $S_{1}=S_{g_{1},n_{1}}$, $S_{2}=S_{g_{2},n_{2}}$ with $g_{1}\geq 3$
and the complexity of $S_{2}$ bounded above by the complexity of $S_{1}$, then
any graph morphism between $\mathcal{C}(S_{1})$ and $\mathcal{C}(S_{2})$ is
induced by a homeomorphism $S_{1}\rightarrow S_{2}$.

Recently, in [12] and [11], Irmak extended this result for the case of
endomorphisms, and proved that any endomorphism of the curve graph of
$S_{g,n}$ with $\kappa(S_{g,n})\geq 2$ is induced by a homeomorphism of
$S_{g,n}$.

In Section 5 we give a short proof of this result using Theorem B and the
results from [7] and [6].

######  Theorem C.

Let $S_{g,n}$ be an orientable, connected surface of finite topological type,
with empty boundary, and $\kappa(S_{g,n})\geq 2$. Let also
$\varphi:\mathcal{C}(S_{g,n})\rightarrow\mathcal{C}(S_{g,n})$ be a graph
(endo)morphism. Then $\varphi$ is an automorphism of $\mathcal{C}(S_{g,n})$.

Also in Section 5 we give an extension of the results in [16] and [6], proving
that with a restriction on the complexity of surfaces $S_{1}$ and $S_{2}$, and
with $S_{1}\neq S_{0,7}$, then any graph morphism between the respective curve
graphs, is induced by a homeomorphism between $S_{1}$ and $S_{2}$.

######  Theorem D.

Let $S_{1}$ and $S_{2}$ be two orientable, connected surfaces of finite
topological type, with empty boundary, such that
$\kappa(S_{1})\geq\kappa(S_{2})\geq 2$, and $S_{1}$ is not homeomorphic to a
7-punctured sphere. Let also
$\varphi:\mathcal{C}(S_{1})\rightarrow\mathcal{C}(S_{2})$ be a graph morphism.
Then, $S_{1}$ is homeomorphic to $S_{2}$ and $\varphi$ is induced by a
homeomorphism $S_{1}\to S_{2}$.

Note that, due to Theorem C, it is sufficient to prove topological rigidity,
i.e. under the hypothesis of Theorem D, $S_{1}$ is homeomorphic to $S_{2}$
(see Theorem 5.4).

Also, the lower bound on the complexity of Theorem C is sharp. Indeed, if
$\kappa(S_{g,n})=1$, there exist infinitely many graph endomorphisms of
$\mathcal{C}(S_{g,n})$ that are surjective, but are NOT injective; thus, they
cannot be induced by an automorphism.

As well, the lower bound on the complexity of Theorem D is also sharp. Indeed,
$\mathcal{C}(S_{0,5})$ and $\mathcal{C}(S_{0,6})$ are isomorphic to
$\mathcal{C}(S_{1,2})$ and $\mathcal{C}(S_{2,0})$, respectively.

Reader’s guide: This work is divided as follows: in Section 1 we give the
necessary preliminaries for this work; in Section 2 we prove Theorems A and B
for the case of genus zero; in Section 3 we prove Theorems A and B for the
case of genus one; in Section 4 we prove Theorems A and B for the case of
genus two; in Section 5 we prove Theorems C and D.
Acknowledgements: The author was supported during the creation of this article
by the UNAM-PAPIIT research grants IA104620 and IN114323. The author was also
supported by the CONAHCYT research grant Ciencia de Frontera 2019 CF 217392.

##  1 Preliminaries

Suppose $S_{g,n}$ is an orientable, connected surface of finite topological
type with empty boundary, genus $g\leq 2$, $n\geq 0$ punctures, and such that
its complexity $\kappa(S_{g,n})=3g-3+n$ is at least $3$. The mapping class
group of $S_{g,n}$, denoted by $\mathrm{Mod}(S_{g,n})$, is the group of all
orientation-preserving self-homeomorphisms of $S_{g,n}$.

In this work, a curve on $S_{g,n}$ is a topological embedding of
$\mathbb{S}^{1}$ into the surface. We often abuse notation and call “curve”
the embedding, its image on $S_{g,n}$, or its isotopy class. The meaning is
clear with the context.

A curve on $S_{g,n}$ is essential if it is neither homotopic to a point, nor
to the boundary curve of a neighbourhood of a puncture. Hereafter we assume
every curve to be essential unless otherwise stated. See Figure 1 for
examples.

The curves on $S_{g,n}$ can be classified as follows: A curve $\alpha$ on
$S_{g,n}$ is separating if $S_{g,n}\backslash\\{\alpha\\}$ is disconnected,
and it is non-separating otherwise. A special case of a separating curve is
when $S_{g,n}\backslash\\{\alpha\\}$ has a connected component homeomorphic to
$S_{0,3}$, in which case we say $\alpha$ is an outer curve. See Figure 1 for
examples.

Figure 1: Examples of curves on surfaces. The red curves are non-essential,
while the rest are essential. The green curves are outer curves, the orange
curves are separating non-outer curves, and the blue curve is non-separating.

Note that there exist non-separating curves on $S_{g,n}$ if and only if
$S_{g,n}$ has positive genus.

Let $\alpha$ and $\beta$ two (isotopy classes of) curves on $S_{g,n}$. The
(geometric) intersection number of $\alpha$ and $\beta$, denoted by
$i(\alpha,\beta)$, is defined as follows:

$i(\alpha,\beta)=\min\\{|a\cap b|:a\in\alpha,b\in\beta\\}.$

Let $\alpha$ and $\beta$ be two curves on $S_{g,n}$. In this work, we say that
$\alpha$ is disjoint from $\beta$ if $i(\alpha,\beta)=0$ and
$\alpha\neq\beta$.

The curve graph of $S_{g,n}$, denoted by $\mathcal{C}(S_{g,n})$, is the
abstract simplicial graph whose vertices are the isotopy classes of essential
curves on $S_{g,n}$, and two vertices span an edge if the corresponding curves
are disjoint. We often abuse notation and denote by $\mathcal{C}(S_{g,n})$
both the curve graph and its vertex set.

###  1.1 Rigid expansions

In [6] we gave a general description for the concept of a rigid subgraph and
rigid expansions of abstract simplicial graphs. Here however, we give the
definitions focused on the case of the curve graph.

Given a curve $\alpha$ on $S_{g,n}$, we define the set $\mathrm{adj}(\alpha)$
as the set of all curves on $S$ that are disjoint from $\alpha$. Note that
$\alpha\notin\mathrm{adj}(\alpha)$.

We say a curve $\alpha\in\mathcal{C}(S_{g,n})$ is uniquely determined by
$A\subset\mathcal{C}(S_{g,n})$, denoted by $\alpha=\langle A\rangle$, if
$\alpha$ is the unique curve on $S_{g,n}$ that is disjoint from every element
in $A$, i.e.

$\\{\alpha\\}=\bigcap_{\gamma\in A}\mathrm{adj}(\gamma).$

See Figure 2 for examples.

Figure 2: In both examples the blue curve is uniquely determined by the set of
red curves.

Let $Y<\mathcal{C}(S_{g,n})$ be a induced subgraph of $\mathcal{C}(S_{g,n})$,
and let $V(Y)$ denote its vertex set. We define the first rigid expansion of
$Y$, denoted by $Y^{1}$ as the induced subgraph of $\mathcal{C}(S_{g,n})$
whose vertex set if the following:

$V(Y^{1})\displaystyle\mathrel{\mathop{:}}=V(Y)\cup\\{\alpha\in\mathcal{C}(S_{g,n}):\alpha=\langle
A\rangle,\text{ for some }A\subset V(Y)\\}.$

We then define the $(n+1)$-th rigid expansion as $(Y^{n})^{1}$, and
$Y^{\omega}$ as the induced subgraph of $\mathcal{C}(S_{g,n})$ induced by the
following vertex set:

$V(Y^{\omega})\displaystyle\mathrel{\mathop{:}}=\bigcup_{n\geq 1}V(Y^{n}).$

Recall that a graph morphism between simplicial graphs is a map between the
corresponding vertex sets that also maps edges to edges.

The following theorem is Theorem B in [6] for the special case of the curve
graph, using the terminology here presented:

######  Theorem 1.1 (B in [6]).

Let $Y<\mathcal{C}(S_{g,n})$ be an induced subgraph, and
$\varphi:Y^{\omega}\rightarrow\mathcal{C}(S_{g,n})$ be a graph morphism such
that there exists an automorphism of $\mathcal{C}(S_{g,n})$ (if
$(g,n)\neq(1,2)$ it is a possibly orientation-reversing homeomorphism of
$S_{g,n}$) $h:\mathcal{C}(S_{g,n})\rightarrow\mathcal{C}(S_{g,n})$ with
$h|_{Y}=\varphi|_{Y}$. Then, $h|_{Y^{\omega}}=\varphi$.

Now, given a curve $\alpha$ on $S_{g,n}$, we define the star of $\alpha$,
denoted by $\mathrm{star}(\alpha)$, as the subgraph whose vertex set is
$\\{\alpha\\}\cup\mathrm{adj}(\alpha)$, and the edges are those from
$\mathcal{C}(S_{g,n})$ that have $\alpha$ as one of its endpoints.

A locally injective map $\varphi:X\rightarrow\mathcal{C}(S_{g,n})$ is a
simplicial map whose restriction to $\mathrm{star}(\alpha)$ is injective for
every $\alpha\in\mathcal{C}(S_{g,n})$.

Now, a induced subgraph $X<\mathcal{C}(S_{g,n})$ is called rigid if every
locally injective map from $X$ to $\mathcal{C}(S_{g,n})$ is induced by an
automorphism of $\mathcal{C}(S_{g,n})$ (again, if $(g,n)\neq(1,2)$ it becomes
a homeomorphism of $S_{g,n}$).

Note that not every supergraph of a rigid subgraph is rigid (see Proposition
3.2 in [2]). However, any rigid expansion of a rigid subgraph is rigid (see
Proposition 3.5 in [2]).

The following corollary is Corollary C in [6] for the special case of the
curve graph, and it comes as a direct consequence of Theorem 1.1 and the
definition of rigidity:

######  Corollary 1.2 (C in [6]).

Let $X<\mathcal{C}(S_{g,n})$ be a rigid subgraph, and
$\varphi:X^{\omega}\rightarrow\mathcal{C}(S_{g,n})$ be an edge-preserving map
such that $\varphi|_{X}$ is locally injective. Then $\varphi$ is the
restriction to $X^{\omega}$ of an automorphism of $\mathcal{C}(S_{g,n})$ (if
$(g,n)\neq(1,2)$, it is a homeomorphism of $S_{g,n}$).

##  2 Genus zero case

In this section, we assume that $S=S_{0,n}$ with $n\geq 5$ (so that
$\kappa(S)\geq 2$), unless otherwise stated.

The structure of this section is as follows: In Section 2.1 we define
$\mathscr{O}$ which is the basis upon which we construct $\mathscr{Y}(S)$; in
Section 2.2 we define the sets of auxiliary curves $\mathscr{D}_{1}$ and
$\mathscr{D}_{2}$, and we also define $\mathscr{Y}(S)$; in Section 2.3 we give
the proof of Theorem A pending the proof of a key lemma (Lemma 2.2); in
Sections 2.4, 2.5, 2.6 and 2.7 we give the proof of the key lemma; finally, in
Section 2.8 we recall from [2] the definition of $\mathfrak{X}(S)$ and prove
Theorem B.

###  2.1 The basis of $\mathscr{Y}(S)$

For $k\geq 2$, let $O=\\{\alpha_{0},\ldots,\alpha_{k}\\}$ be a set of outer
curves. We say $O$ is an outer chain if $i(\alpha_{i},\alpha_{j})=2$ if
$|i-j|=1$ and $\alpha_{i}$ is disjoint from $\alpha_{j}$ otherwise. This
notation comes as an analogue for a chain in higher genus. We say $O$ is a
closed outer chain if (counting the subindices modulo $k+1$)
$i(\alpha_{i},\alpha_{j})=2$ if $|i-j|=1$ and $\alpha_{i}$ is disjoint from
$\alpha_{j}$ otherwise. See Figure 3 for examples. In both cases, we set the
length of $O$ as its cardinality.

Figure 3: On the left, an example of an outer chain. On the right, an example
of a closed outer chain.

Note that a closed outer chain of maximal length has cardinality $n$, and its
regular neighbourhood is homeomorphic to a $n$-punctured annulus.

Let $\mathscr{O}=\\{\alpha_{0},\ldots,\alpha_{n-1}\\}$ be a closed outer chain
of maximal length. While these curves seem to be a good candidate for
$\mathscr{Y}(S)$, they are not enough, i.e. they eventually stabilize. So, we
have to add some auxiliary curves.

###  2.2 Auxiliary curves

To define the auxiliary curves, we must first number the punctures of $S$ in
such a way that $\alpha_{i}\in\mathscr{O}$ bounds the $i$-th and $(i+1)$-th
punctures (with indices modulo $n$). See Figure 4 for an example.

2pt $\alpha_{i}$ [bl] at 80 165 $i+1$ at 45 258 $i$ [br] at 0 130

Figure 4: The $i$-th and $(i+1)$-th puncture along with the curve
$\alpha_{i}$.

Now, let $\alpha$ be an outer curve on $S$. We call the arc corresponding to
$\alpha$, to the unique arc “joining” the punctures bounded by $\alpha$.
Similarly, let $a$ be an arc on $S$ with different endpoints; we call the
outer curve corresponding to $a$, to the unique outer curve that bounds the
endpoints of $a$ and contains $a$ in its interior. See Figure 5 for examples.

Figure 5: The blue and purple arcs have the green and orange curves as their
(respective) corresponding outer curves.

Note that there is a bijective correspondance between the set of all outer
curves on $S$, and the set of all arcs on $S$ with differing endpoints.

Let $A=\\{a_{i}\\}$ be the set of arcs $a_{i}$ corresponding to
$\alpha_{i}\in\mathscr{O}$. Let also $H_{1}$ and $H_{2}$ be two identical
$n$-gons with numbered vertices such that $S$ can be identified with the
result of gluing $H_{1}$ and $H_{2}$ together by their sides (and eliminating
the vertices), and the labeling of the vertices in $H_{1}$ and $H_{2}$
coincides with the labeling of the punctures on $S$. Moreover, we particularly
choose the identification, so that the (glued together) sides of $H_{1}$ and
$H_{2}$ are identified with the elements of $A$. See Figure 6.

$H_{1}$ at 146 133 $H_{2}$ at 518 133 $1$ [br] at 70 255 $2$ [bl] at 229 255
$3$ [l] at 295 130 $4$ [tl] at 225 5 $5$ [tr] at 72 5 $6$ [r] at 5 130 $1$
[br] at 435 255 $2$ [bl] at 594 255 $3$ [l] at 665 130 $4$ [tl] at 590 5 $5$
[tr] at 437 5 $6$ [r] at 370 130 $\rightsquigarrow$ at 730 133 $\alpha_{1}$ at
924 205 $\alpha_{6}$ [bl] at 852 160

Figure 6: The $n$-gons $H_{1}$ and $H_{2}$ (left), glued together to obtain
the surface $S$ (right).

For $0\leq i<j<n$, let $b_{i,j}$ be the arc on $S$ joining the $i$-th and
$j$-th punctures of $S$ that is identified with the diagonal on $H_{1}$
joining the $i$-th and $j$-th vertices. Similarly, for $0\leq i<j<n$, let
$c_{i,j}$ be the arc on $S$ joining the $i$-th and $j$-th punctures of $S$
that is identified with the diagonal on $H_{2}$ joining the $i$-th and $j$-th
vertices. See Figure 7 for an example.

$b_{1,5}$ [l] at 80 130 $c_{1,6}$ [b] at 517 135 $\beta_{1,5}$ [bl] at 885 200
$\gamma_{1,6}$ [bl] at 760 45

Figure 7: The arcs $b_{1,5}$ and $c_{1,6}$ that induce the curves
$\beta_{1,5}$ and $\gamma_{1,6}$ respectively.

Then, let $\mathscr{D}_{1}=\\{\beta_{i,j}\\}$ be the set of outer curves
$\beta_{i,j}$ corresponding to the arc $b_{i,j}$, and let
$\mathscr{D}_{2}=\\{\gamma_{i,j}\\}$ be the set of outer curves $\gamma_{i,j}$
corresponding to the arc $c_{i,j}$.

Finally, we define:

$\mathscr{Y}(S)\displaystyle\mathrel{\mathop{:}}=\mathscr{O}\cup\mathscr{D}_{1}\cup\mathscr{D}_{2}.$

###  2.3 Proof of Theorem A

For the proof of Theorem A, we first recall some definitions and properties
around outer curves on $S$ and half-twists.

Let $\alpha$ be a curve on $S$. We denote by $\eta_{\alpha}$ the (left) half-
twist along $\alpha$ (see Figure 8); recall that this is defined if and only
if $\alpha$ is an outer curve. Also recall that there is exactly one half-
twist along $\alpha$ if $S\ncong S_{0,4}$.

