# Automorphism groups of simplicial complexes of infinite type surfaces

Jesús Hernández Hernández  Aix-Marseille Université, 39 rue F. Joliot Curie,
13453 Marseille Cedex 13, France
_e-mail:_<EMAIL_ADDRESS>José Ferrán Valdez Lorenzo  Centro de Ciencias
Matemáticas, UNAM, Campus Morelia
58190, Morelia, Michoacán, México
_e-mail:_<EMAIL_ADDRESS>
###### Abstract

Let $S$ be any orientable surface of infinite genus with a finite number of
boundary components. In this work we consider the curve complex
$\mathcal{C}(S)$, the nonseparating curve complex $\mathcal{N}(S)$ and the
Schmutz graph $\mathcal{G}(S)$ of $S$. When all the topological ends of $S$
carry genus, we show that all elements in the automorphism groups
$\mathrm{Aut}(\mathcal{C}(S))$, $\mathrm{Aut}(\mathcal{N}(S))$ and
$\mathrm{Aut}(\mathcal{G}(S))$ are _geometric_ , _i.e._ these groups are
naturally isomorphic to the _extended_ mapping class group
$\mathrm{MCG}^{*}(S)$ of the infinite surface $S$. Finally, we study rigidity
phenomena within $\mathrm{Aut}(\mathcal{C}(S))$ and
$\mathrm{Aut}(\mathcal{N}(S))$.

##  1 Introduction

The mapping class group and the extended mapping class group of a given
surface $S$, that we will denote by $\mathrm{MCG}(S)$ and
$\mathrm{MCG}^{*}(S)$ respectively, have been studied mostly when $S$ has
_finite topological type_ , that is, when its fundamental group is finitely
generated. The main purpose of this article is the study of the natural
(simplicial) action of the group $\mathrm{MCG}^{*}(S)$ on two abstract
simplicial complexes and one simplicial graph associated to $S$ when the
surface $S$ has infinite genus. These complexes and graph are:

  1. 1.

The _curve complex_ $\mathcal{C}(S)$. This is the abstract simplicial complex
whose vertices are the (isotopy classes of) essential curves in $S$, and whose
simplexes are multicurves of finite cardinality. It was introduced by Harvey
in 1978.

  2. 2.

The _nonseparating curve complex_ $\mathcal{N}(S)$. This is the simplicial
subcomplex of $\mathcal{C}(S)$ formed by all nonseparating curves, that is,
all the (isotopy classes of) essential curves $\alpha$ such that
$S\setminus\alpha$ is connected. This was first introduced by Schmutz in
[Sch].

  3. 3.

The _Schmutz graph_ $\mathcal{G}(S)$. Introduced by Paul Schmutz Schaller in
[Sch], this is the simplicial graph whose vertex set is the same as the vertex
set of $\mathcal{N}(S)$, and two vertices span an edge whenever their
geometric intersection number is 1. It is also known as _a modified complex of
nonseparating curves_ (see [Farb]), for it can be thought as a 1-dimensional
simplicial complex.

Recall that any _orientable_ surface of infinite topological type is
completely determined, up to homeomorphism, by its genus
$g(S)\in\mathbb{N}\cup\\{\infty\\}$, and a nested pair of topological spaces
$\mathrm{Ends}^{*}(S)\subset\mathrm{Ends}(S)$. Roughly speaking,
$\mathrm{Ends}(S)$ are the _topological ends_ of $S$ and
$\mathrm{Ends}^{*}(S)$ is formed by those that carry (infinite) genus. We will
focus our attention on infinite genus surfaces $S$ for which the boundary
$\partial S$ has finitely many connected components (possibly none) and
$\mathrm{Ends}^{*}(S)=\mathrm{Ends}(S)$.
To each of the aforementioned simplicial complexes one can associate its
automorphism group. We denote these groups by $\mathrm{Aut}(\mathcal{C}(S))$,
$\mathrm{Aut}(\mathcal{N}(S))$ and $\mathrm{Aut}(\mathcal{G}(S))$
respectively. For surfaces of finite topological type of positive genus (with
the exception of the two-holed torus), every element in
$\mathrm{Aut}(\mathcal{C}(S))$, $\mathrm{Aut}(\mathcal{N}(S))$ and
$\mathrm{Aut}(\mathcal{G}(S))$ is _geometric_. That is, if
$X=\mathcal{C}(S),\mathcal{N}(S)$ or $\mathcal{G}(S)$, then the natural map:

$\Psi_{X}:$ | $\mathrm{MCG}^{*}(S)$ | $\longrightarrow$ | $\mathrm{Aut}(X)$
---|---|---|---
| $[h]$ | $\mapsto$ | $h_{*}$
(1)

where $h_{*}$ is given by $h_{*}([\alpha])=[h(\alpha)]$ is an isomorphism.
This result is due to Ivanov [Ivanov] for $X=\mathcal{C}(S)$, to Irmak and
Schmutz [Irmak], [Sch] for $X=\mathcal{N}(S)$ and to Schmutz [Sch] when
$X=\mathcal{G}(S)$. The main purpose of this article is to extend this result
for surfaces of infinite genus:

######  Theorem 1.

Let $S$ be an infinite genus surface with finitely many boundary components
such that $\mathrm{Ends}^{*}(S)=\mathrm{Ends}(S)$. Then the natural map
$\Psi_{X}:\mathrm{MCG}^{*}(S)\longrightarrow\mathrm{Aut}(X)$ is an isomorphism
for $X=\mathcal{C}(S),\mathcal{N}(S)$ or $\mathcal{G}(S)$.

The techniques that we use to prove this result rely heavily on the hypothesis
$\mathrm{Ends}^{*}(S)=\mathrm{Ends}(S)$. However, we suspect that this theorem
remains valid for surfaces with arbitrarily many planar ends. In addition to
the study of the action of $\mathrm{MCG}^{*}(S)$ on simplicial complexes, we
study rigidity phenomena within the curve complex and the nonseparating curve
complex. More precisely:

######  Theorem 2.

Let $S_{1}$ and $S_{2}$ be infinite genus surfaces with finitely many boundary
components, such that $\mathrm{Ends}(S_{i})=\mathrm{Ends}^{*}(S_{i})$ for
$i=1,2$ and let $\phi:\mathcal{C}(S_{1})\rightarrow\mathcal{C}(S_{2})$ be an
isomorphism. Then $S_{1}$ is homeomorphic to $S_{2}$.

As we will see in section §4.2 this result is not valid if we allow the
infinite genus surface $S$ to have planar ends. In §4.2 we will also see that
the tools used in the proof of this theorem work for nonseparating curves.
Hence, we have the following:

######  Corollary 1.

Let $S_{1}$ and $S_{2}$ be infinite genus surfaces with finitely many boundary
components, such that $\mathrm{Ends}(S_{i})=\mathrm{Ends}^{*}(S_{i})$ for
$i=1,2$ and let $\phi:\mathcal{N}(S_{1})\rightarrow\mathcal{N}(S_{2})$ be an
isomorphism. Then $S_{1}$ is homeomorphic to $S_{2}$.