$\beta$ [bl] at 135 60 $\alpha$ [bl] at 165 210 $\mapsto$ at 360 130
$\eta_{\alpha}(\beta)$ [bl] at 550 160

Figure 8: On the left, the outer curve $\alpha$ and the curve $\beta$; on the
right, the curve $\eta_{\alpha}(\beta)$.

######  Remark 2.1.

Note that if $\alpha$ and $\beta$ are two outer curves that intersect twice,
then we have that $\eta_{\alpha}(\beta)=\eta_{\beta}^{-1}(\alpha)$. This fact
can also be found in Proposition 3.9 of [2].

Let $O$ be a set of outer curves and $A$ be a set of curves on $S$. We denote
by $\eta_{O}(A)$ the following set

$\eta_{O}(A)\displaystyle\mathrel{\mathop{:}}=\bigcup_{\varepsilon\in
O}\eta_{\varepsilon}(A).$

Let $\mathscr{G}\displaystyle\mathrel{\mathop{:}}=\\{\eta_{\alpha_{i}}^{\pm
1}:\alpha_{i}\in\mathscr{O}\\}$. It is a well-known fact (see Theorem 4.9 in
[4]) that $\mathscr{G}$ is a symmetric generating set of $\mathrm{Mod}(S)$.
With this, if $A$ is a set of curves, we denote by $\mathscr{G}\cdot A$ the
following set $\mathscr{G}\cdot
A\displaystyle\mathrel{\mathop{:}}=\eta_{\mathscr{O}}^{\pm 1}(A)$.

Now, similarly to the proofs in [7], we state a key lemma for the proof of
Theorem A. The proof of this lemma is discussed and given after the proof of
Theorem A (see Subsections 2.5 and 2.6).

######  Lemma 2.2.

Let $\mathscr{G}$ be as above. Then
$\mathscr{G}\cdot\mathscr{Y}(S)\subset\mathscr{Y}(S)^{3}$.

###### Proof of Theorem A:.

We divide this proof into two parts: first we prove that all outer curves on
$S$ are elements in $\mathscr{Y}(S)^{\omega}$; then we argue that any other
curve on $S$ is uniquely determined by a finite number of outer curves and
therefore we obtain the desired result.

First part: Let $\beta$ be an outer curve on $S$. Since all outer curves on
$S$ have the same topological type, then there exists a element
$\alpha_{i}\in\mathscr{O}$ and a homeomorphism $h\in\mathrm{Mod}(S)$ such that
$h(\alpha_{i})=\beta$. Given that $\mathscr{G}$ is a symmetric generating set
of $\mathrm{Mod}(S)$, then there exist elements
$f_{1},\ldots,f_{k}\in\mathscr{G}$ and powers $n_{1},\ldots,n_{k}\geq 0$ such
that

$f_{1}^{n_{1}}\circ\cdots\circ f_{k}^{n_{k}}(\alpha_{i})=\beta.$

By an iterative use of Lemma 2.2, this implies that
$\beta\in\mathscr{Y}(S)^{3(n_{1}+\cdots+n_{k})}$. Thus, every outer curve on
$S$ is an element of $\mathscr{Y}(S)^{\omega}$.

Second part: Let $\gamma$ be a curve on $S$ that is not an outer curve. Since
there are only finitely many topological types of such curves, which depend on
how they separate the punctures of $S$, up to homeomorphism $\gamma$ is as in
the example on Figure 9. Then, there exists two (finite) outer chains, namely
$A$ and $B$, that uniquely determine $\gamma$ (see Figure 9 for an example).
Given that by the first part of this proof $A\cup B\subset\mathscr{Y}(S)^{k}$
for some $k\geq 0$, we have that $\gamma\in\mathscr{Y}(S)^{k+1}$. Therefore,
every curve on $S$ that is not an outer curve, is also an element of
$\mathscr{Y}(S)^{\omega}$.

Figure 9: The set $A$ (red curves) and the set $B$ (green curves) uniquely
determining the curve $\gamma$ (in blue).

∎

As for the proof of Lemma 2.2, we first prove the lemma for the complexity $2$
case in Subsection 2.4; then we prove them for complexity at least $3$ in more
generality. The reason for this division is that in the complexity $2$ we “do
not have enough space” for the general method to work.

For the complexity at least $3$ case, we divide the proof into the following
claims:
Claim 1: $\eta_{\mathscr{O}}^{\pm 1}(\mathscr{O})\subset\mathscr{Y}(S)$
Claim 2: $\eta_{\mathscr{O}}^{\pm
1}(\mathscr{D}_{1})\subset\mathscr{Y}(S)^{2}$
Claim 3: $\eta_{\mathscr{O}}^{\pm
1}(\mathscr{D}_{2})\subset\mathscr{Y}(S)^{3}$

###  2.4 Complexity $2$ case

Let $S$ be a $5$-punctured sphere, $\alpha\in\mathscr{O}$ and
$\delta\in\mathscr{Y}(S)$. Note that if $\alpha$ and $\delta$ are disjoint, we
have that $\eta_{\alpha}^{\pm 1}(\delta)=\delta$. Thus, we focus solely on the
cases when they intersect.

We can easily check that

$\eta_{\alpha_{i+1}}(\alpha_{i})=\gamma_{i,i+2}\in\mathscr{Y}(S),$
$\eta_{\alpha_{i-1}}(\alpha_{i})=\beta_{i,i+2}\in\mathscr{Y}(S).$

This and Remark 2.1 imply that

$\eta_{\alpha_{i+1}}^{-1}(\alpha_{i})=\eta_{\alpha_{i}}(\alpha_{i+1})\in\mathscr{Y}(S),$
$\eta_{\alpha_{i-1}}^{-1}(\alpha_{i})=\eta_{\alpha_{i}}(\alpha_{i-1})\in\mathscr{Y}(S).$

Hence, $\eta_{\mathscr{O}}^{\pm 1}(\mathscr{O})\subset\mathscr{Y}(S)$.

Since $S$ is a $5$-punctured sphere, then every $\beta\in\mathscr{D}_{1}$ is
of the form either $\beta_{i,i+2}$ or $\beta_{i,i+3}$ for some $0\leq i<n$
(with the indices modulo $5$).

If $\beta=\beta_{i,i+3}$ we have that

$\eta_{\alpha_{i-1}}(\beta_{i,i+3})=\alpha_{i-2}\in\mathscr{Y}(S),$
$\eta_{\alpha_{i}}(\beta_{i,i+3})=\langle\\{\beta_{i,i+2},\gamma_{i-1,i+2}\\}\rangle\in\mathscr{Y}(S)^{1}.$

See Figure 10 for an example.

Figure 10: On the left, the curves $\gamma_{i-1,i+2}$ (orange) and
$\beta_{i,i+2}$; on the right the curve they uniquely determine,
$\eta_{\alpha_{i}}(\beta_{i,i+3})$, in green.

If $\beta=\beta_{i,i+2}$, on one hand we have that

$\eta_{\alpha_{i-1}}(\beta_{i,i+2})=\beta_{i-1,i+2}\in\mathscr{Y}(S).$

On the other hand, for $\eta_{\alpha_{i}}(\beta_{i,i+2})$ we need an auxiliary
curve, which we define as follows (using indices modulo $5$):

$\delta=\langle\\{\beta_{i-1,i+2},\gamma_{i-1,i+1}\\}\rangle.$

See Figure 11.

Figure 11: On the left, the curves $\beta_{i-1,i+2}$ (blue) and
$\gamma_{i-1,i+1}$ (green); on the right the auxiliary curve $\delta$ (orange)
which they uniquely determine.

Then, we have that

$\eta_{\alpha_{i}}(\beta_{i,i+2})=\langle\\{\alpha_{i-2},\delta\\}\rangle\in\mathscr{Y}(S)^{2}.$

See Figure 12.

Figure 12: On the left, the curves $\alpha_{i-2}$ (blue) and the auxiliary
curve $\delta$ (orange); on the right, $\eta_{\alpha_{i}}(\beta_{i,i+2})$,
which they uniquely determine.

Given that the rest of the cases (with $\alpha_{i+1}$, $\alpha_{i+2}$ and
$\alpha_{i+3}$) are analogous to the cases above, we get that
$\eta_{\mathscr{O}}(\mathscr{D}_{1})\subset\mathscr{Y}(S)^{2}$.

Again, since $S$ is a $5$-punctured sphere, then every
$\gamma\in\mathscr{D}_{2}$ is of the form either $\gamma_{i,i+2}$ or
$\gamma_{i,i+3}$ for some $0\leq i<n$ (with the indices modulo $5$).

If $\gamma=\gamma_{i,i+2}$, we have that

$\eta_{\alpha_{i}}(\gamma_{i,i+2})=\alpha_{i+1}\in\mathscr{Y}(S),$

and that the case for $\eta_{\alpha_{i-1}}(\gamma_{i,i+2})$ is analogous to
the case for $\eta_{\alpha_{i}}(\beta_{i,i+3})$; hence

$\eta_{\alpha_{i-1}}(\gamma_{i,i+2})\in\mathscr{Y}(S)^{1}.$

If $\gamma=\gamma_{i,i+3}$, we have that

$\eta_{\alpha_{i}}(\gamma_{i,i+3})=\gamma_{i+1,i+3}\in\mathscr{Y}(S),$

and that the case for $\eta_{\alpha_{i-1}}(\gamma_{i,i+3})$ is analogous to
the case for $\eta_{\alpha_{i}}(\beta_{i,i+2})$; hence

$\eta_{\alpha_{i-1}}(\gamma_{i,i+3})\in\mathscr{Y}(S)^{2}.$

Analogously to the cases for $\mathscr{D}_{1}$, the cases above imply that
$\eta_{\mathscr{O}}(\mathscr{D}_{2})\subset\mathscr{Y}(S)^{2}$.

Finally, the cases for $\eta_{\mathscr{O}}^{-1}(\mathscr{D}_{1})$ and
$\eta_{\mathscr{O}}^{-1}(\mathscr{D}_{2})$ mirror the cases for
$\eta_{\mathscr{O}}(\mathscr{D}_{2})$ and
$\eta_{\mathscr{O}}(\mathscr{D}_{1})$, respectively. Therefore,
$\eta_{\mathscr{O}}^{\pm 1}(\mathscr{Y}(S))\subset\mathscr{Y}(S)^{2}$.

###  2.5 Proof of Claim 1: $\eta_{\mathscr{O}}^{\pm
1}(\mathscr{O})\subset\mathscr{Y}(S)$

Let $\alpha_{i},\alpha_{j}\in\mathscr{O}$. Note that if $\alpha_{i}$ is
disjoint from $\alpha_{j}$, then $\eta_{\alpha_{i}}^{\pm
1}(\alpha_{j})=\alpha_{j}\in\mathscr{Y}(S)$. So, we only focus on the cases
where $\alpha_{i}$ intersects $\alpha_{j}$, i.e. $|i-j|=1$.

For the proof of Claim 1, let $0\leq i<n$, then we have that (using indices
modulo $n$):

$\eta_{\alpha_{i+1}}(\alpha_{i})=\gamma_{i,i+2}.$

Also, we have

$\eta_{\alpha_{i+1}}^{-1}(\alpha_{i})=\beta_{i,i+2}.$

With this and Remark 2.1 we have that $\eta_{\mathscr{O}}^{\pm
1}(\mathscr{O})\subset\mathscr{Y}(S)$.

###  2.6 Proof of Claim 2: $\eta_{\mathscr{O}}^{\pm
1}(\mathscr{D}_{1})\subset\mathscr{Y}(S)^{2}$

Let $\beta_{i,j}\in\mathscr{D}_{1}$ and $\alpha_{k}\in\mathscr{O}$. Note that
if $\alpha_{k}$ is disjoint from $\beta_{i,j}$ then $\eta_{\alpha_{k}}^{\pm
1}(\beta_{i,j})=\beta_{i,j}\in\mathscr{Y}(S)$. So, we only focus on the cases
where $\alpha_{k}$ intersects $\beta_{i,j}$.

Now, we divide the proof of this claim into three cases according to the
triple $(i,j,k)$:

First case: We have the following subcases:

  1. (i)

Suppose $j\neq i+2$ and $i\neq j+2$ (both modulo $n$).

  2. (ii)

Suppose $j=i+2$ and $k\neq i,i+1$.

  3. (iii)

Suppose $i=j+2$ and $k\neq j,j+1$.

Then, there exists non-empty sets $O\subset\mathscr{O}$ and
$D\subset\mathscr{D}_{2}$ such that $\beta_{i,j}=\langle O\cup D\rangle$ and
$\alpha_{k}$ is disjoint from every element in $D$. Thus,
$\eta_{\alpha_{k}}^{\pm 1}(\beta_{i,j})=\langle\eta_{\alpha_{k}}^{\pm
1}(O)\cup\eta_{\alpha_{k}}^{\pm 1}(D)\rangle=\langle\eta_{\alpha_{k}}^{\pm
1}(O)\cup D\rangle$. From this, by Claim 1, we have that
$\eta_{\alpha_{k}}^{\pm 1}(\beta_{i,j})\in\mathscr{Y}(S)^{1}$.

The existence of these sets for each subcase is inferred from the following
examples (see Figure 13):

  1. (i)

If $j\neq i+2$ and $i\neq j+2$ (both modulo $n$), and if $k=i-1$ or $k=i$, we
set
$O=\\{\alpha_{i+1},\ldots,\alpha_{j-2}\\}\cup\\{\alpha_{j+1},\ldots,\alpha_{i-2}\\}$
and $D=\\{\gamma_{j-1,j+1}\\}$. The case for $k=j-1$ or $k=j$ is analogous.

  2. (ii)

If $j=i+2$ and $k\neq i,i+1$, we set
$O=\\{\alpha_{j+1},\ldots,\alpha_{i-2}\\}$ and $D=\\{\gamma_{i-1,i+1}\\}$.

  3. (iii)

The case for $i=j+2$ and $k\neq j,j+1$ is analogous to the subcase (ii).

$i$ [bl] at 230 250 $k$ [br] at 70 250 $j$ [r] at 70 0
$i$ [bl] at 230 250 $j$ [l] at 225 0

Figure 13: On top, the curves corresponding to the subcase $(i)$ and the curve
$\beta_{i,j}$; on the bottom, the curves corresponding to the subcase $(ii)$.
On the right, the curves $\beta_{i,j}$ they uniquely determine.

Second case: Suppose $j=i+2$ and $k=i$ (the case $i=j+2$ and $k=j$ is
completely analogous). Then
$\eta_{\alpha_{i}}^{-1}(\beta_{i,i+2})=\alpha_{i+1}$.

For $\eta_{\alpha_{i}}(\beta_{i,i+2})$ we need an auxiliary curve, which we
define as follows (recall the indices are modulo $n$):

$\delta\displaystyle\mathrel{\mathop{:}}=\langle\\{\alpha_{i+4},\ldots\alpha_{i-2}\\}\cup\\{\beta_{i+2,i+4}\\}\cup\\{\gamma_{i-1,i+1}\\}\rangle.$

See Figure 14 for an example.

Then, we have that

$\eta_{\alpha_{i}}(\beta_{i,i+2})=\langle\\{\alpha_{i+3},\ldots,\alpha_{i-2}\\}\cup\\{\delta\\}\rangle\in\mathscr{Y}(S)^{2}.$

See Figure 14 for an example.

$i=k$ [bl] at 230 250 $j$ [l] at 225 5 $\delta$ [bl] at 625 220
$i$ [bl] at 230 245 $\eta_{\alpha_{i}}(\beta_{i,i+2})$ [l] at 660 170

Figure 14: On top, the curves needed to uniquely determine the auxiliary curve
$\delta$. On the bottom, the curves needed to uniquely determine
$\eta_{\alpha_{i}}(\beta_{i,i+2})$ in the second case.

Third case: Suppose $j=i+2$ and $k=i+1$ (the case $i=j+2$ and $k=j+1$ is
completely analogous). Then $\eta_{\alpha_{i+1}}(\beta_{i,i+2})=\alpha_{i}$.

Now, for $\eta_{\alpha_{i+1}}^{-1}(\beta_{i,i+2})$ we need again an auxiliary
curve, which we define as follows (recall the indices are modulo $n$):

$\delta\displaystyle\mathrel{\mathop{:}}=\langle\\{\alpha_{i+3},\ldots,\alpha_{i-3}\\}\cup\\{\beta_{i-2,i}\\}\cup\\{\gamma_{i+1,i+3}\\}\rangle.$

See Figure 15 for an example.

Then, we have that

$\eta_{\alpha_{i+1}}^{-1}(\beta_{i,i+2})=\langle\\{\alpha_{i+3},\ldots,\alpha_{i-2}\\}\cup\\{\delta\\}\rangle\in\mathscr{Y}(S)^{2}.$

See Figure 15 for an example.

$i$ [l] at 300 130 $\delta$ [bl] at 625 220
$i$ [l] at 300 130 $\eta_{\alpha_{i+1}}^{-1}(\beta_{i,i+2})$ [l] at 660 170

Figure 15: On top, the curves needed to uniquely determine the auxiliary curve
$\delta$. On the bottom, the curves needed to uniquely determine
$\eta_{\alpha_{i+1}}^{-1}(\beta_{i,i+2})$ in the second case.

Thus, $\eta_{\mathscr{O}}(\mathscr{D}_{1})\subset\mathscr{Y}(S)^{2}$.

###  2.7 Proof of Claim 3: $\eta_{\mathscr{O}}^{\pm
1}(\mathscr{D}_{2})\subset\mathscr{Y}(S)^{3}$

The proof of this claim follows from Claim 1 and Claim 2: If
$\gamma\in\mathscr{D}_{2}$, then there exist sets $O\subset\mathscr{O}$ and
$D\subset\mathscr{D}_{1}$ such that $\gamma=\langle O\cup D\rangle$. Thus, for
all $\alpha_{i}\in\mathscr{O}$, $\eta_{\alpha_{i}}^{\pm
1}(\gamma)=\langle\eta_{\alpha_{i}}^{\pm 1}(O)\cup\eta_{\alpha_{i}}^{\pm
1}(D)\rangle$. By Claims 1 and 2, this implies that
$\eta_{\alpha_{i}}(\gamma)\in\mathscr{Y}(S)^{3}$.

The existence of these sets can be easily inferred from the following examples
(see Figure 16):

  * •

If $j\neq i+2$ and $i\neq j+2$ (both modulo $n$), we set
$O=\\{\alpha_{i+1},\ldots,\alpha_{j-2}\\}\cup\\{\alpha_{j+1},\ldots,\alpha_{i-2}\\}$
and $D=\\{\beta_{i+1,j+1}\\}$.

  * •

If $j=i+2$ (modulo $n$), we set $O=\\{\alpha_{j+1},\ldots,\alpha_{i-2}\\}$ and
$D=\\{\beta_{i+1,j+1}\\}$.

$i$ [bl] at 225 250 $j$ [r] at 75 0
$i$ [bl] at 225 250 $j$ [l] at 220 0

Figure 16: On the left, the sets $O$ (red) and $D$ (blue); on the right, the
curve $\gamma_{i,j}$ they uniquely determine.

The rest of the cases are analogous to one of the above.

###  2.8 Proof of Theorem B

For the purposes of this work, we define the set $\mathfrak{X}(S)$ as the set
$\mathfrak{X}_{A}$ from Section 4 in [2], which was proved to be rigid (see
Theorem 4.2 in [2]).

According to the notation above, we can define $\mathfrak{X}(S)$ as follows:

$\mathfrak{X}(S)\displaystyle\mathrel{\mathop{:}}=\mathscr{O}^{1}\cup\mathscr{G}\cdot\mathscr{O}.$

###### Proof of Theorem B.

To prove Theorem B we only need to prove that
$\mathscr{Y}(S)\subset\mathfrak{X}(S)^{1}$, since this would imply that (by
Theorem A)
$\mathcal{C}(S)=\mathscr{Y}(S)^{\omega}\subset\mathfrak{X}(S)^{\omega}$.

For $\gamma\in\mathscr{O}$, it is obvious that $\gamma\in\mathfrak{X}(S)$.
Thus, we suppose that $\gamma\in\mathscr{D}_{1}\cup\mathscr{D}_{2}$.

If $\gamma\in\mathscr{D}_{1}$, there exists $0\leq i<j<n$ such that
$\gamma=\beta_{i,j}$. Thus we have two possible cases:

  1. 1.

Suppose that $j\neq i+2$ (modulo $n$). Then (see Figure 17)

$\beta_{i,j}=\langle\\{\alpha_{i+1},\ldots,\alpha_{j-2}\\}\cup\\{\alpha_{j+1},\ldots\alpha_{i-2}\\}\cup\\{\eta_{\alpha_{i}}(\alpha_{i-1})\\}\rangle.$

  2. 2.

Suppose that $j=i+2$ (modulo $n$). Then
$\beta_{i,j}=\eta_{\alpha_{i+1}}^{-1}(\alpha_{i})$.

$i$ [bl] at 225 250 $j$ [r] at 75 0

Figure 17: On the left, the curves needed to uniquely determine the curve
$\beta_{i,j}$ on the right.

If $\gamma\in\mathscr{D}_{2}$, there exists $0\leq i<j<n$ such that
$\gamma=\gamma_{i,j}$. Thus we have two possible cases, which are analogous to
(1) and (2) above, simply substituting $\eta_{\alpha_{i}}(\alpha_{i-1})$ for
$\eta_{\alpha_{i}}^{-1}(\alpha_{i-1})$ in case (1) and
$\eta_{\alpha_{i+1}}^{-1}(\alpha_{i})$ for $\eta_{\alpha_{i+1}}(\alpha_{i})$
in case (2).

Then, $\mathscr{Y}(S)\subset\mathfrak{X}(S)^{1}$, and therefore
$\mathfrak{X}(S)^{\omega}=\mathcal{C}(S)$. ∎

##  3 Genus one case

Given that $\mathcal{C}(S_{1,2})\cong\mathcal{C}(S_{0,5})$, Subsections 2.4
and 2.8 already prove Theorems B and A for the twice-punctured torus. So, in
this section we assume that $S=S_{1,n}$ with $n\geq 3$ (so that $\kappa(S)\geq
3$), unless otherwise stated.

The structure of this section is as follows: In Section 3.1 we define
$\mathscr{C}$ and $\mathscr{D}$ which are the basis upon which we construct
$\mathscr{Y}(S)$; in Section 3.2 we define the set of auxiliary curves
$\mathscr{A}$, and we also define $\mathscr{Y}(S)$; in Section 3.3 we give the
proof of Theorem A pending the proof of a key lemma (Lemma 3.2); in
Subsections 3.4, 3.5, 3.6 and 3.7, we give the proof of the key lemma;
finally, in Section 3.8 we recall from [2] the definition of $\mathfrak{X}(S)$
and prove Theorem B.

###  3.1 The basis of $\mathscr{Y}(S)$

Let
$\mathscr{C}\displaystyle\mathrel{\mathop{:}}=\\{\alpha_{0}^{0},\ldots,\alpha_{0}^{n-1},\alpha_{1}\\}$
be a set of curves such that $\alpha_{0}^{i}$ is disjoint from
$\alpha_{0}^{j}$ if $i\neq j$, and $i(\alpha_{1},\alpha_{0}^{i})=1$ for all
$0\leq i<n$. See Figure 18.

Using $\mathscr{C}$, we number the punctures in $S$ in the following way: the
$i$-th puncture is in the annulus bounded by $\alpha_{0}^{i-1}$ and
$\alpha_{0}^{i}$. In Figure 18, this means that the $i$-th puncture is on the
left of $\alpha_{0}^{i}$.

$1$ [b] at 107 140 $2$ [b] at 206 140 $3$ [b] at 305 140 $\alpha_{1}$ [l] at
405 70 $\alpha_{0}^{0}$ [t] at 50 7 $\alpha_{0}^{1}$ [t] at 149 7
$\alpha_{0}^{2}$ [t] at 249 7

Figure 18: The curve $\alpha_{1}$ in red, and the curves
$\alpha_{0}^{0},\alpha_{0}^{1},\alpha_{0}^{2}$ in blue.

With this, we can define for each $0\leq i<n$ the following curve (see Figure
19):

$\beta_{i}\displaystyle\mathrel{\mathop{:}}=\langle\mathscr{C}\backslash\\{\alpha_{0}^{i}\\}\rangle.$

$\beta_{1}$ [b] at 705 165

Figure 19: On the left, the curves needed to uniquely determine the curve
$\beta_{1}$ on the right.

Now, we define
$\mathscr{D}\displaystyle\mathrel{\mathop{:}}=\\{\beta_{i}\\}_{i=0}^{n-1}$.
Note that $\mathscr{D}$ is a closed outer chain.

Ideally we would like $\mathscr{Y}(S)$ to be equal to the union of
$\mathscr{C}$ and $\mathscr{D}$; however,
$(\mathscr{C}\cup\mathscr{D})^{2}=(\mathscr{C}\cup\mathscr{D})^{1}$. So, we
need some auxiliary curves.

###  3.2 Auxiliary curves

To define the auxiliary curves, first we must recall (from Subsection 2.3)
that if $\alpha$ is an outer curve on $S$, then $\eta_{\alpha}$ denotes the
half-twist along $\alpha$.

For each $0\leq i<n$, we now define the following curves (with indices modulo
$n$):

$\varepsilon_{i}^{+}\displaystyle\mathrel{\mathop{:}}=\eta_{\beta_{i}}(\beta_{i-1})\hskip
56.9055pt\varepsilon_{i}^{-}\displaystyle\mathrel{\mathop{:}}=\eta_{\beta_{i}}^{-1}(\beta_{i-1}).$

See Figure 20.

$i$ [tr] at 100 130 $i$ [br] at 580 143 $\varepsilon_{i}^{+}$ [t] at 260 130
$\varepsilon_{i}^{-}$ [b] at 740 140

Figure 20: On the left the curve $\varepsilon_{i}^{+}$; on the right the curve
$\varepsilon_{i}^{-}$.

Thus, we can define the sets
$\mathscr{A}^{+}\displaystyle\mathrel{\mathop{:}}=\\{\varepsilon_{i}^{+}\\}_{i=0}^{n-1}$,
$\mathscr{A}^{-}\displaystyle\mathrel{\mathop{:}}=\\{\varepsilon_{i}^{-}\\}_{i=0}^{n-1}$,
and
$\mathscr{A}\displaystyle\mathrel{\mathop{:}}=\mathscr{A}^{+}\cup\mathscr{A}^{-}$.

Finally, we define

$\mathscr{Y}(S)\displaystyle\mathrel{\mathop{:}}=\mathscr{C}\cup\mathscr{D}\cup\mathscr{A}.$

###  3.3 Proof of Theorem A

For the proof of Theorem A, we first recall some definitions and properties of
Dehn twists along curves on $S$.

Let $\alpha$ be a curve on $S$; we denote by $\tau_{\alpha}$ the (left) Dehn
twist along $\alpha$. Similarly to Remark 2.1, we have the following property.

######  Remark 3.1.

Note that if $\alpha$ and $\beta$ are curves that intersect once, then we have
that $\tau_{\alpha}(\beta)=\tau_{\beta}^{-1}(\alpha)$. Once again, this fact
can also be found in Proposition 3.9 in [2].

Similarly to the notation around half-twists, if $C$ and $A$ are sets of
curves on $S$, we denote by $\tau_{C}(A)$ the following set

$\tau_{C}(A)=\bigcup_{\gamma\in C}\tau_{\gamma}(A).$

Now, let $\mathscr{G}\displaystyle\mathrel{\mathop{:}}=\\{\tau_{\alpha}^{\pm
1}:\alpha\in\mathscr{C}\\}\cup\\{\eta_{\beta}^{\pm 1}:\beta\in\mathscr{D}\\}$.
It is a well-known fact (Theorem 4.14 in [4]) that $\mathscr{G}$ is a
symmetric generating set of $\mathrm{Mod}(S)$ (often called the Humphries-
Lickorish generating set). As in the previous section if $A$ is a set of
curves on $S$, we denote by $\mathscr{G}\cdot A$ the following set
$\mathscr{G}\cdot A\displaystyle\mathrel{\mathop{:}}=\tau_{\mathscr{C}}^{\pm
1}(A)\cup\eta_{\mathscr{D}}^{\pm 1}(A)$.

As in the previous section, we state a key lemma for the proof of Theorem A,
leaving its proof to the following subsections (see Subsections 3.4, 3.5, 3.6
and 3.7).

######  Lemma 3.2.

Let $\mathscr{G}$ be as above. Then
$\mathscr{G}\cdot\mathscr{Y}(S)\subset\mathscr{Y}(S)^{6}$.

###### Proof of Theorem A.

We divide the proof into two parts: first we prove that all outer curves and
non-separating curves on $S$ are elements in $\mathscr{Y}(S)^{\omega}$; then
we argue that any curve on $S$ is uniquely determined by a finite set of non-
separating curves and outer curves, thus obtaining the desired result.

First part: Let $\gamma$ be either an outer curve or a non-separating curve on
$S$. Recall that all outer curves on $S$ have the same topological type, and
it is the same case for all non-separating curves. Then, there exists a curve
$\delta\in\mathscr{Y}(S)$ and a homeomorphism $h\in\mathrm{Mod}(S)$ such that
$h(\delta)=\gamma$. Since $\mathscr{G}$ is a symmetric generating set of
$\mathrm{Mod}(S)$, there exist elements $f_{1},\ldots,f_{k}\in\mathscr{G}$ and
powers $n_{1},\ldots,n_{k}\geq 0$ such that

$f_{1}^{n_{1}}\circ\cdots\circ f_{k}^{n_{k}}(\delta)=\gamma.$

By an iterated use of Lemma 3.2, this implies that
$\gamma\in\mathscr{Y}(S)^{6(n_{1}+\cdots+n_{k})}$. Thus, every outer curve and
every non-separating curve on $S$ is an element of $\mathscr{Y}(S)^{\omega}$.

Second part: Let $\gamma$ be a separating curve on $S$ that is not an outer
curve. The topological type of $\gamma$ is then determined the punctures it
bounds. Thus, up to homeomorphism, $\gamma$ is as in the example on Figure 21
(right). Then, there exist finite sets $N$ and $O$ of non-separating and outer
curves respectively, that uniquely determine $\gamma$. See Figure 21 for an
example. The first part of this proof implies that there exists $k\geq 0$ such
that $N\cup O\subset\mathscr{Y}(S)^{k}$. Thus, $\gamma\in\mathscr{Y}(S)^{k+1}$
and therefore every curve on $S$ that is neither an outer curve nor a non-
separating curve, is also an element of $\mathscr{Y}(S)^{\omega}$.

Figure 21: On the left, the set $N$ (red) and the set $O$ (green); on the
right the non-outer separating curve they uniquely determine.

∎

Similarly to the previous section, we prove Lemma 3.2 by proving the following
claims:
Claim 1: $\tau_{\mathscr{C}}^{\pm
1}(\mathscr{C}\cup\mathscr{D})\subset\mathscr{Y}(S)^{3}$.
Claim 2: $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{A})\subset\mathscr{Y}(S)^{6}$.
Claim 3: $\eta_{\mathscr{D}}^{\pm
1}(\mathscr{C}\cup\mathscr{D})\subset\mathscr{Y}(S)^{1}$.
Claim 4: $\eta_{\mathscr{D}}^{\pm 1}(\mathscr{A})\subset\mathscr{Y}(S)^{3}$.

###  3.4 Proof of Claim 1: $\tau_{\mathscr{C}}^{\pm
1}(\mathscr{C}\cup\mathscr{D})\subset\mathscr{Y}(S)^{3}$

We divide this proof into two parts:

First part $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{D})\subset\mathscr{Y}(S)^{2}$:
Note that $\tau_{\alpha}(\beta)=\beta$ if $\alpha$ and $\beta$ are disjoint
from each other. So, we only need to prove that $\tau_{\alpha_{0}^{i}}^{\pm
1}(\beta_{i})\in\mathscr{Y}(S)^{2}$; to prove this we need some auxiliary
curves.

For each $0\leq i<n$ we define the following curves (recall the subindices are
modulo $n$)

$\gamma_{i}^{+}\displaystyle\mathrel{\mathop{:}}=\langle\\{\beta_{i+1},\ldots,\beta_{i-2}\\}\cup\\{\varepsilon_{i}^{+}\\}\cup\\{\alpha_{1}\\}\rangle\in\mathscr{Y}(S)^{1},$
$\gamma_{i}^{-}\displaystyle\mathrel{\mathop{:}}=\langle\\{\beta_{i+1},\ldots,\beta_{i-2}\\}\cup\\{\varepsilon_{i}^{-}\\}\cup\\{\alpha_{1}\\}\rangle\in\mathscr{Y}(S)^{1}.$

See Figure 22 for examples.

$i$ [t] at 206 127 $i$ [t] at 683 127 $\gamma_{i}^{+}$ [l] at 884 65  $i$ [b]
at 206 147 $i$ [b] at 683 147 $\gamma_{i}^{-}$ [l] at 884 195

Figure 22: On the left, the curves needed to uniquely determine the curves
$\gamma_{i}^{+}$ and $\gamma_{i}^{-}$.

Now, we have the following (see Figure 23):

$\tau_{\alpha_{0}^{i}}(\beta_{i})=\langle\\{\gamma_{i}^{-},\gamma_{i+1}^{+}\\}\cup\\{\alpha_{0}^{i+1},\ldots,\alpha_{0}^{i-1}\\}\rangle\in\mathscr{Y}(S)^{2},$
$\tau_{\alpha_{0}^{i}}^{-1}(\beta_{i})=\langle\\{\gamma_{i}^{+},\gamma_{i+1}^{-}\\}\cup\\{\alpha_{0}^{i+1},\ldots,\alpha_{0}^{i-1}\\}\rangle\in\mathscr{Y}(S)^{2}.$

$i$ [b] at 206 144 $\tau_{\alpha_{0}^{i}}(\beta_{i})$ [b] at 782 145
$i$ [t] at 206 124 $\tau_{\alpha_{0}^{i}}^{-1}(\beta_{i})$ [t] at 782 125

Figure 23: On the left, the curves needed to uniquely determine the curves on
the right $\tau_{\alpha_{0}^{i}}(\beta_{i})$ (top) and
$\tau_{\alpha_{0}^{i}}^{-1}(\beta_{i})$ (bottom).