As we will see in section §5, contrary to the compact case, this kind of
rigidity results cannot be extended to injective simplicial maps when $S$ is
an infinite genus surface.
We must remark that while the results of this article are highly inspired by
those of the compact case, many proofs have been either modified or outright
rewritten to accommodate for the infinite type surfaces. We have also
stablished new results and techniques on which the main results of this
article rely. Among these we underline the relation between $\mathrm{Ends}(S)$
and the space of ends of the adjacency graph of a pants decomposition of $S$
(see §3, theorem 4) and a variant of the Alexander method for infinite type
surfaces (see §5, theorem 4).
We refer the reader to [Fou1], [Fou2] and [Fuji] for previous work on groups
formed by mapping classes of infinite type surfaces. We want to stress,
however, that the cited authors focus their work on several subgroups of what
we here call the mapping class group (_e.g._ those with assymptotic qualities
for a specific surface or quasiconformal automorphisms of a Riemann surface)
and on their action on the Teichmüller space.
_Acknowledgements_. We want to thank Camilo Ramírez Maluendas for the question
that lead to the creation of this article. We are greateful to Hamish Short
and Javier Aramayona for carefully reading preliminary versions of this text.
The first author would like to thanks Daniel Juan Pineda for his support
during the realization of this project. The second author was generously
supported by LAISLA, CONACYT CB-2009-01 127991 and PAPIIT projects IN103411 &
IB100212 during the realization of this project.

##  2 Preliminaries

###  2.1 Topological invariants for infinite type surfaces

Let $X$ be a locally compact, locally connected, connected Hausdorff space.

######  Definition 2.1.

[F] Let $U_{1}\supseteq U_{2}\supseteq\ldots$ be an infinite sequence of non-
empty connected open subsets of $X$ such that for each $i\in\mathbb{N}$ the
boundary $\partial U_{i}$ is compact and
$\bigcap\limits_{i\in\mathbb{N}}\overline{U_{i}}=\emptyset$. Two such
sequences $U_{1}\supseteq U_{2}\supseteq\ldots$ and $U^{\prime}_{1}\supseteq
U^{\prime}_{2}\supseteq\ldots$ are said to be equivalent if for every
$i\in\mathbb{N}$ there exist $j,k$ such that $U_{i}\supseteq U^{\prime}_{j}$
and $U^{\prime}_{i}\supseteq U_{k}$. The corresponding equivalence class is
called a topological end of $X$.

The set of ends ${\rm Ends}(X)$ of $X$ can be endowed with a topology in the
following way. For any set $U$ in $X$ whose boundary is compact, we define
$U^{*}$ to be the set of all ends $[U_{1}\supseteq U_{2}\supseteq\ldots]$ for
which there is a representative such that $U_{n}\subset U$ for $n$
sufficiently large. With respect to this topology, ${\rm Ends}(X)$ is a
compact, closed, totally disconnected space without interior points (see for
example Theorem 1.5, [Ray]).
The genus of a surface is the maximum of the genera of its compact
subsurfaces. A surface is said to be planar if all of its compact subsurfaces
are of genus zero. We define $\mathrm{Ends}^{*}(S)\subset{\rm Ends}(S)$ as the
set of all ends which are not planar. As stated in the following theorem, any
orientable surface is determined, up to homeomorphism, by its genus, boundary
and space of ends. Henceforth all surfaces in this text are connected.

######  Theorem 3.

Let $S$ and $S^{\prime}$ be two orientable surfaces of the same genus. Then
$S$ and $S^{\prime}$ are homeomorphic if and only if they have the same number
of boundary components, and $\mathrm{Ends}^{*}(S)\subset{\rm Ends}(S)$ and
$\mathrm{Ends}^{*}(S^{\prime})\subset{\rm Ends}(S^{\prime})$ are homeomorphic
as nested topological spaces.

The proof of this theorem for the case when $S$ and $S^{\prime}$ have no
boundary can be found in [R]. The case for surfaces with boundary was proven
in [PM].

###  2.2 Complexes and graphs of curves

There are several curve complexes that one can associate to a surface of
finite genus, with finitely many boundary components and punctures. In this
section we extend the definitions of these complexes to noncompact surfaces of
infinite topological type and explore some of their basic properties.
Abusing language and notation, we will call curve, a topological embedding
$S^{1}\hookrightarrow S$, the isotopy class of this embedding and its image on
$S$. A curve is said to be _essential_ if it is neither homotopic to a point
nor to a boundary component. Hereafter all curves are considered essential
unless otherwise stated. An essential curve is said to be _separating_ if the
surface obtained by cutting $S$ along its image is disconnected. It is said to
be nonseparating otherwise. A separating curve $\alpha$ is said to be an
_outer separating_ curve if by cutting $S$ along $\alpha$ one of the resulting
connected components is a pair of pants (_i.e._ a genus 0 surface with three
boundary components). A non-outer separating curve is a separating curve which
is not an outer separating curve. Two curves are _disjoint_ if they are
distinct and their (geometric) intersection number is $0$.