Thus, $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{D})\subset\mathscr{Y}(S)^{2}$.

Second part $\tau_{\mathscr{C}}^{\pm
1}(\mathscr{C})\subset\mathscr{Y}(S)^{3}$: Recalling Remark 3.1 and that
$\tau_{\alpha}(\beta)=\beta$ if $\alpha$ and $\beta$ are disjoint from each
other, we only need to prove that $\tau_{\alpha_{0}^{i}}^{\pm
1}(\alpha_{1})\in\mathscr{Y}(S)^{3}$ for each $0\leq i<n$.

Note that $\alpha_{1}=\langle\mathscr{D}\rangle$ (see Figure 24), thus
$\tau_{\alpha_{0}^{i}}^{\pm 1}(\alpha_{1})=\langle\tau_{\alpha_{0}^{i}}^{\pm
1}(\mathscr{D})\rangle$. This implies that $\tau_{\alpha_{0}^{i}}^{\pm
1}(\alpha_{1})\in\mathscr{Y}(S)^{3}$ for each $0\leq i<n$.

Figure 24: On the left, the set $\mathscr{D}$; on the right the curve
$\alpha_{1}$.

Therefore, $\tau_{\mathscr{C}}^{\pm
1}(\mathscr{C}\cup\mathscr{D})\subset\mathscr{Y}(S)^{3}$.

###  3.5 Proof of Claim 2: $\tau_{\mathscr{C}}^{\pm
1}(\mathscr{A})\subset\mathscr{Y}(S)^{6}$

As was done in the previous subsection, we only need to prove that for each
$0\leq i<n$ we have that $\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\varepsilon_{i}^{\pm}),\tau_{\alpha_{0}^{i}}^{\pm
1}(\varepsilon_{i}^{\pm})\in\mathscr{Y}(S)^{6}$, since any other curve in
$\mathscr{C}$ is disjoint from both $\varepsilon_{i}^{+}$ and
$\varepsilon_{i}^{-}$.

To do this, we divide the proof into two parts, according to the number of
punctures in $S$.

First part, $n=3$: Since we do not have enough punctures for this part, we
uniquely determine $\varepsilon_{i}^{\pm}$ by using the curves
$\gamma_{i}^{\pm}$ from Subsection 3.4 twice.

In the case for $\alpha_{0}^{i-1}$ recall and check that:

$\gamma_{i+1}^{\pm}:=\langle\\{\beta_{i+2}\\}\cup\\{\varepsilon_{i+1}^{\pm}\\}\cup\\{\alpha_{1}\\}\rangle,\text{
(see Figure \ref{fig:Sec3-4Fig1y2})}$
$\gamma_{i}^{\pm}=\langle\\{\beta_{i+1}\\}\cup\\{\alpha_{1}\\}\cup\\{\gamma_{i+1}^{\mp}\\}\rangle,\text{
(see Figure \ref{fig:Sec3-5Fig1y2} top)}$
$\varepsilon_{i}^{\pm}=\langle\\{\alpha_{0}^{i-2}\\}\cup\\{\alpha_{1}\\}\cup\\{\gamma_{i}^{\pm}\\}\rangle.\text{
(see Figure \ref{fig:Sec3-5Fig1y2} bottom)}$

$i$ [tr] at 200 130
$i$ [tr] at 200 130

Figure 25: On the left, the curves needed to uniquely determine the curves
$\gamma_{i}^{+}$ (top) and $\varepsilon_{i}^{+}$ (bottom) on the right.

Thus, by Claim 1 we have that

$\begin{array}[]{rl}\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\gamma_{i+1}^{\pm})&=\langle\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\\{\beta_{i+2}\\})\cup\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\\{\varepsilon_{i+1}^{\pm}\\})\cup\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\\{\alpha_{1}\\})\rangle\\\ &=\langle\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\\{\beta_{i+2}\\})\cup\\{\varepsilon_{i+1}^{\pm}\\}\cup\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\\{\alpha_{1}\\})\rangle\in\mathscr{Y}(S)^{4},\end{array}$

hence

$\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\gamma_{i}^{\pm})=\langle\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\\{\beta_{i+1}\\})\cup\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\\{\alpha_{1}\\})\cup\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\\{\gamma_{i+1}^{\mp})\\}\rangle\in\mathscr{Y}(S)^{5},$

and finally

$\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\varepsilon_{i}^{\pm})=\langle\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\\{\alpha_{0}^{i-2}\\})\cup\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\\{\alpha_{1}\\})\cup\tau_{\alpha_{0}^{i-1}}^{\pm
1}(\\{\gamma_{i}^{\pm}\\})\rangle\in\mathscr{Y}(S)^{6}.$

The case for $\alpha_{0}^{i}$ is completely analogous, substituting
$\gamma_{i+1}^{\pm}$ by $\gamma_{i-1}^{\pm}$.

Second part, $n\geq 4$: We proceed similarly to the first case in Section 2.6
to prove that $\tau_{\alpha}(\varepsilon_{i}^{\pm})\in\mathscr{Y}(S)^{4}$ with
$\alpha=\alpha_{0}^{i-1},\alpha_{0}^{i}$. We can uniquely determine
$\varepsilon_{i}^{\pm}$ as follows: We define the set

$C=\\{\alpha_{0}^{i+1},\ldots,\alpha_{0}^{i-2}\\}\cup\\{\alpha_{1}\\},$

and the curve

$\varepsilon^{\pm}=\left\\{\begin{array}[]{cl}\varepsilon_{i+1}^{\pm}&\text{
for the case $\alpha_{0}^{i-1}$,}\\\ \varepsilon_{i-1}^{\pm}&\text{ for the
case $\alpha_{0}^{i}$.}\end{array}\right.$

Then we check that (see Figure 26)

$\varepsilon_{i}^{\pm}=\langle C\cup\\{\varepsilon^{\mp}\\}\rangle.$

Figure 26: On the left, the curves needed to uniquely determine
$\varepsilon_{i}^{+}$ on the right, for the cases $\alpha_{0}^{i-1}$ (top) and
$\alpha_{0}^{i}$ (bottom).

Thus, by Claim 1 we have that

$\begin{array}[]{rl}\tau_{\alpha_{0}^{i-1}}(\varepsilon_{i}^{\pm})&=\langle\tau_{\alpha_{0}^{i-1}}(C)\cup\tau_{\alpha_{0}^{i-1}}(\\{\varepsilon_{i+1}^{\mp}\\})\rangle\\\
&=\langle\tau_{\alpha_{0}^{i-1}}(C)\cup\\{\varepsilon_{i+1}^{\mp}\\}\rangle\in\mathscr{Y}(S)^{4},\end{array}$
$\begin{array}[]{rl}\tau_{\alpha_{0}^{i}}(\varepsilon_{i}^{\pm})&=\langle\tau_{\alpha_{0}^{i}}(C)\cup\tau_{\alpha_{0}^{i}}(\\{\varepsilon_{j}\\})\rangle\\\
&=\langle\tau_{\alpha_{0}^{i}}(C)\cup\\{\varepsilon_{j}\\}\rangle\in\mathscr{Y}(S)^{4}.\end{array}$

The cases for
$\tau_{\alpha_{0}^{i-1}}^{-1}(\varepsilon_{i}^{\pm}),\tau_{\alpha_{0}^{i}}^{-1}(\varepsilon_{i}^{\pm})\in\mathscr{Y}(S)^{4}$
are analogous.

Therefore $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{A})\subset\mathscr{Y}(S)^{6}$.

###  3.6 Proof of Claim 3: $\eta_{\mathscr{D}}^{\pm
1}(\mathscr{C}\cup\mathscr{D})\subset\mathscr{Y}(S)^{1}$

Recall that $\varepsilon_{i}^{\pm}=\eta_{\beta_{i}}^{\pm 1}(\beta_{i-1})$.
Using this and Remark 2.1 we have that $\eta_{\mathscr{D}}^{\pm
1}(\mathscr{D})\subset\mathscr{Y}(S)$.

Now, similarly to the previous subsections, we only need to prove that
$\eta_{\beta_{i}}^{\pm 1}(\alpha_{0}^{i})\in\mathscr{Y}(S)^{1}$. This can be
done as follows: We define a set $B\subset\mathscr{D}$ as

$B=\left\\{\begin{array}[]{cl}\varnothing&\text{ if $n=3$,}\\\
\\{\beta_{i+2},\ldots,\beta_{i-2}\\}&\text{if $n\geq 4$}.\end{array}\right.$

Then we have (see Figure 27)

$\eta_{\beta_{i}}(\alpha_{0}^{i})=\langle\\{\varepsilon_{i}^{+},\varepsilon_{i+1}^{-}\\}\cup
B\cup\\{\alpha_{0}^{i+1},\ldots,\alpha_{0}^{i-1}\\}\rangle,$
$\eta_{\beta_{i}}^{-1}(\alpha_{0}^{i})=\langle\\{\varepsilon_{i}^{-},\varepsilon_{i+1}^{+}\\}\cup
B\cup\\{\alpha_{0}^{i+1},\ldots,\alpha_{0}^{i-1}\\}\rangle.$

$i$ at 167 115 $i$ at 645 155
$i$ at 167 155 $i$ at 640 115

Figure 27: On the left, the curves needed to uniquely determine the curves
$\eta_{\beta_{i}}(\alpha_{0}^{i})$ (top) and
$\eta_{\beta_{i}}^{-1}(\alpha_{0}^{i})$ (bottom) on the right.

Therefore, $\eta_{\mathscr{D}}^{\pm
1}(\mathscr{C}\cup\mathscr{D})\subset\mathscr{Y}(S)^{1}$.

###  3.7 Proof of Claim 4: $\eta_{\mathscr{D}}^{\pm
1}(\mathscr{A})\subset\mathscr{Y}(S)^{3}$

For this claim, similarly to the previous subsections, we only need to prove
for $\varepsilon=\varepsilon_{i}^{+},\varepsilon_{i}^{-}$, that
$\eta_{\beta_{i-2}}^{\pm 1}(\varepsilon),\eta_{\beta_{i-1}}^{\pm
1}(\varepsilon),\eta_{\beta_{i}}^{\pm 1}(\varepsilon),\eta_{\beta_{i+1}}^{\pm
1}(\varepsilon)\in\mathscr{Y}(S)^{3}$.

First note that by definition
$\eta_{\beta_{i-1}}(\varepsilon_{i}^{+})=\eta_{\beta_{i-1}}^{-1}(\varepsilon_{i}^{-})=\beta_{i}\in\mathscr{Y}(S)$
and
$\eta_{\beta_{i}}^{-1}(\varepsilon_{i}^{+})=\eta_{\beta_{i}}(\varepsilon_{i}^{-})=\beta_{i-1}\in\mathscr{Y}(S)$.

Also note that we can uniquely determine $\varepsilon_{i}^{\pm}$ as in
Subsection 3.5 as follows (see Figure 28):

$\varepsilon_{i}^{\pm}=\langle\\{\alpha_{1}\\}\cup\\{\alpha_{0}^{i+1},\ldots,\alpha_{0}^{i-2}\\}\cup\\{\gamma_{i}^{\pm}\\}\rangle,$

hence, by Claim 3 and using that $\eta_{\beta_{i-2}}^{\pm
1}(\gamma_{i}^{+})=\eta_{\beta_{i+1}}^{\pm 1}(\gamma_{i}^{+})=\gamma_{i}^{+}$
and $\eta_{\beta_{i-2}}^{\pm 1}(\gamma_{i}^{-})=\eta_{\beta_{i+1}}^{\pm
1}(\gamma_{i}^{-})=\gamma_{i}^{-}$, we get that $\eta_{\beta_{i-2}}^{\pm
1}(\varepsilon_{i}^{+}),\eta_{\beta_{i-2}}^{\pm
1}(\varepsilon_{i}^{-}),\eta_{\beta_{i+1}}^{\pm
1}(\varepsilon_{i}^{+}),\eta_{\beta_{i+1}}^{\pm
1}(\varepsilon_{i}^{-})\in\mathscr{Y}(S)^{2}.$

$i$ [t] at 87 125 $i$ [t] at 563 125

Figure 28: On the left, the curve needed to uniquely determine the curve
$\varepsilon_{i}^{+}$ on the right. The case for $\varepsilon_{i}^{-}$ is
analogous.

Thus we only need to prove that
$\eta_{\beta_{i-1}}^{-1}(\varepsilon_{i}^{+})$,
$\eta_{\beta_{i}}(\varepsilon_{i}^{+})$,
$\eta_{\beta_{i-1}}(\varepsilon_{i}^{-})$,
$\eta_{\beta_{i}}^{-1}(\varepsilon_{i}^{-})\in\mathscr{Y}(S)^{3}$.

All of these cases can be uniquely determined by an auxiliary curve
$\delta\in\mathscr{Y}(S)^{2}$ (a different one for each case) and the set
$\mathscr{C}\backslash\\{\alpha_{0}^{i-1},\alpha_{0}^{i}\\}$. See Figure 29
for an example.

$i$ [br] at 194 120 $i$ [br] at 675 100

Figure 29: An example of the curve
$\eta_{\beta_{i-1}}^{-1}(\varepsilon_{i}^{+})$ (right) being uniquely
determined by the set
$\mathscr{C}\backslash\\{\alpha_{0}^{i-1},\alpha_{0}^{i}\\}$ and the curve
$\delta$ (left).

The subsets of $\mathscr{Y}(S)^{1}$ used to uniquely determine $\delta$ are
different for each case and again we make use of the curves of the form
$\gamma_{j}^{\pm}$ from the first part of Subsection 3.4, but they are
completely analogous to each other and can be inferred from the following
examples and Figure 30:

  * •

If $n=3$, then we use the following curve to obtain
$\eta_{\beta_{i-1}}^{-1}(\varepsilon_{i}^{+})$:

$\delta\displaystyle\mathrel{\mathop{:}}=\langle\\{\alpha_{1}\\}\cup\\{\varepsilon_{i-1}^{-}\\}\cup\\{\gamma_{i+1}^{-}\\}\rangle.$

  * •

If $n\geq 4$, then we use the following curve to obtain
$\eta_{\beta_{i-1}}^{-1}(\varepsilon_{i}^{+})$:

$\delta\displaystyle\mathrel{\mathop{:}}=\langle\\{\alpha_{1}\\}\cup\\{\varepsilon_{i-1}^{-}\\}\cup\\{\beta_{i+1},\ldots,\beta_{i-3}\\}\cup\\{\gamma_{i+1}^{-}\\}\rangle.$

$i$ [r] at 200 135 $i$ [r] at 675 135
$i$ [r] at 158 130 $i$ [r] at 635 135

Figure 30: On the left, the curves needed to uniquely determine the curve
$\delta$ on the right.

Then, $\eta_{\mathscr{D}}^{\pm 1}(\mathscr{A})\subset\mathscr{Y}(S)^{3}$

###  3.8 Proof of Theorem B

For the purposes of this work, we define the set $\mathfrak{X}(S)$ as the set
$\mathfrak{X}_{1}$ from Section 6 in [2], which was proved to be rigid (see
Proposition 6.2 in [2]). For this definition we need the curves
$\gamma_{i}^{\pm}$ from Subsection 3.4 and the following auxiliary curves.

If $n=3$, for each $0\leq i<n$ we define the following curves
$\gamma_{i,i+1}^{+}\displaystyle\mathrel{\mathop{:}}=\gamma_{i+2}^{-}$ and
$\gamma_{i,i+1}^{-}\displaystyle\mathrel{\mathop{:}}=\gamma_{i+2}^{+}$.

If $n\geq 4$, for each $0\leq i<n$ we define the following curves (see Figure
31):

$\gamma_{i,i+1}^{+}\displaystyle\mathrel{\mathop{:}}=\langle\\{\alpha_{1}\\}\cup\\{\gamma_{i}^{+},\gamma_{i+1}^{+},\gamma_{i+2}^{-}\\}\cup\\{\beta_{i+2},\ldots,\beta_{i-2}\\}\rangle,$
$\gamma_{i,i+1}^{-}\displaystyle\mathrel{\mathop{:}}=\langle\\{\alpha_{1}\\}\cup\\{\gamma_{i}^{-},\gamma_{i+1}^{-},\gamma_{i+2}^{-}\\}\cup\\{\beta_{i+2},\ldots,\beta_{i-2}\\}\rangle.$

$i$ [t] at 165 130 $i$ [t] at 642 130
$i$ [b] at 167 145 $i$ [b] at 642 145

Figure 31: On the left, the curves needed to uniquely determine the curves
$\gamma_{i,i+1}^{+}$ (top) and $\gamma_{i,i+1}^{-}$ (bottom) on the right.

According to the notation above, we can define $\mathfrak{X}(S)$ as follows:

$\mathfrak{X}(S)\displaystyle\mathrel{\mathop{:}}=(\mathscr{C}\cup\mathscr{D})^{1}\cup\\{\gamma_{i}^{\pm}\\}_{i=0}^{n-1}\cup\\{\gamma_{i,i+1}^{\pm}\\}_{i=0}^{n-1}.$

###### Proof of Theorem B.

For this proof, we proceed as in Subsection 2.8, so we only need to prove that
$\mathscr{Y}(S)\subset\mathfrak{X}(S)^{1}$.

Given that $\mathscr{C}\cup\mathscr{D}\subset\mathfrak{X}(S)$, we only need to
prove that $\mathscr{A}\subset\mathfrak{X}(S)^{1}$. For this, as in Subsection
3.7 note that for each $0\leq i<n$ we have that (see Figure 28):

$\varepsilon_{i}^{+}=\langle\\{\gamma_{i}^{+}\\}\cup(\mathscr{C}\backslash\\{\alpha_{i-1},\alpha_{i}\\})\rangle,$
$\varepsilon_{i}^{-}=\langle\\{\gamma_{i}^{-}\\}\cup(\mathscr{C}\backslash\\{\alpha_{i-1},\alpha_{i}\\})\rangle.$

Thus, $\mathscr{Y}(S)\subset\mathfrak{X}(S)^{1}$, and therefore
$\mathfrak{X}(S)^{\omega}=\mathcal{C}(S)$. ∎

##  4 Genus two case

Given that $\mathcal{C}(S_{2,0})$ is isomorphic to $\mathcal{C}(S_{0,6})$,
Subsections 2.3 and 2.8 prove Theorems A and B (respectively) for $S_{2,0}$.
Thus, in this section we assume that $S=S_{2,n}$ with $n\geq 1$ (then
$\kappa(S)\geq 4$).

The structure of this section is as follows: In Subsection 4.1 we define
$\mathscr{C}$, $\mathscr{B}$ and $\mathscr{Y}(S)$; in Subsection 4.2 we define
the sets of auxiliary curves $\mathscr{D}$ used for the proof of Theorem A; in
Subsection 4.3 we give the proof of Theorem A pending the proof of a key lemma
(Lemma 4.1); in Subsection 4.4 we define the auxiliary curves $\mathscr{E}$
and $\mathscr{Z}$ used to prove the key lemma; in Subsections 4.5, 4.6, 4.7
and 4.8, we give the proof of the key lemma; finally, in Subsection 4.9 we
recall from [2] the definition of $\mathfrak{X}(S)$ and prove Theorem B.

###  4.1 The definition of $\mathscr{Y}(S)$

Let $C=\\{\gamma_{1},\ldots,\gamma_{k}\\}$ be a set of curves for $k>1$. We
say $C$ is a chain if $i(\gamma_{i},\gamma_{j})=1$ for $|i-j|=1$, and
$\gamma_{i}$ is disjoint from $\gamma_{j}$ otherwise. Note the similarities
with an outer chain (defined in Subsection 2.1). We define the length of $C$
as its cardinality. See Figure 32 for an example.

Figure 32: On the left, a chain of length 4; on the right, a chain of length 3
and its corresponding bounding pair.

Note that if a chain $C$ has length $2j$, any (open) regular neighbourhood of
$C$ is homeomorphic to $S_{j,1}$, while if $C$ has length $2j+1$, then any
(open) regular neighbourhood of $C$ is homeomorphic to $S_{j,2}$.

Given a chain $C$ of odd length, the bounding pair associated to $C$ is
defined as the set containing the boundary curves of a closed regular
neighbourhood of $C$. See Figure 32 for an example.

Let
$\mathscr{C}_{0}\displaystyle\mathrel{\mathop{:}}=\\{\alpha_{1},\ldots,\alpha_{5}\\}$
be the chain depicted in Figure 33, and let
$\mathscr{C}_{f}\displaystyle\mathrel{\mathop{:}}=\\{\alpha_{0}^{0},\ldots,\alpha_{0}^{n}\\}$
be the multicurve also depicted in Figure 33. Note that for all $0\leq i\leq
n$ we have that $i(\alpha_{0}^{i},\alpha_{j})=1$ if $j=1,5$, and
$\alpha_{0}^{i}$ is disjoint from $\alpha_{j}$ otherwise. Then, we define the
set
$\mathscr{C}\displaystyle\mathrel{\mathop{:}}=\mathscr{C}_{0}\cup\mathscr{C}_{f}$.

$\alpha_{1}$ [bl] at 165 155 $\alpha_{2}$ [b] at 250 135 $\alpha_{3}$ [bl] at
305 155 $\alpha_{4}$ [bl] at 400 125 $\alpha_{5}$ [tl] at 110 238
$\alpha_{0}^{0}$ [l] at 670 215 $\alpha_{0}^{1}$ [bl] at 515 205
$\alpha_{0}^{2}$ [bl] at 490 130 $\alpha_{0}^{3}$ [tl] at 512 65
$\alpha_{0}^{4}$ [l] at 670 25

Figure 33: The curves in $\mathscr{C}_{0}$ on the right and the curves in
$\mathscr{C}_{f}$ on the left.

We can then fix several subsurfaces of $S$ using $\mathscr{C}$ as follows (see
Figure 34):

  * •

Let $\Sigma_{o}^{+}$ be the closed subsurface of $S$ bounded by
$\\{\alpha_{1},\alpha_{3},\alpha_{5}\\}$ that is not compact, and
$\Sigma_{o}^{-}$ be the compact subsurface of $S$ bounded by
$\\{\alpha_{1},\alpha_{3},\alpha_{5}\\}$.

  * •

Define $S_{0}^{+}$ as the compact subsurface of $S$ bounded by
$\\{\alpha_{0}^{0},\alpha_{2},\alpha_{4}\\}$, and $S_{0}^{-}$ as the
noncompact subsurface of $S$ bounded by
$\\{\alpha_{0}^{0},\alpha_{2},\alpha_{4}\\}$.

  * •

For each $0<i\leq n$, let $S_{i}^{+}$ be the closed subsurface of $S$ bounded
by $\\{\alpha_{0}^{i},\alpha_{2},\alpha_{4}\\}$ that contains $S_{0}^{+}$. Let
$S_{0}^{-}$ be the closed subsurface of $S$ bounded by
$\\{\alpha_{0}^{i},\alpha_{2},\alpha_{4}\\}$ that does not contain
$S_{0}^{+}$.

$\Sigma_{o}^{+}$ at 250 165 $\Sigma_{o}^{-}$ [bl] at 367 220 $S_{0}^{+}$ at
780 200 $S_{0}^{-}$ at 735 60
$S_{2}^{+}$ at 250 190 $S_{2}^{-}$ at 250 70

Figure 34: On the top left, the subsurfaces $\Sigma_{o}^{+}$ (punctured pair
of pants at the front) and $\Sigma_{o}^{-}$ (pair of pants at the back); on
the top right, the subsurfaces $S_{0}^{+}$ (pair of pants at the top right)
and $S_{0}^{-}$ (punctured pair of pants at the bottom); on the bottom, the
subsurfaces $S_{i}^{+}$ (punctured pair of pants at the top) and $S_{i}^{-}$
(possibly punctured pair of pants at the bottom).

Let $C\subset\mathscr{C}$ be a chain of odd length such that
$|C\cap\mathscr{C}_{f}|\leq 1$, and let $B(C)$ be the bounding pair associated
to $C$. Then, note that the elements of $B(C)$ are either contained in
$\mathrm{int}(S_{i}^{+})\cup\mathrm{int}(S_{i}^{-})$ for some $i$, or
contained in $\mathrm{int}(\Sigma_{o}^{+})\cup\mathrm{int}(\Sigma_{o}^{-})$.
We denote by $\beta_{C}^{+}$ the element of $B(C)$ contained in either
$\mathrm{int}(S_{i}^{+})$ for some $i$ or $\Sigma_{o}^{+}$; we denote by
$\beta_{C}^{-}$ the element of $B(C)$ contained in either
$\mathrm{int}(S_{i}^{-})$ for some $i$ or $\Sigma_{o}^{-}$. See Figure 35.

Figure 35: On the left, the set $C=\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}$ and
the curves $\beta_{C}^{+}$ (front) and $\beta_{C}^{-}$ (back); on the right,
the set $C=\\{\alpha_{2},\alpha_{3},\alpha_{4}\\}$ and the curves
$\beta_{C}^{+}$ (top) and $\beta_{C}^{-}$ (bottom).

We define the set
$\displaystyle\mathscr{B}\displaystyle\mathrel{\mathop{:}}=\\{\beta_{C}^{+},\beta_{C}^{-}:C\subset\mathscr{C}$
is a chain such that $|C\cap\mathscr{C}_{f}|\leq 1\\}$.

Note that since we are considering only chains that contain at most one
element of $\mathscr{C}_{f}$, then every curve in $\mathscr{B}$ is non-
separating.

Finally, we define

$\mathscr{Y}(S)\displaystyle\mathrel{\mathop{:}}=\mathscr{C}\cup\mathscr{B}.$

###  4.2 Auxiliary curves for Theorem A

As a difference with the previous sections, here we use the auxiliary curves
for the key lemma and the proof of Theorem A, instead of using them for the
definition of $\mathscr{Y}(S)$.

For each $0<i<n$, we define the following outer curves (see Figure 36):

$\delta_{i}=\langle\mathscr{C}\backslash\\{\alpha_{0}^{i}\\}\rangle\in\mathscr{C}^{1}.$

Then, we define the following curve (see Figure 36):

$\Delta=\langle\mathscr{C}_{0}\cup\\{\delta_{j}\\}_{j=1}^{n-1}\rangle\in\mathscr{C}^{2}.$

Figure 36: On the left, the curves needed to uniquely determine the curves
$\delta_{1}$ (top) and $\Delta$ (bottom).

With this, we can define the following set

$\mathscr{D}\displaystyle\mathrel{\mathop{:}}=\\{\delta_{i}:0<i<n\\}\cup\\{\Delta\\}\subset\mathscr{C}^{2}$

Note that if $n=1$, then $\mathscr{D}=\varnothing$, and if $n=2$ then
$\mathscr{D}=\\{\delta_{1}\\}$.

###  4.3 Proof of Theorem A

Similarly to the previous sections, let
$\mathscr{G}\displaystyle\mathrel{\mathop{:}}=\\{\tau_{\alpha}^{\pm
1}:\alpha\in(\mathscr{C}\backslash\\{\alpha_{5}\\})\\}\cup\\{\eta_{\delta_{i}}^{\pm
1}:0<i<n\\}$. It is a well-known fact (see Corollary 4.15 in [4]) that
$\mathscr{G}$ is a symmetric generating set of $\mathrm{Mod}(S)$ (often called
the Humphries-Lickorish generating set). As before, if $A$ is a set of curves
on $S$, we denote by $\mathscr{G}\cdot A$ the following set $\mathscr{G}\cdot
A\displaystyle\mathrel{\mathop{:}}=\tau_{(\mathscr{C}\backslash\\{\alpha_{5}\\})}^{\pm
1}(A)\cup\eta_{\\{\delta_{i}:0<i<n\\}}^{\pm 1}(A)$.

Now, we state the key lemma of this section, whose proof we leave to the
following subections (see Subsections 4.5, 4.6, 4.7 and 4.8)

######  Lemma 4.1.

Let $\mathscr{G}$ be as above. Then
$\mathscr{G}\cdot\mathscr{Y}(S)\subset\mathscr{Y}(S)^{11}$.

Recalling Proposition 3.7 in [7] (restated for our purposes), we obtain an
immediate consequence.

######  Proposition 4.2 (3.7 in [7]).

Let $h\in\mathrm{Aut}(\mathcal{C}(S))$ and $Y\subset\mathcal{C}(S)$. If
$h(Y)\subset\mathscr{Y}(S)^{k}$ for some $k\geq 0$, then
$h(Y^{m})\subset\mathscr{Y}(S)^{k+m}$ for all $m\geq 0$.

######  Corollary 4.3.

For $\mathscr{G}$ and $\mathscr{D}$ as above, we have that
$\mathscr{G}\cdot(\mathscr{Y}(S)\cup\mathscr{D})\subset\mathscr{Y}(S)^{13}$.

###### Proof of Theorem A.

The proof of Theorem A for this case is the same as the one for the case
$g=1$, simply substituting “Lemma 3.2” for “Corollary 4.3”, “Figure 21” for
“Figure 37”, and fixing the rigid expansions to multiples of $13$ instead of
$6$. ∎

Figure 37: On the left, the curves needed to uniquely determine the non-outer
separating curves on the right.

As before, we divide the proof of Lemma 4.1 into four claims:
Claim 1: $\tau_{(\mathscr{C}\backslash\\{\alpha_{5}\\})}^{\pm
1}(\mathscr{C})\subset\mathscr{Y}(S)^{8}$.
Claim 2: $\tau_{(\mathscr{C}\backslash\\{\alpha_{5}\\})}^{\pm
1}(\mathscr{B})\subset\mathscr{Y}(S)^{11}$.
Claim 3: $\eta_{\\{\delta_{i}:0<i<n\\}}^{\pm
1}(\mathscr{C})\subset\mathscr{Y}(S)^{3}$.
Claim 4: $\eta_{\\{\delta_{i}:0<i<n\\}}^{\pm
1}(\mathscr{B})\subset\mathscr{Y}(S)^{6}$.
In the following section we define the auxiliary curves used to prove these
claims.

###  4.4 Auxiliary curves for the claims

There are two types of auxiliary curves that are used throughout this section.

The first type of curves (which are only defined for surfaces with $n\geq 2$
punctures) are defined for each $k=2,4$ as follows (see Figure 38):

$\varepsilon_{k}\displaystyle\mathrel{\mathop{:}}=\langle\\{\alpha_{0}^{i}\\}_{i=1}^{n-1}\cup(\mathscr{C}_{0}\backslash\\{\alpha_{k}\\})\rangle\in\mathscr{C}^{1}.$

Figure 38: On the left, the curves needed to uniquely determine the curves
$\varepsilon_{2}$ (top) and $\varepsilon_{4}$, on the right.

To define the second type of curve we must first define for each $0\leq j\leq
n$ the following chain:

$C_{j}\displaystyle\mathrel{\mathop{:}}=\\{\alpha_{0}^{j},\alpha_{1},\alpha_{2}\\}.$

Now, the second type of curve is defined for $i=0$ as follows:

$\zeta^{0}\displaystyle\mathrel{\mathop{:}}=\langle\\{\alpha_{1},\alpha_{3},\alpha_{5}\\}\cup\\{\beta_{C_{j}}^{+}:0<j\leq
n\\}\rangle,$

for each $0<i<n+1$ as follows:

$\zeta^{i}=\langle\\{\alpha_{1},\alpha_{3},\alpha_{4},\alpha_{5}\\}\cup\\{\beta_{C_{j}}^{+}:j<i\\}\cup\\{\beta_{C_{k}}^{-}:i\leq
k\\}\rangle,$

and finally for $i=n+1$ as follows:

$\zeta^{n+1}\displaystyle\mathrel{\mathop{:}}=\langle\\{\alpha_{1},\alpha_{3},\alpha_{5}\\}\cup\\{\beta_{C_{j}}^{-}:0\leq
j<n\\}\rangle.$

See Figure 39 for examples.

Figure 39: On the left, the curves needed to uniquely determine the curves
$\zeta^{0}$ (top), $\zeta^{1}$ (upper middle), $\zeta^{3}$ (lower middle) and
$\zeta^{n+1}$, on the right.

Note that for all $0\leq i\leq n+1$ we have that
$\zeta^{i}\in\mathscr{Y}(S)^{1}$.

Finally, we define the sets

$\mathscr{E}:=\left\\{\begin{array}[]{cl}\mathscr{D}&\text{if $n=1$},\\\
\mathscr{D}\cup\\{\varepsilon_{2},\varepsilon_{4}\\}&\end{array}\right.$

(thus $\mathscr{E}\subset\mathscr{C}^{2}$), and

$\mathscr{Z}:=\\{\zeta^{j}\\}_{j=0}^{n+1}$

(thus $\mathscr{Z}\subset\mathscr{Y}(S)^{1}$).

###  4.5 Proof of Claim 1:
$\tau_{(\mathscr{C}\backslash\\{\alpha_{5}\\})}^{\pm
1}(\mathscr{C})\subset\mathscr{Y}(S)^{8}$

Given that $S$ has genus 2, the strategy used in [7] for genus at least 3
(find two “standard” cases of the claim and then use the action of a subgroup
of $\mathrm{Mod}(S)$) is actually more complicated than listing exactly each
case of $\tau_{\alpha}^{\pm 1}(\alpha^{\prime})$ for
$\alpha,\alpha^{\prime}\in\mathscr{C}$ (there is too little space in a genus 2
surface). Also, using the fact that $\tau_{\alpha}^{\pm
1}(\alpha^{\prime})=\alpha^{\prime}$ if and only if
$i(\alpha,\alpha^{\prime})=0$, and recalling Remark 3.1, we only need to prove
that $\tau_{\alpha_{i}}^{\pm 1}(\alpha_{i-1})\in\mathscr{Y}(S)^{k}$ for $0\leq
i\leq 5$ (modulo 6), where $\alpha_{0}=\alpha_{0}^{i}$ for each case of $0\leq
i\leq n$ and $k\leq 8$.