######  Definition 2.2 (Multicurves).

A multicurve is either a set of just one curve, or a pairwise disjoint and
locally finite set of curves of $S$. We allow multicurves to consist of an
infinite set of curves. If $M$ is a multicurve of $S$, the surface obtained by
cutting $S$ along pairwise disjoint representatives of the elements of $M$
will be denoted by $S_{M}$.

Infinite _countable_ multicurves arise in surfaces with nonfinitely generated
fundamental group. Take for example the Loch Ness Monster, that is, a surface
with infinite genus and one end. If $S$ is a compact surface of genus $g$ with
$n$ boundary components, the _complexity_ of $S$, denoted by $\kappa(S)$, is
equal to $3g-3+n$. This is the cardinality of a maximal multicurve in $S$.

######  Definition 2.3 (The Curve Complex).

The Curve complex of $S$, $\mathcal{C}(S)$, is the abstract simplicial complex
whose vertices are the isotopy classes of essential curves in $S$, and whose
simplexes are multicurves of finite cardinality. We denote the set of vertices
of $\mathcal{C}(S)$ by $\mathcal{V}(\mathcal{C}(S))$.

The $1$-skeleton of $\mathcal{C}(S)$ will be denoted by $\mathcal{C}^{1}(S)$.
Since every automorphism of $\mathcal{C}(S)$ is determined uniquely by a
function of its vertices, and the same statement is true for automorphisms of
$\mathcal{C}^{1}(S)$, then the groups $\mathrm{Aut}(\mathcal{C}(S))$ and
$\mathrm{Aut}(\mathcal{C}^{1}(S))$ are isomorphic.

######  Definition 2.4 (The Nonseparating Curve Complex).

The Nonseparating curve complex of $S$, $\mathcal{N}(S)$, is the subcomplex of
$\mathcal{C}(S)$ whose vertices are the isotopy classes of essential
_nonseparating_ curves in $S$. We denote the set of vertices of
$\mathcal{N}(S)$ by $\mathcal{V}(\mathcal{N}(S))$.

######  Definition 2.5 (The Schmutz graph).

The Schmutz graph of $S$, $\mathcal{G}(S)$, is the simplicial graph whose
vertices are the isotopy classes of essential nonseparating curves in $S$, and
two vertices span an edge if their geometric intersection number is $1$.

######  Proposition 2.6.

Let $S$ be a surface of infinite genus. Then $\mathcal{C}(S)$,
$\mathcal{N}(S)$ and $\mathcal{G}(S)$ are connected. In particular
$\mathcal{C}^{1}(S)$ and $\mathcal{N}^{1}(S)$ have diameter 2 while
$\mathcal{G}(S)$ has diameter 4.

###### Proof.

Given any two distinct curves $\alpha$ and $\beta$ (either in
$\mathcal{V}(\mathcal{C}(S))$ or in $\mathcal{V}(\mathcal{N}(S))$), we can
always find a compact (finite genus) subsurface $S^{\prime}$ such that
contains $\alpha$ and $\beta$. Hence we can take an essential nonseparating
curve $\gamma$ on $S$ contained in $S\backslash S^{\prime}$ and not isotopic
to $\alpha$ and $\beta$. Therefore $\mathcal{C}^{1}(S)$ and
$\mathcal{N}^{1}(S)$ are connected,
$\mathrm{diam}(\mathcal{C}^{1}(S))=\mathrm{diam}(\mathcal{N}^{1}(S))=2$.
If $\alpha$ and $\beta$ are two distinct nonseparating curves, as in the
paragraph above, we can always find a curve $\gamma$ such that
$i(\alpha,\gamma)=i(\gamma,\beta)=0$; then we can always find curves
$\delta_{1}$ and $\delta_{2}$ such that
$i(\alpha,\delta_{1})=i(\delta_{1},\gamma)=i(\gamma,\delta_{2})=i(\delta_{2},\beta)=1$.
Hence $\mathcal{G}(S)$ is connected, $\mathrm{diam}(\mathcal{G}(S))\leq 4$. ∎