The general strategy to prove that
$\tau_{\alpha_{i}}(\alpha_{i-1})\in\mathscr{Y}(S)^{8}$ (and
$\tau_{\alpha_{i}}^{-1}(\alpha_{i-1})\in\mathscr{Y}(S)^{8}$) is the following:

  * •

For $n=1$, we first prove that $\tau^{\pm
1}_{\alpha_{5}}(\alpha_{4})\in\mathscr{Y}(S)^{2}$. We do this is a similar
fashion as in [7]. Then, for $i<5$ we use elements in $\mathscr{Y}(S)$ to fill
the complement of the one-holed torus induced by $\alpha_{i}$ and
$\alpha_{i-1}$, and then use $\tau^{\mp 1}_{\alpha_{i+1}}(\alpha_{i})$ to
uniquely determine $\tau^{\pm 1}_{\alpha_{i}}(\alpha_{i-1})$. This yields that
$\tau^{\pm 1}_{\alpha_{i}}(\alpha_{i-1})\in\mathscr{Y}(S)^{7}$ for all $i$.

  * •

For $n\geq 2$, the strategy is almost the same with the exception of needing
particular auxiliary curves and $\mathscr{D}$, hence getting that $\tau^{\pm
1}_{\alpha_{5}}(\alpha_{4})\in\mathscr{Y}(S)^{3}$. This causes that in the end
$\tau^{\pm 1}_{\alpha_{i}}(\alpha_{i-1})\in\mathscr{Y}(S)^{8}$ for all $i$.

######  Remark 4.4.

Note that, since $\mathscr{E}\subset\mathscr{C}^{2}$, due to Proposition 4.2
we have that

$\tau_{(\mathscr{C}\backslash\\{\alpha_{5}\\})}^{\pm
1}(\mathscr{E})\subset\mathscr{Y}(S)^{10}.$

Now, for the purpose of facilitating the exposition to the reader, we
illustrate the case $n=1$ and for the case $n\geq 2$ we introduce the
auxiliary curves and instruct the necessary changes from the previous case.

####  4.5.1 Case $n=1$

First, let $\gamma^{\pm}$ be the following curves (see Figure 40):

$\gamma^{+}=\langle\\{\alpha_{0}^{0},\alpha_{0}^{1},\alpha_{3}\\}\cup\\{\beta_{\\{\alpha_{5},\alpha_{0}^{1},\alpha_{1}\\}}^{+}\\}\rangle\in\mathscr{Y}(S)^{1},$
$\gamma^{-}=\langle\\{\alpha_{0}^{0},\alpha_{0}^{1},\alpha_{3}\\}\cup\\{\beta_{\\{\alpha_{5},\alpha_{0}^{0},\alpha_{1}\\}}^{+}\\}\rangle\in\mathscr{Y}(S)^{1}.$

Figure 40: On the left, the curves needed to uniquely determine the curves
$\gamma^{+}$ (top) and $\gamma^{-}$ (bottom), on the right.

With these curves we get (see Figure 41)

$\tau^{\pm
1}_{\alpha_{5}}(\alpha_{4})=\langle\\{\alpha_{1},\alpha_{2}\\}\cup\\{\beta_{\\{\alpha_{3},\alpha_{4},\alpha_{5}\\}}^{+}\\}\cup\\{\gamma^{\pm}\\}\rangle\in\mathscr{Y}(S)^{2}.$

Figure 41: On the left, the curves needed to uniquely determine the curve
$\tau_{\alpha_{5}}(\alpha_{4})$, on the right.

For $\tau^{\pm 1}_{\alpha_{i}}(\alpha_{i-1})$ with $i<5$, let $A$ be as in
Table 1 and we get

$\tau^{\pm 1}_{\alpha_{i}}(\alpha_{i-1})=\langle A\cup\\{\tau^{\mp
1}_{\alpha_{i+1}}(\alpha_{i})\\}\rangle\in\mathscr{Y}(S)^{j},$

where $i+j=7$ and $\alpha_{0}=\alpha_{0}^{k}$ for each $k=0,1$. See Figure 42
for examples.

i | $A$
---|---
4 | $\\{\alpha_{0}^{0},\alpha_{0}^{1},\alpha_{1}\\}$
3 | $\\{\alpha_{0}^{0},\alpha_{0}^{1},\alpha_{5}\\}$
2 | $\\{\alpha_{4},\alpha_{5},\beta_{\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}}^{+}\\}$
1 |  $\\{\alpha_{3},\alpha_{4},\beta_{\\{\alpha_{5},\alpha_{0}^{0},\alpha_{1}\\}}^{+}\\}$ for $k=0$,
|
$\\{\alpha_{3},\alpha_{4},\beta_{\\{\alpha_{1},\alpha_{0}^{1},\alpha_{5}\\}}^{+}\\}$
for $k=1$,
0 |  $\\{\alpha_{2},\alpha_{3},\beta_{\\{\alpha_{1},\alpha_{0}^{0},\alpha_{5}\\}}^{+}\\}$ for $k=0$,
|
$\\{\alpha_{2},\alpha_{3},\beta_{\\{\alpha_{1},\alpha_{0}^{1},\alpha_{5}\\}}^{+}\\}$
for $k=1$.
Table 1: The set $A$ for uniquely determining $\tau^{\pm
1}_{\alpha_{i}}(\alpha_{i-1})$ for $n=1$.

Figure 42: On the left, the curves needed to uniquely determine the curves
$\tau^{-1}_{\alpha_{4}}(\alpha_{3})$ (top) and $\tau_{\alpha_{3}}(\alpha_{2})$
(bottom), on the right.

####  4.5.2 Case $n\geq 2$

For this case we first define the following auxiliary curves. For let
$\Delta_{1,n}=\Delta\in\mathscr{Y}(S)^{2}$ (see Subsection 4.2); then for
$1<i<n$ we define $\Delta_{1,i}$ and $\Delta_{i,n}$ as in Table 2 (see Figure
43).

$\Delta_{1,2}$ | $\delta_{1}\in\mathscr{C}^{1}$
---|---
$\Delta_{n-1,n}$ | $\delta_{n-1}\in\mathscr{C}^{1}$
$\Delta_{1,i}$ |  $\langle(\mathscr{C}\backslash\\{\alpha_{0}^{1},\ldots,\alpha_{0}^{i-1}\\})\cup\\{\delta_{1},\ldots\delta_{i-1}\\}\rangle\in\mathscr{Y}(S)^{2}$ for $2<i<n-1$,
$\Delta_{i,n}$ |  $\langle(\mathscr{C}\backslash\\{\alpha_{0}^{i},\ldots,\alpha_{0}^{n-1}\\})\cup\\{\delta_{i},\ldots\delta_{n-1}\\}\rangle\in\mathscr{Y}(S)^{2}$ for $2<i<n-1$,
Table 2: The auxiliary curves $\Delta_{1,i}$ and $\Delta_{i,n}$.

Note that $\mathscr{D}_{1,i},\mathscr{D}_{i,n}\subset\mathscr{Y}(S)^{2}$.

Figure 43: On the left, the curves needed to uniquely determine the curves
$\Delta_{1,3}$ (top) and $\Delta_{2,4}$ (bottom).

With these curves, we define the sets

$\mathscr{D}_{1,i}:=\left\\{\begin{array}[]{cl}\varnothing&\text{if }i=1,\\\
\\{\delta_{1}\\}&\text{if }i=2,\\\
\\{\delta_{1},\ldots,\delta_{i-1}\\}\cup\\{\Delta_{1,i}\\}&\text{if }i\geq
3.\end{array}\right.$
$\mathscr{D}_{i,n}:=\left\\{\begin{array}[]{cl}\varnothing&\text{if }i=n,\\\
\\{\delta_{i-1}\\}&\text{if }i=n-1,\\\
\\{\delta_{i},\ldots,\delta_{n-1}\\}\cup\\{\Delta_{i,n}\\}&\text{if }i\leq
n-2.\end{array}\right.$

Using these sets, we define the following auxiliary curves (see Figure 44)

$\beta_{i}^{+}=\langle(\mathscr{C}\backslash\\{\alpha_{0}^{0},\ldots,\alpha_{0}^{i-1},\alpha_{2},\alpha_{4}\\})\cup\mathscr{D}_{1,i}\rangle\in\mathscr{Y}(S)^{3},$
$\beta_{i}^{-}=\langle(\mathscr{C}\backslash\\{\alpha_{0}^{i},\ldots,\alpha_{0}^{n},\alpha_{2},\alpha_{4}\\})\cup\mathscr{D}_{i,n}\rangle\in\mathscr{Y}(S)^{3}.$

Figure 44: On the left, the curves needed to uniquely determine
$\beta_{2}^{+}$, on the right.

Now we can prove the claim for this case. Note that we need only substitute
the set $\\{\alpha_{0}^{0},\alpha_{0}^{1}\\}$ for $\mathscr{C}_{f}$ in the
unique determination of the curves $\gamma^{\pm}$ of Case $n=1$, to obtain
analogous curves (again denoted by $\gamma^{\pm}$) for Case $n\geq 2$. Then,
for the unique determination of $\tau_{\alpha_{5}}^{\pm 1}(\alpha_{4})$ we
proceed as in Case $n=1$, adding the set $\mathscr{D}$ and using these new
$\gamma^{\pm}$. This implies that $\tau_{\alpha_{5}}^{\pm
1}(\alpha_{4})\in\mathscr{Y}(S)^{3}$

Similarly, to uniquely determine $\tau_{\alpha_{4}}^{\pm 1}(\alpha_{3})$ and
$\tau_{\alpha_{3}}^{\pm 1}(\alpha_{2})$ we also substitute in the set $A$ (see
Table 1) the set $\\{\alpha_{0}^{0},\alpha_{0}^{1}\\}$ for $\mathscr{C}_{f}$.
Thus, $\tau_{\alpha_{4}}^{\pm 1}(\alpha_{3})\in\mathscr{Y}(S)^{4}$ and
$\tau_{\alpha_{3}}^{\pm 1}(\alpha_{2})\in\mathscr{Y}(S)^{5}$. For
$\tau_{\alpha_{2}}^{\pm 1}(\alpha_{1})$ we add in $A$ the set $\mathscr{D}$
and get $\tau_{\alpha_{2}}^{\pm 1}(\alpha_{1})\in\mathscr{Y}(S)^{6}$.

For $\tau_{\alpha_{1}}^{\pm 1}(\alpha_{0}^{0})$ and $\tau_{\alpha_{1}}^{\pm
1}(\alpha_{0}^{n})$, we substitute in $A$ the curves $\beta_{C}^{+}$ for
$\beta_{1}^{-}$ and $\beta_{n}^{+}$ respectively, and add the set
$\mathscr{D}$. Hence, $\tau_{\alpha_{1}}^{\pm
1}(\alpha_{0}^{0}),\tau_{\alpha_{1}}^{\pm
1}(\alpha_{0}^{n})\in\mathscr{Y}(S)^{7}$. The curves $\tau_{\alpha_{1}}^{\pm
1}(\alpha_{0}^{i})$ for $1\leq i\leq n-1$ are a little trickier: we substitute
$\beta_{C}^{+}$ for the set $\\{\beta_{i}^{+},\beta_{i+1}^{-}\\}$ and add the
sets $\mathscr{D}_{1,i}$ and $\mathscr{D}_{i+1,n}$. See Figure 45 for an
example. Then, $\tau_{\alpha_{1}}^{\pm
1}(\alpha_{0}^{i})\in\mathscr{Y}(S)^{7}$ for all $0\leq i\leq n$.

Figure 45: On the left, the curves needed to uniquely determine
$\tau_{\alpha_{1}}(\alpha_{0}^{2})$, on the right.

Proceeding as above, for the curves $\tau_{\alpha_{0}^{0}}^{\pm
1}(\alpha_{5})$ and $\tau_{\alpha_{0}^{n}}^{\pm 1}(\alpha_{5})$ we substitute
for $\beta_{1}^{-}$ and $\beta_{n}^{+}$ respectively while adding
$\mathscr{D}$ to $A$ for all of them, and for the curves
$\tau_{\alpha_{0}^{i}}^{\pm 1}(\alpha_{5})$ (with $0<i<n)$ we substitute in
$A$ the curve $\beta_{C}^{+}$ for the set
$\\{\beta_{i}^{+},\beta_{i+1}^{-}\\}$ and add the sets $\mathscr{D}_{1,i}$ and
$\mathscr{D}_{i+1,n}$. See Figure 46. Therefore, $\tau_{\alpha_{0}^{i}}^{\pm
1}(\alpha_{5})\in\mathscr{Y}(S)^{8}$ for all $0\leq i\leq n$.

Figure 46: On the left, the curves needed to uniquely determine
$\tau_{\alpha_{0}^{2}}(\alpha_{5})$, on the right.

###  4.6 Proof of Claim 2:
$\tau_{(\mathscr{C}\backslash\\{\alpha_{5}\\})}^{\pm
1}(\mathscr{B})\subset\mathscr{Y}(S)^{11}$

Similarly to Subsections 2.6 (first case), 2.7 and 3.5, the proof of this
claim follows from the results in Subsection 4.5: Let $\beta\in\mathscr{B}$
and $\alpha\in\mathscr{C}\backslash\\{\alpha_{5}\\}$. Since $S$ has genus $2$,
we can uniquely determine $\beta$ using sets $C\subset\mathscr{C}$ and
$E\subset\mathscr{E}$. Then we have $\tau_{\alpha}^{\pm
1}(\beta)=\langle\tau_{\alpha}^{\pm 1}(C\cup E)\rangle$, which by the results
in Subsection 4.5 and Remark 4.4 is an element in $\mathscr{Y}(S)^{11}$.

The existence of these sets can be easily inferred from the following examples
(see Figure 47):

  * •

For $C=\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}$: Let $D=\varnothing$ if $n=1$
and $D=\\{\delta_{j}\\}_{j=1}^{n-1}\cup\\{\varepsilon_{4}\\}$ if $n\geq 2$.
Then we have

$\beta_{C}^{+}=\langle\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}\cup\\{\alpha_{5}\\}\cup
D\rangle,$ $\beta_{C}^{-}=\alpha_{5}.$

  * •

For $C=\\{\alpha_{0}^{i},\alpha_{1},\alpha_{2}\\}$ with $0<i<n$:

$\beta_{C}^{+}=\langle\\{\alpha_{0}^{i},\alpha_{1},\alpha_{2}\\}\cup\\{\alpha_{0}^{j}:i<j\\}\cup\\{\alpha_{4}\\}\cup\\{\delta_{j}:j<i\\}\cup\\{\varepsilon_{4}\\}\rangle,$
$\beta_{C}^{-}=\langle\\{\alpha_{0}^{i},\alpha_{1},\alpha_{2}\\}\cup\\{\alpha_{0}^{j}:j<i\\}\cup\\{\alpha_{4}\\}\cup\\{\delta_{j}:i<j\\}\cup\\{\varepsilon_{4}\\}\rangle.$

  * •

For $C=\\{\alpha_{0}^{n},\alpha_{1},\alpha_{2}\\}$:

$\beta_{C}^{+}=\langle\\{\alpha_{0}^{n},\alpha_{1},\alpha_{2}\\}\cup\\{\alpha_{4}\\}\cup\mathscr{D}\rangle,$
$\beta_{C}^{-}=\alpha_{4}.$

  * •

For $C=\\{\alpha_{5},\alpha_{0}^{i},\alpha_{1}\\}$ with $0<i<n$:

$\beta_{C}^{+}=\langle\\{\alpha_{5},\alpha_{0}^{i},\alpha_{1}\\}\cup\\{\alpha_{3}\\}\cup\\{\delta_{j}:j\neq
i\\}\cup\\{\varepsilon_{2},\varepsilon_{4}\\}\rangle,$
$\beta_{C}^{-}=\alpha_{3}.$

  * •

For $C=\\{\alpha_{5},\alpha_{0}^{0},\alpha_{1}\\}$:

$\beta_{C}^{+}=\langle\\{\alpha_{5},\alpha_{0}^{0},\alpha_{1}\\}\cup\\{\alpha_{3}\\}\cup\mathscr{D}\rangle,$
$\beta_{C}^{-}=\alpha_{3}.$

  * •

For $C=\\{\alpha_{4},\alpha_{5},\alpha_{0}^{i}\\}$ with $0<i<n$:

$\beta_{C}^{+}=\langle\\{\alpha_{4},\alpha_{5},\alpha_{0}^{i}\\}\cup\\{\alpha_{0}^{j}:i<j\\}\cup\\{\alpha_{2}\\}\cup\\{\delta_{j}:j<i\\}\cup\\{\varepsilon_{2}\\}\rangle,$
$\beta_{C}^{-}=\langle\\{\alpha_{4},\alpha_{5},\alpha_{0}^{i}\\}\cup\\{\alpha_{0}^{j}:j<i\\}\cup\\{\alpha_{2}\\}\cup\\{\delta_{j}:i<j\\}\cup\\{\varepsilon_{2}\\}\rangle.$

  * •

For $C=\\{\alpha_{4},\alpha_{5},\alpha_{0}^{n}\\}$:

$\beta_{C}^{+}=\langle\\{\alpha_{4},\alpha_{5},\alpha_{0}^{n}\\}\cup\\{\alpha_{2}\\}\cup\mathscr{D}\rangle,$
$\beta_{C}^{-}=\alpha_{2}.$

  * •

For $C=\\{\alpha_{3},\alpha_{4},\alpha_{5}\\}$:

$\beta_{C}^{+}=\langle\\{\alpha_{3},\alpha_{4},\alpha_{5}\\}\cup\\{\alpha_{1}\\}\cup\\{\delta_{j}\\}_{j=1}^{n-1}\cup\\{\varepsilon_{2}\\}\rangle,$
$\beta_{C}^{-}=\alpha_{1}.$

Figure 47: On the left, the curves needed to uniquely determine the curves
$\beta_{C}^{+}$ for $C=\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}$ (top),
$C=\\{\alpha_{0}^{2},\alpha_{1},\alpha_{1},\alpha_{2}\\}$ (upper middle),
$C=\\{\alpha_{5},\alpha_{0}^{2},\alpha_{1}\\}$ (lower middle) and
$\beta_{C}^{-}$ for $C=\\{\alpha_{4},\alpha_{5},\alpha_{0}^{2}\\}$ (bottom),
on the right.

###  4.7 Proof of Claim 3: $\eta_{\\{\delta_{i}:0<i<n\\}}^{\pm
1}(\mathscr{C})\subset\mathscr{Y}(S)^{3}$

In this subsection we assume that $n\geq 2$ (otherwise there would be no outer
curves).

To prove this claim, first note that we only need to prove that
$\eta_{\delta_{i}}^{\pm 1}(\alpha_{0}^{i})\in\mathscr{Y}(S)^{3}$ (all other
curves in $\mathscr{C}$ are disjoint from $\delta_{i}$). To do so, we define
the following auxiliar curves.

For each $1\leq i\leq n$, we define the following curves (see Figure 48):

$\gamma_{+}^{i}=\langle\\{\alpha_{0}^{j}:j<i\\}\cup\\{\alpha_{1},\alpha_{2}\\}\cup\\{\alpha_{4}\\}\cup\\{\zeta^{k}:i<k\leq
n\\}\rangle,$
$\gamma_{-}^{i}=\langle\\{\alpha_{0}^{j}:i<j\\}\cup\\{\alpha_{1},\alpha_{2}\\}\cup\\{\alpha_{4}\\}\cup\\{\zeta^{k}:1\leq
k\leq i\\}\rangle.$

Note that for all $1\leq i\leq n$ we have that
$\gamma_{+}^{i},\gamma_{-}^{i}\in\mathscr{Y}(S)^{2}$.

Figure 48: On the left, the curves needed to uniquely determine the curves
$\gamma_{+}^{2}$ (top) and $\gamma_{-}^{2}$ (bottom), on the right.

Finally, for each $1\leq i<n$, we have that (see Figure 49):

$\eta_{\delta_{i}}(\alpha_{i})=\langle(\mathscr{C}\backslash\\{\alpha_{0}^{i},\alpha_{1},\alpha_{5}\\})\cup\\{\gamma_{+}^{i}\\}\rangle,$
$\eta_{\delta_{i}}^{-1}(\alpha_{i})=\langle(\mathscr{C}\backslash\\{\alpha_{0}^{i},\alpha_{1},\alpha_{5}\\})\cup\\{\gamma_{-}^{i}\\}\rangle.$

Therefore, $\eta_{\\{\delta_{i}:0<i<n\\}}^{\pm
1}(\mathscr{C})\subset\mathscr{Y}(S)^{3}$

Figure 49: On the left, the curves needed to uniquely determine
$\eta_{\delta_{i}}(\alpha_{2})$, on the right.

######  Remark 4.5.

Again (as in Subsection 4.5), since $\mathscr{E}\subset\mathscr{C}^{2}$, we
have that

$\eta_{\\{\delta_{i}:0<i<n\\}}^{\pm 1}(\mathscr{E})\subset\mathscr{Y}(S)^{5}.$

###  4.8 Proof of Claim 4: $\eta_{\\{\delta_{i}:0<i<n\\}}^{\pm
1}(\mathscr{B})\subset\mathscr{Y}(S)^{6}$

In this subsection we assume that $n\geq 2$ (otherwise there would be no outer
curves).

Analogously to Subsection 4.6, this claim follows from the results in
Subsection 4.7: Let $\beta\in\mathscr{B}$ and
$\delta\in\\{\delta_{i}:0<i<n\\}$; since $\beta=\langle C\cup E\rangle$ with
$C\subset\mathscr{C}$ and $E\subset\mathscr{E}$ (as in Subsection 4.6), we
have that $\eta_{\delta}^{\pm 1}(\beta)=\langle\eta_{\delta}^{\pm 1}(C\cup
E)\rangle$, which by the results in Subsection 4.7 and Remark 4.5 is contained
in $\mathscr{Y}(S)^{6}$.

For the existence of these sets, we refer the reader to the examples and
figures in Subsection 4.6.

###  4.9 Proof of Theorem B

Similarly to Subsections 2.8 and 3.8, here we define the set $\mathfrak{X}(S)$
as the set $\mathfrak{X}_{1}$ from Section 5 in [2], which was proved to be
rigid (see Lemma 5.2 in [2]).

Let $X$ be the set defined as $\mathscr{C}$ union the set of all boundary
curves of regular neighbourhoods of sets $A$ with one of the following forms:

  1. 1.

$A=\\{\alpha_{0}^{i},\alpha_{0}^{j},\alpha_{k}\\}$ where $0\leq i\leq j\leq n$
and $k=1,5$.

  2. 2.

$A=\\{\alpha_{0}^{i},\alpha_{0}^{j},\alpha_{k},\alpha_{k+1}\\}$ where $0\leq
i\leq j\leq n$ and $k=1,4$.

  3. 3.

$A=\\{\alpha_{i},\alpha_{i+1},\ldots,\alpha_{j}\\}$, where
$I=\\{i,i+1,\ldots,j\\}$ is an integer subinterval of $\\{0,\ldots,5\\}$
modulo $6$ (We assume that $\alpha_{0}$ denotes $\alpha_{0}^{k}$ for some
$k$). If the length of the interval $I$ is odd, then we require that both $i$
and $j$ are even.

Then, we define $\mathfrak{X}(S)$ as the union of $X^{1}$ and the set of all
boundary curves of regular neighbourhoods of sets
$A=\\{\alpha_{i},\alpha_{i+1},\ldots,\alpha_{j}\\}$, where $0<i\leq j\leq 3$,
and both $i$ and $j$ are odd.

Finally, note that $\mathscr{Y}(S)\subset\mathfrak{X}(S)$, which immediatly
implies Theorem B.

##  5 Rigidity

In this section we prove Theorems C and D, i.e. we prove that using only
topological restrictions on the surfaces, all graph morphisms between the
corresponding curve graphs are geometric. The ideas and techniques used in
this section are essentially those used in [6].

To prove Theorems C and D, we first need the following definitions and lemmas.

Let $S=S_{g,n}$ be such that $\kappa(S)=3g-3+n\geq 2$. A multicurve $M$ on $S$
is a set of curves on $S$ corresponding in $\mathcal{C}(S)$ to a complete
subgraph. Note that $|M|\leq\kappa(S)$. If the equality holds, then $M$ is
called a pants decomposition of $S$, since $S\backslash M$ would then be equal
to the disjoint union of surfaces homeomorphic to $S_{0,3}$.

######  Lemma 5.1.

Let $S_{1}=S_{g_{1},n_{1}}$ and $S_{2}=S_{g_{2},n_{2}}$ be such that
$\kappa(S_{1}),\kappa(S_{2})\geq 2$, and let
$\varphi:\mathcal{C}(S_{1})\to\mathcal{C}(S_{2})$ be a graph morphism. Then
$\varphi$ maps multicurves to multicurves of the same cardinality. Moreover,
$\kappa(S_{1})\leq\kappa(S_{2})$.

###### Proof.

The first part follows directly from $\varphi$ being a graph morphism, and a
multicurve $M$ being a complete graph in $\mathcal{C}(S_{1})$. The second part
follows from a pants decomposition $P$ of $S$ having maximal cardinality, and
the first part of the lemma. ∎

######  Remark 5.2.

Note that this lemma implies that if $\kappa(S_{1})\geq\kappa(S_{2})$, and the
hypothesis of the lemma are satisfied, then $\kappa(S_{1})=\kappa(S_{2})$. In
particular, pants decompositions of $S_{1}$ would then be mapped to pants
decompositions of $S_{2}$.

Let $\alpha$ and $\beta$ be curves on $S=S_{g,n}$. We say $\alpha$ and $\beta$
are Farey neighbours if their open regular neighbourhood $N(\alpha,\beta)$ on
$S$ has complexity one, i.e. is homeomorphic to either $S_{0,4}$ or $S_{1,1}$.
In particular, we say $\alpha$ and $\beta$ are spherical (respectively
toroidal) Farey neighbours if $N(\alpha,\beta)$ is homeomorphic to $S_{0,4}$
(respectively $S_{1,1}$). This terminology is inspired by the fact that both
$\mathcal{C}(S_{0,4})$ and $\mathcal{C}(S_{1,1})$ (with a slightly modified
definition, see [4]) are isomorphic to the Farey graph (the 1-skeleton of the
ideal triangulation of the hyperbolic plane); thus, if $\alpha$ and $\beta$
are Farey neighbours, then they are adjacent (neighbours) in
$\mathcal{C}(N(\alpha,\beta))$.

Now, let $\varphi:\mathcal{C}(S_{1})\to\mathcal{C}(S_{2})$ be a graph
morphism. We say $\varphi$ is spherical Farey (respectively toroidal Farey) if
for all curves $\alpha$ and $\beta$ on $S_{1}$ that are spherical
(respectively toroidal) Farey neighbours, they are mapped to intersecting
curves, i.e. $i(\varphi(\alpha),\varphi(\beta))\neq 0$. We say $\varphi$ is a
Farey map either if it is both spherical Farey and toroidal Farey when
$g_{1}\geq 1$, or if it is sphereical Farey when $g_{1}=0$. Note that being
spherical/toroidal Farey (or both) is a generalisation of the concept of
superinjectivity, see [8], [10], [3] and [9].

######  Lemma 5.3.

Let $S_{1}=S_{g_{1},n_{1}}$ and $S_{2}=S_{g_{2},n_{2}}$ be such that
$\kappa(S_{1})\geq\kappa(S_{2})\geq 2$, and let
$\varphi:\mathcal{C}(S_{1})\to\mathcal{C}(S_{2})$ be a graph morphism. If
$S_{1}$ is not homeomorphic to $S_{1,2}$ nor $S_{2,0}$, then $\varphi$ is a
Farey map.

###### Proof.

Let $\alpha$ and $\beta$ be curves on $S_{1}$ such that they are Farey
neighbours. Then, there exists a multicurve $M$ on $S_{1}$ such that
$M_{\alpha}=M\cup\\{\alpha\\}$ and $M_{\beta}=M\cup\\{\beta\\}$ are pants
decompositions of $S_{1}$. By Lemma 5.1 and Remark 5.2, we know that
$\kappa(S_{1})=\kappa(S_{2})$, and that $\varphi(M_{\alpha})$ and
$\varphi(M_{\beta})$ are pants decompositions of $S_{2}$. Since
$\varphi(M_{\alpha})$ and $\varphi(M_{\beta})$ differ by exactly one curve, a
complexity argument can be used to conclude that $\varphi(\alpha)$ and
$\varphi(\beta)$ are contained in a subsurface $N$ of $S_{2}$ with
$\kappa(N)=1$.

Thus, either $\varphi(\alpha)=\varphi(\beta)$ or
$i(\varphi(\alpha),\varphi(\beta))\neq 0$.

To prove that $\varphi(\alpha)\neq\varphi(\beta)$ we use an auxiliary curve:
Let $\gamma$ be a curve on $S_{1}$ such that $\alpha$ and $\gamma$ are
disjoint, and $\beta$ is a Farey neighbour of $\gamma$. Using the same
argument as above, we can deduce that either $\varphi(\beta)=\varphi(\gamma)$
or $i(\varphi(\beta),\varphi(\gamma))\neq 0$. But none of these options can be
true if $\varphi(\alpha)=\varphi(\gamma)$. Hence,
$i(\varphi(\alpha),\varphi(\beta))\neq 0$. ∎

Note that, if $S_{1}$ is homeomorphic to either $S_{1,2}$ or $S_{2,0}$, while
we can prove that $\varphi$ is toroidal Farey using the same proof, in both
cases there exist curves $\alpha$ and $\beta$ that are spherical Farey
neighbours, such that any curve $\gamma$ disjoint from one of them, cannot be
a Farey neighbour of the other. See Figure 50.

$\beta$ [b] at 222 218 $\alpha$ [tr] at 185 175  $\beta$ [b] at 220 208
$\alpha$ [tr] at 180 180

Figure 50: Examples of curves $\alpha$ and $\beta$ on $S_{1,2}$ (left) and
$S_{2,2}$ (right) such that $\alpha$ and $\beta$ are spherical Farey
neighbours, and any $\gamma$ disjoint from any of them cannot be a Farey
neighbour of the other.

Armed with Lemma 5.3, to prove Theorem C we only need Corollary C in [6],
which for the sake of completeness we rewrite here for the particular case of
$\mathcal{C}(S_{g,n})$.

######  Corollary (C in [6]).

Let $Y<\mathcal{C}(S_{g,n})$ be an rigid subgraph of $\mathcal{C}(S_{g,n})$,
and $\phi:Y^{\omega}\to\mathcal{C}(S_{g,n})$ a graph morphism such that
$\phi|_{Y}$ is locally injective. Then $\phi$ is the restriction to
$Y^{\omega}$ of an automorphism of $\mathcal{C}(S_{g,n})$, unique up to the
pointwise stabilizer of $Y$ in $\mathrm{Aut}(\mathcal{C}(S_{g,n}))$.

We recall Theorem C from the Introduction:

######  Theorem (C).

Let $S=S_{g,n}$ with $\kappa(S)\geq 2$ and let
$\varphi:\mathcal{C}(S)\to\mathcal{C}(S)$ be a graph morphism. Then $\varphi$
is an automorphism of $\mathcal{C}(S)$.

###### Proof of Theorem C.

As was mentioned in the Introduction, since Theorem C has already been proved
by the author for the case $g\geq 3$, we assume that $g\leq 2$. For now, let
$S$ be different from $S_{1,2}$ and $S_{2,0}$. Due to Theorem B (from this
work) and Corollary C (from [6]), we simply need to prove that
$\varphi|_{\mathfrak{X}(S)}$ is locally injective.

Let $\alpha$ and $\beta$ be different curves in $\mathfrak{X}(S)$. By
construction, one of the following situations holds:

  1. 1.

$\alpha$ is disjoint from $\beta$.

  2. 2.

$\alpha$ and $\beta$ are Farey neighbours.

  3. 3.

There exists a curve $\gamma$ on $S$ such that $\alpha$ is disjoint from
$\gamma$, and $\beta$ is a Farey neighbour of $\gamma$.

Since $\varphi$ is a graph morphism, by Lemma 5.3 $\varphi$ is a Farey map,
and thus $\varphi(\alpha)\neq\varphi(\beta)$ if either (1) or (2) holds. If
(3) holds, then $\varphi(\alpha)$ is disjoint from $\varphi(\gamma)$, and
$i(\varphi(\beta),\varphi(\gamma))\neq 0$, thus
$\varphi(\alpha)\neq\varphi(\beta)$. This implies that
$\varphi|_{\mathfrak{X}(S)}$ is injective (and in particular locally
injective) as desired.

If $S$ is homeomorphic to either $S_{1,2}$ or $S_{2,0}$, let $\Sigma$ be
homeomorphic to either $S_{0,5}$ or $S_{0,6}$, respectively. Then there exists
an isomorphism $\psi:\mathcal{C}(S)\to\mathcal{C}(\Sigma)$ (induced by the
hyperelliptic involution). Hence $\psi\circ\varphi\circ\psi^{-1}$ is a graph
endomorphism of $\mathcal{C}(\Sigma)$, and by the argument above is an
automorphism of $\mathcal{C}(\Sigma)$. Thus $\varphi$ is an automorphism of
$\mathcal{C}(S)$. ∎

Now, as was mentioned in the Introduction, to prove Theorem D we simply need
to prove the prove the following theorem.

######  Theorem 5.4.

Let $S_{1}=S_{g_{1},n_{1}}$ and $S_{2}=S_{g_{2},n_{2}}$ be such that
$\kappa(S_{1})\geq\kappa(S_{2})\geq 4$ with $(g_{1},n_{1})\neq(0,7)$, and
$\varphi:\mathcal{C}(S_{1})\to\mathcal{C}(S_{2})$ be a graph morphism. Then,
$S_{1}$ is homeomorphic to $S_{2}$.

To prove this theorem, we are going to use (almost standard) techniques and
geometric results. These techniques are heavily inspired by those in [16] and
are practically identical to those in [6] (to the point that in many cases we
simply give a sketch of the proof and a reference).

Let $S=S_{g,n}$ with $\kappa(S)\geq 4$, and $P$ be a pants decomposition of
$S$. We say a closed subsurface $\Sigma$ of $S$ is induced by $P$ if all its
boundary curves are pairwise non-isotopic, and all its boundary curves are
elements of $P$. We say $\alpha,\beta\in P$ are adjacent with respect to $P$
if there exists a closed subsurface $\Sigma$ induced by $P$, whose interior is
homeomorphic to $S_{0,3}$ and such that $\alpha$ and $\beta$ are boundary
curves of $\Sigma$. With this, we define the adjacency graph of $P$, denoted
by $\mathcal{A}(P)$, as the simplicial graph whose vertex set is $P$, and two
vertices span an edge if they are adjacent to each other with respect to $P$.