######  Remark 1.

Number 2 as diameter for $\mathcal{C}^{1}(S)$ and $\mathcal{N}^{1}(S)$ is
optimal, but 4 as diameter for $\mathcal{G}(S)$ is not necessarily optimal.

###  2.3 Mapping Class Group

Through this article, we will be working with the mapping class group of a
surface $S$. When $S$ is compact, this group has different (equivalent)
definitions, see for example [Farb], §2.1. In this paper we will be working
with the following definition.

######  Definition 2.7 (Mapping Class Group).

Let $S$ be a surface. Then $\mathrm{Homeo}^{+}(S,\partial S)$ is the group of
orientation-preserving homeomorphisms of $S$ that restrict to the identity on
the boundary, and $\mathrm{Homeo}(S)$ is the group of _all_ homeomorphisms of
$S$. The mapping class group of $S$, $\mathrm{MCG}(S)$ is the group
$\mathrm{Homeo}^{+}(S)/\thicksim$, where $\thicksim$ represents the isotopy
relation relative to the boundary. The extended mapping class group of $S$ is
the group $\mathrm{MCG}^{*}(S)\coloneqq\mathrm{Homeo}(S)/\thicksim$, where
$\thicksim$ represents the isotopy relation.

The group $\mathrm{MCG}^{*}(S)$ is incredibly big. As evidence for this we
have the following lemma and corollaries.

######  Lemma 2.8.

Let $S$ be an infinite genus surface and $F$ a subsurface of $S$ such that
$S\backslash F$ has genus at least $1$ and the boundary components of $F$ are
either boundary components of $S$ or essential curves of $S$. Then there
exists a subgroup of $\mathrm{MCG}^{*}(S)$ isomorphic to $\mathrm{MCG}(F)$,
with infinite index in $\mathrm{MCG}^{*}(S)$.

###### Proof.

The subgroup of $\mathrm{MCG}^{*}(S)$ formed by those orientation-preserving
elements $[h]\in\mathrm{MCG}^{*}(S)$ that have a representative $h$ with
support on $F$, is isomorphic to $\mathrm{MCG}(F)$. This subgroup will have
index greater or equal to the number of different elements in
$\mathrm{MCG}^{*}(S)$ that have its support in the interior of the complement
of $F$, thus it will have infinite index. ∎

######  Corollary 2.

Let $S$ be an infinite genus surface and $S_{g,n}$ be a compact surface of
genus $g$ and $n$ boundary components. Then
$\mathrm{MCG}^{*}(S_{g,n})<\mathrm{MCG}^{*}(S)$.

######  Corollary 3.

Let $S$ be an infinite genus surface,
$\\{(g_{i},n_{i})\\}_{i\in\mathbb{N}}\subset(\mathbb{N}\times\mathbb{Z}^{+})\backslash\\{(0,1)\\}$
be a sequence and $S_{i}$ be a compact orientable surface of genus $g_{i}$ and
$n_{i}$ boundary components. Then $\mathrm{MCG}^{*}(S)$ contains a subgroup
isomorphic to $\prod_{i\in\mathbb{N}}\mathrm{MCG}^{*}(S_{i})$.

##  3 Ends of adjacency graphs and surfaces

In this section we prove that, under the hypotheses
$\mathrm{Ends}(S)=\mathrm{Ends}^{*}(S)$, one can determine topologically
$\mathrm{Ends}(S)$ using the adjacency graph of a pants decomposition of $S$.

######  Definition 3.1 (Pants decomposition and the adjacency graph).

A _pants decomposition_ is a multicurve $P$ of maximal cardinality. We say
$\alpha,\beta\in P$ are adjacent with respect to $P$ if they bound the same
pair of pants in $S_{P}$. The adjacency graph of $P$, $\mathcal{A}(P)$, is the
simplicial graph whose vertex set is $P$ and two vertices span an edge if and
only if they are adjacent with respect to $P$. We say two nonseparating curves
form a _peripheral pair_ if they bound, along with a boundary component of
$S$, a pair of pants.

If $P$ is a pants decomposition, $S_{P}$ is the disjoint union of surfaces
homeomorphic to a pair of pants, for otherwise we contradict maximality. As an
abstract graph, $\mathcal{A}(P)$ is a subgraph of $\mathcal{C}^{1}(S)$, but we
have to keep in mind that adjacency of vertices in $\mathcal{A}(P)$ and
$\mathcal{C}^{1}(S)$ means different things for the corresponding curves in
$S$.

######  Remark 2.

It can be easily checked that the only cut points of an adjacency graph
$\mathcal{A}(P)$ are non-outer separating curves, and non-outer separating
curves are always cut points of any adjacency graph in which they are
vertices. Also, we can easily check outer separating curves always have degree
less or equal to two.

######  Theorem 4.

Let $S$ be an infinite genus surface such that
$\mathrm{Ends}(S)=\mathrm{Ends}^{*}(S)$ and $P$ be a pants decomposition of
$S$. Then $\mathrm{Ends}(\mathcal{A}(P))$ is homeomorphic to
$\mathrm{Ends}(S)$.

###### Proof.

For every pants decomposition there is a natural, but not canonical,
topological embedding:

$f:\mathcal{A}(P)\hookrightarrow S$ (2)

This embedding is illustrated in figure 1. Let $\Gamma$ be a subgraph of
$\mathcal{A}(P)$ whose boundary $\partial\Gamma$ is compact. We define
$S(\Gamma)$ as the subsurface of $S$ formed by all pants in $S$ (defined by
the multicurve $P$) that intersect $f(\Gamma)$, _deprived of its boundary_. By
definition $S(\Gamma)$ is an open subsurface of $S$ whose boundary is formed
by a finite collection of curves $\\{C_{1},\ldots,C_{n}\\}\subset P$. Remark
that if the graph $\Gamma$ is connected, so is $S(\Gamma)$. Moreover if
$\Gamma\supset\Gamma^{\prime}$ are two connected subgraphs of $\mathcal{A}(P)$
with compact boundaries we have that $S(\Gamma)\supset S(\Gamma^{\prime})$.
For every $[\Gamma_{1}\supseteq\Gamma_{2}\supseteq\ldots]$ in
$\mathrm{Ends}(\mathcal{A}(P))$ we define:

$f_{*}[\Gamma_{1}\supseteq\Gamma_{2}\supseteq\ldots]=[S(\Gamma_{1})\supseteq
S(\Gamma_{2})\supseteq\ldots]\in\mathrm{Ends}(S)$ (3)

Figure 1: A natural embedding of $\mathcal{A}(P)$ into $S$.

It follows directly from definition 2.1 that $f_{*}$ is well defined. We claim
that $f_{*}:\mathrm{Ends}(\mathcal{A}(P))\to\mathrm{Ends}(S)$ is an
homeomorphism. The injectivity of $f_{*}$ follows from the following general
lemma:

######  Lemma 3.2.

[F] Let $[U_{1}\supseteq U_{2}\supseteq\ldots]$ and $[U^{\prime}_{1}\supseteq
U^{\prime}_{2}\supseteq\ldots]$ be two different points in $\mathrm{Ends}(X)$.
Then there exists $i\in\mathbb{N}$ such that $U_{i}\cap
U^{\prime}_{i}=\emptyset$.