Note that if $S_{1}=S_{g_{1},n_{1}}$ and $S_{2}=S_{g_{2},n_{2}}$ is such that
$\kappa(S_{1})\geq\kappa(S_{2})\geq 4$, and
$\varphi:\mathcal{C}(S_{1})\to\mathcal{C}(S_{2})$ is a graph morphism, then
for every pants decomposition $P$ of $S_{1}$, by Remark 5.2, there exists an
induced map $\varphi_{P}:\mathcal{A}(P)\to\mathcal{A}(\varphi(P))$. The
following Lemma proves that $\varphi_{P}$ is a graph isomorphism.

######  Lemma 5.5 (cf. Lemma 7 in [16] and Lemma 3.6 in [6]).

Let $S_{1}=S_{g_{1},n_{1}}$ and $S_{2}=S_{g_{2},n_{2}}$ be such that
$\kappa(S_{1})\geq\kappa(S_{2})\geq 4$,
$\varphi:\mathcal{C}(S_{1})\to\mathcal{C}(S_{2})$ be a graph morphism, and $P$
be a pants decomposition of $S_{1}$. Then, for all $\alpha,\beta\in P$,
$\alpha$ and $\beta$ are adjacent with respect to $P$ if and only if
$\varphi(\alpha)$ and $\varphi(\beta)$ are adjacent with respect to
$\varphi(P)$. In particular, the induced map $\varphi_{P}$ is a graph
isomorphism.

###### Sketch of proof.

Let $\gamma_{\alpha}$ and $\gamma_{\beta}$ be Farey neighbours of $\alpha$ and
$\beta$, respectively, such that
$P_{\alpha}=(P\backslash\\{\alpha\\})\cup\\{\gamma_{\alpha}\\}$ and
$P_{\beta}=(P\backslash\\{\beta\\})\cup\\{\gamma_{\beta}\\}$ are pants
decompositions. Then by Remark 5.2 $\varphi(P)$, $\varphi(P_{\alpha})$ and
$\varphi(P_{\beta})$ are also pants decompositions.

If $\alpha$ and $\beta$ are adjacent with respect to $P$, we can choose
$\gamma_{\alpha}$ and $\gamma_{\beta}$ as Farey neighbours of each other.
Then, $i(\varphi(\alpha),\varphi(\gamma_{\alpha}))\neq 0$,
$i(\varphi(\beta),\varphi(\gamma_{\beta}))\neq 0$, and
$i(\varphi(\gamma_{\alpha},\varphi(\gamma_{\beta})\neq 0$ by Lemma 5.3. But
this would be impossible if $\varphi(\alpha)$ and $\varphi(\beta)$ were not
adjacent with respect to $\varphi(P)$. Thus, $\varphi(\alpha)$ and
$\varphi(\beta)$ are adjacent with respect to $\varphi(P)$.

If $\alpha$ and $\beta$ are not adjacent with respect to $P$, then
$\gamma_{\alpha}$ and $\gamma_{\beta}$ are disjoint. Then,
$i(\varphi(\alpha),\varphi(\gamma_{\alpha}))\neq 0$,
$i(\varphi(\beta),\varphi(\gamma_{\beta}))\neq 0$, and
$\varphi(\gamma_{\alpha})$ and $\varphi(\gamma_{\beta})$ are disjoint, by
Lemma 5.3 and $\varphi$ being a graph morphism. But this would be impossible
if $\varphi(\alpha)$ and $\varphi(\beta)$ were adjacent with respect to
$\varphi(P)$. Thus, $\varphi(\alpha)$ and $\varphi(\beta)$ are not adjacent
with respect to $\varphi(P)$. ∎

Recall that an outer curve $\alpha$ is a separating curve on $S$ such that
$S\backslash\\{\alpha\\}$ has a connected component homeomorphic to $S_{0,3}$.

######  Lemma 5.6 (cd. Lemma 9 in [16], and Lemma 3.7 in [6]).

Let $S_{1}=S_{g_{1},n_{1}}$ and $S_{2}=S_{g_{2},n_{2}}$ be such that
$\kappa(S_{1})\geq\kappa(S_{2})\geq 4$, and
$\varphi:\mathcal{C}(S_{1})\to\mathcal{C}(S_{2})$ be a graph morphism. Then,
$\varphi$ maps non-outer separating curves to non-outer separating curves.

###### Sketch of proof.

This follows from the fact that $\varphi_{P}$ is an isomorphism and noting
that if $P$ is a pants decomposition and $\alpha\in P$, then $\alpha$ is an
non-outer separating curve if and only if the vertex corresponding to $\alpha$
in $\mathcal{A}(P)$ is a cut vertex. ∎

######  Lemma 5.7 (cf. Lemma 10 in [16], and Lemma 3.8 in [6]).

Let $S_{1}=S_{g_{1},n_{1}}$ and $S_{2}=S_{g_{2},n_{2}}$ be such that
$\kappa(S_{1})\geq\kappa(S_{2})\geq 4$, and
$\varphi:\mathcal{C}(S_{1})\to\mathcal{C}(S_{2})$ be a graph morphism. Then,
$\varphi$ maps non-separating curves to non-separating curves.

###### Sketch of proof.

Again, this follows from Lemma 5.6, the fact that $\varphi_{P}$ is an
isomorphism, and noting two things:

  1. 1.

If $\alpha$ is an outer curve, then for all pants decompositions $P$ with
$\alpha\in P$, $\alpha$ can adjacent to at most 2 different curves with
respect to $P$.

  2. 2.

If $\alpha$ is a non-separating curve, twe can always choose a pants
decomposition $P$ with $\alpha\in P$, such that $\alpha$ is adjacen to at
least three different curves with respect to $P$.

∎

Let $S=S_{g,n}$ be such that $\kappa(S)\geq 4$, and $\alpha$ and $\beta$ be
disjoint curves on $S$. The set $\\{\alpha,\beta\\}$ is a peripheral pair if
$\alpha$ and $\beta$ together bound a closed subsurface of $S$ homeomorphic to
a once-punctured annulus. Let $\gamma$ be a curve disjoint from $\alpha$ and
$\beta$; we say $\alpha$, $\beta$ and $\gamma$ _bound a pair of pants_ if they
are the only boundary curves of a closed subsurface of $S$, whose interior is
homeomorphic to $S_{0,3}$.

######  Remark 5.8.

Let $S=S_{g,n}$ be such that $\kappa(S)\geq 4$, $P$ be a pants decomposition
of $S$, and $\alpha,\beta,\gamma\in P$. Note that we have the following:

  1. 1.

If $\alpha$ is adjacent with respect to $P$ to only $\beta$ and $\gamma$, and
these three curves are non-separating, then $\\{\alpha,\beta\\}$ and
$\\{\alpha,\gamma\\}$ are peripheral pairs.

  2. 2.

If $\alpha$, $\beta$ and $\gamma$ are adjacent to each other with respect to
$P$, then either together they bound a pair of pants, or they are non-
separating curves that look up to homeomorphism like Figure 51.

  3. 3.

(Using previous point) If $\alpha$, $\beta$ and $\gamma$ are adjacent to each
other with respect to $P$, and two of them are separating curves, then the
three of them are separating curves.

$\alpha$ [t] at 155 208 $\beta$ [bl] at 65 85 $\gamma$ [br] at 250 90

Figure 51: Curves $\alpha$, $\beta$, $\gamma$ adjacent to each other with
respect to a pants decomposition, as in Remark 5.8 (2).

######  Lemma 5.9 (cf. Lemma 11 in [16], and Lemma 3.10 in [6]).

Let $S_{1}=S_{g_{1},n_{1}}$ and $S_{2}=S_{g_{2},n_{2}}$ be such that
$\kappa(S_{1})\geq\kappa(S_{2})\geq 4$, and
$\varphi:\mathcal{C}(S_{1})\to\mathcal{C}(S_{2})$ be a graph morphism. Then,
$\varphi$ maps outer curves to outer curves.

###### Proof.

Let $\alpha$ be an outer curve of $S_{1}$. By the argument in the proof of
Lemma 5.6, we know that $\varphi(\alpha)$ cannot be a non-outer separating
curve. This implies that if $\varphi(\alpha)$ were not an outer curve it would
have to be non-separating. Now, we divide the proof into two cases:

Case $(g_{1},n_{1})\neq(1,4)$: Let $\beta$ and $\gamma$ be non-outer
separating curves such that $\alpha$, $\beta$ and $\gamma$ bound a pair of
pants. Let also $P$ be a pants decomposition of $S_{1}$ with
$\alpha,\beta,\gamma\in P$. By Lemma 5.6 we know that $\varphi(\beta)$ and
$\varphi(\gamma)$ are non-outer separating curves, and by Lemma 5.5 we know
that $\varphi(\alpha)$, $\varphi(\beta)$ and $\varphi(\gamma)$ are adjacent to
each other with respect to $\varphi(P)$. Finally by Remark 5.8 (3) we know
that $\varphi(\alpha)$ has to be a separating curve. Thus, $\varphi(\alpha)$
has to be an outer curve.

Case $(g_{1},n_{1})=(1,4)$: Let $P$ be a pants decomposition of $S_{1}$ such
that $\alpha\in P$, $P\backslash\\{\alpha\\}$ is composed solely of non-
separating curves, and $\alpha$ has exactly two curves adjacent to it with
respect to $P$, say $\beta$ and $\gamma$. If we suppose that $\varphi(\alpha)$
is a non-separating curve, by Remark 5.8 (1), we have that
$\\{\varphi(\alpha),\varphi(\beta)\\}$ and
$\\{\varphi(\alpha),\varphi(\gamma)\\}$ are peripheral pairs. In particular
$\varphi(\beta)$ and $\varphi(\gamma)$ bound a twice-punctured annulus. But
$\beta$ and $\gamma$ are adjacent to each other with respect to $P$, thus by
Lemma 5.5 we have that $\varphi(\beta)$ and $\varphi(\gamma)$ are adjacent
with respect to $\varphi(P)$. This implies that there exists a closed
subsurface $\Sigma$ with interior homeomorphic to $S_{0,3}$, induced by
$\varphi(P)$ with $\varphi(\beta)$ and $\varphi(\gamma)$ as two of its
boundary curves. Note that $\Sigma$ cannot be a once-punctured annulus (this
would imply that $\kappa(S_{2})=3$), thus there exists a curve $\delta\in
P\backslash\\{\alpha,\beta,\gamma\\}$ such that $\varphi(\delta)$ is the other
boundary curve of $\Sigma$. However, this construction would imply that
$\varphi(\delta)$ is a separating curve, which contradicts Lemma 5.7.
Therefore, $\varphi(\alpha)$ has to be an outer curve. ∎

######  Lemma 5.10.

Let $S=S_{g,n}$ be such that $\kappa(S)\geq 4$, and $P$ be a pants
decomposition composed solely of non-separating curves. Then, $g=1$ if and
only if $\mathcal{A}(P)$ is a cycle graph.

###### Proof.

If $g=1$, then $P=\\{\alpha_{0},\ldots,\alpha_{n-1}\\}$. Then we can have a
homeomorphism $f:S\to A/\sim$ where $A$ is an $n$-punctured annulus and $\sim$
identifies its boundaries, such that the image of the boundaries under the
quotient is $\alpha_{0}$. This implies that every $\alpha_{i}$ with $i\neq 0$
is mapped to an essential curve in $A$ (that is not a boundary curve);
moreover, since for all $i\in\mathbb{Z}/n\mathbb{Z}$ we have that $\alpha_{i}$
is a non-separating curve, it follows that $f(\alpha_{i})$ cannot be a
separating curve of $A$. Thus, for all $i\neq 0$, $f(\alpha_{i})$ and
“$f(\alpha_{0})$” bound a $k$-punctured annulus in $A$ for some $k>0$. Hence
(up to relabelling), for all $i\in\mathbb{Z}/n\mathbb{Z}$, $f(\alpha_{i})$ and
$f(\alpha_{i+1})$ bound a once-punctured annulus. This implies that (up to
relabelling) for all $i\in\mathbb{Z}/n\mathbb{Z}$,
$\\{\alpha_{i},\alpha_{i+1}\\}$ is a peripheral pair, and then $P$ is (up to
homeomorphism) the pants decomposition from Figure 52, for which
$\mathcal{A}(P)$ is a cycle graph.

$\alpha_{0}$ [t] at 35 5 $\alpha_{1}$ [t] at 104 5 $\alpha_{2}$ [t] at 169 5
$\alpha_{3}$ [t] at 235 5 $\alpha_{4}$ [t] at 297 5 $\alpha_{0}$ [t] at 728
250 $\alpha_{1}$ [r] at 837 171 $\alpha_{2}$ [br] at 793 43 $\alpha_{3}$ [bl]
at 661 43 $\alpha_{4}$ [l] at 617 171

Figure 52: On the left, a punctured torus with a pants decomposition composed
solely of non-separating curves; on the right its corresponding adjacency
graph (a cycle graph).

If $\mathcal{A}(P)$ is a cycle graph, we can label the curves in $P$ so that
$P=\\{\alpha_{0},\ldots,\alpha_{\kappa}\\}$ (with $\kappa=\kappa(S)$), and for
all $i\in\mathbb{Z}/\kappa\mathbb{Z}$, the vertices corresponding to
$\alpha_{i}$ and $\alpha_{i+1}$ are adjacent in $\mathcal{A}(P)$ (see the
right side of Figure 52). By Remark 5.8 (1), this implies that for all
$i\in\mathbb{Z}/\kappa\mathbb{Z}$, $\\{\alpha_{i},\alpha_{i+1}\\}$ is a
peripheral pair, and thus bound a once-punctured annulus $A_{i}$ in $S$. The
result follows from reconstructing $S$ using the annuli $A_{i}$. ∎

###### Proof of Theorem 5.4.

Since this theorem was already proved by the author in the case $g_{1}\geq 3$
(Lemma 3.16 in [6]), let $g_{1}\leq 2$. Then, we divide the proof into three
cases depending on the genus of $S_{1}$:
Case $g_{1}=0$: Let $P$ be a pants decomposition of $S_{1}$. Since $g_{1}=0$,
this implies that $P$ is composed solely of separating curves. By Remark 5.2
and Lemmas 5.6 and 5.9, we have that $\varphi(P)$ is a pants decomposition of
$S_{2}$ composed solely of separating curves, which is only possible if
$g_{2}=0$. The result follows from the fact that $\kappa(S_{1})=\kappa(S_{2})$
and the Classification of Surfaces.
Case $g_{1}=1$: Let $P$ be a pants decomposition of $S_{1}$ composed solely of
non-separating curves. By Lemma 5.10 we have that $\mathcal{A}(P)$ is a cycle
graph. This is preserved by $\varphi$ due to Lemma 5.5. Hence, by Remark 5.2
and Lemmas 5.5 and 5.7, $\varphi(P)$ is a pants decomposition of $S_{2}$
composed solely of non-separating curves, such that $\mathcal{A}(\varphi(P))$
is a cycle graph. Then, by Lemma 5.10, $g_{2}=1$. As before, the result
follows from the fact that $\kappa(S_{1})=\kappa(S_{2})$ and the
Classification of Surfaces.
Case $g_{1}=2$: Let $P$ be a pants decomposition composed solely of non-
separating curves, whose adjacency graph is not a cycle graph by Lemma 5.10.
By Remark 5.2 and Lemmas 5.5 and 5.7, we have that $\varphi(P)$ is a pants
decomposition of $S_{2}$ composed solely of non-separating curves, with
$\mathcal{A}(\varphi(P))$ not a cycle graph. This implies that $g_{2}\geq
2=g_{1}$. Since $\kappa(S_{1})=\kappa(S_{2})$, this implies that $n_{1}\geq
n_{2}$.

If $n_{1}=1$, the result then follows from $\kappa(S_{1})=\kappa(S_{2})$ and
the Classification of Surfaces.

If $n_{1}>1$, let $M$ be a multicurve of cardinality
$\lfloor\frac{n_{1}}{2}\rfloor$, composed solely of outer curves. By Lemmas
5.1 and 5.9, $\varphi(M)$ is a multicurve of $S_{2}$ with cardinality
$\lfloor\frac{n_{1}}{2}\rfloor$ and composed solely of outer curves. This
implies that $2\lfloor\frac{n_{1}}{2}\rfloor\leq n_{2}$. Thus, either
$n_{1}=n_{2}$, or $n_{1}=n_{2}+1$; but the latter case leads us to a
contradiction when substituting in $\kappa(S_{1})=\kappa(S_{2})$. Therefore
$n_{1}=n_{2}$ and the result follows from the fact that
$\kappa(S_{1})=\kappa(S_{2})$ and the Classification of Surfaces. ∎

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Jesús Hernández Hernández
<EMAIL_ADDRESS>https://sites.google.com/site/jhdezhdez/ Centro de
Ciencias Matemáticas Universidad Nacional Autónoma de México Morelia, Mich.
58190 México