Indeed, let us suppose that
$[\Gamma_{1}\supseteq\Gamma_{2}\supseteq\ldots]\neq[\Gamma^{\prime}_{1}\supseteq\Gamma^{\prime}_{2}\supseteq\ldots]$.
Then there exists an $i\in\mathbb{N}$ such that
$\Gamma_{i}\cap\Gamma^{\prime}_{i}=\emptyset$. Let us suppose that
$[S(\Gamma_{1})\supseteq
S(\Gamma_{2})\supseteq\ldots]=[S(\Gamma^{\prime}_{1})\supseteq
S(\Gamma^{\prime}_{2})\supseteq\ldots]$. Hence, for the previous
$i\in\mathbb{N}$ there exist $l,k\in\mathbb{N}$ such that
$S(\Gamma_{i})\supseteq S(\Gamma^{\prime}_{l})$ and
$S(\Gamma^{\prime}_{i})\supseteq S(\Gamma_{k})$. Without loss of generality
suppose that $S(\Gamma^{\prime}_{l})\supseteq S(\Gamma^{\prime}_{i})$, hence
$S(\Gamma_{i})\cap S(\Gamma^{\prime}_{i})=S(\Gamma^{\prime}_{i})$ and, since
both $\partial\Gamma_{i}$ and $\partial\Gamma^{\prime}_{i}$ have compact
boundary we conclude that $\Gamma_{i}\cap\Gamma^{\prime}_{i}\neq\emptyset$.
This contradicts our initial assumption. The case where
$S(\Gamma^{\prime}_{i})\supseteq S(\Gamma^{\prime}_{l})$ is analogous.
We address now surjectivity. Let $[S_{1}\supseteq
S_{2}\supseteq\ldots]\in\mathrm{Ends}(S)$. Since there are no planar ends,
that is $\mathrm{Ends}(S)=\mathrm{Ends}^{*}(S)$, we can consider, for each
$S_{i}$ the surface $\mathfrak{S}_{i}$ formed by all pants in the pants in the
decomposition defined by $P$ that intersect $S_{i}$. Since $S_{i}$ is
connected, then $\mathfrak{S}_{i}$ must be connected. Also, since $\partial
S_{i}$ is compact, so is $\partial\mathfrak{S}_{i}$. Moreover, by definition,
if $i\leq j$ then $\mathfrak{S}_{i}\supseteq\mathfrak{S}_{j}$. Hence we have a
well defined end
$[\mathfrak{S}_{1}\supseteq\mathfrak{S}_{2}\supseteq\ldots]\in\mathrm{Ends}(S)$.
By construction, for every $i\in\mathbb{N}$ we have that
$S_{i}\subset\mathfrak{S}_{i}$. On the other hand, given $i\in\mathbb{N}$ we
can find $S_{j}$ such that $S_{i}\setminus S_{j}$ contains (properly) a
connected surface formed by pants in the pant decomposition defined by $P$.
This implies that there exists $k\in\mathbb{N}$ such that
$S_{j}\subset\mathfrak{S_{k}}$. Therefore
$[\mathfrak{S}_{1}\supseteq\mathfrak{S}_{2}\supseteq\ldots]=[S_{1}\supseteq
S_{2}\supseteq\ldots]$. Now define $\Gamma_{i}$ as the maximal subgraph of $S$
such that $f(\Gamma_{i})\subset\mathfrak{S}_{i}$. The graph $\Gamma_{i}$ has
compact boundary for $\mathfrak{S}_{i}$ has compact boundary and by definition
$f_{*}[\Gamma_{1}\supset\Gamma_{2}\supset\ldots]=[S_{1}\supseteq
S_{2}\supseteq\ldots]$. This proves that $f_{*}$ is a bijection.
Now we prove that $f_{*}$ is an homeomorphism. Let $\Gamma$ be a subgraph of
$\mathcal{A}(P)$ with compact boundary as before. We define

$\Gamma^{*}\coloneqq\\{[\Gamma_{1}\supseteq\Gamma_{2}\supseteq\ldots]\hskip
2.84526pt|\hskip 2.84526pt\Gamma\supseteq\Gamma_{i}\hskip 2.84526pt\text{for i
sufficiently big}\\}$ (4)

The collection of all $\Gamma^{*}$’s generates the topology of
$\mathrm{Ends}(\mathcal{A}(P))$. On the other hand we know from [R] that the
topology of $\mathrm{Ends}(S)$ is generated by

$U^{*}:=\\{[\hat{S}_{1}\supseteq\hat{S}_{2}\supseteq\ldots]\hskip
2.84526pt|\hskip 2.84526ptU\supseteq\hat{S}_{i}\hskip 2.84526pt\text{for i
sufficiently big}\\},$ (5)

where $U\subset S$ is an open subset with compact boundary. Clearly
$f_{*}\Gamma_{*}=S(\Gamma)^{*}$ , hence $f_{*}$ is open. From [Ray] we know
that both $\mathrm{Ends}(\mathcal{A}(P))$ and $\mathrm{Ends}(S)$ are compact
Hausdorff topological spaces. Hence $f_{*}$ is an homeomorphism. ∎

######  Remark 3.

We can think of punctures on a surface as planar ends, and hence the preceding
result is not true if we allow the surface $S$ to have them.

##  4 Proof of main results.

###  4.1 Injectivity.

In this section we will prove the following result:

######  Theorem 5.

Let $S$ be an infinite genus surface such that
$\mathrm{Ends}(S)=\mathrm{Ends}^{*}(S)$. The natural map:

$\Psi_{\mathcal{C}(S)}:{\rm MCG}^{*}(S)\to{\rm Aut}(\mathcal{C}(S))$ (6)

is injective.

Most of the proof of this theorem will rely in the following lemma and a
variant of the Alexander method (see [Farb] for details on this method).

######  Lemma 4.1.

Let $S$ be an infinite genus surface possibly with marked points and possibly
a finite number of boundary components. Let $\gamma_{1},\ldots,\gamma_{n}$ be
a collection of simple closed curves and simple proper arcs in $S$ such that
satisfy the three following properties:

  1. 1.

The $\gamma_{i}$ are in pairwise minimal position. That is, for $i\neq j$, the
(geometric) intersection of $\gamma_{i}$ with $\gamma_{j}$ is minimal within
their homotopy classes.

  2. 2.

The $\gamma_{i}$ are pairwise nonisotopic.

  3. 3.

For distinct $i,j,k$, at least one of $\gamma_{i}\cap\gamma_{j}$,
$\gamma_{i}\cap\gamma_{k}$, or $\gamma_{j}\cap\gamma_{k}$ is empty.

If $\gamma_{1}^{\prime},\ldots,\gamma_{n}^{\prime}$ is another such collection
so that $\gamma_{i}$ is isotopic to $\gamma_{i}^{\prime}$ for each $i$, then
there is an isotopy of $S$ that takes $\gamma_{i}^{\prime}$ to $\gamma_{i}$
for all $i$ simultaneously, and hence takes $\cup\gamma_{i}$ to
$\cup\gamma_{i}^{\prime}$.

A collection of curves $\gamma_{1},\ldots,\gamma_{n}$ satisfying (1)-(3) in
the preceding lemma will be called an _Alexander system_ in S. The proof of
this lemma is exactly the same as the proof of lemma 2.9 in [Farb].
Proof theorem 5. Let $h:S\rightarrow S$ be an homeomorphism such that
$h(\alpha)$ is isotopic to $\alpha$ for all
$\alpha\in\mathcal{V}(\mathcal{C}(S))$. For every infinite genus surface such
that $\mathrm{Ends}(S)=\mathrm{Ends}^{*}(S)$ we can find a family of compact
subsurfaces $\\{K_{i}\\}_{i\in\mathbf{N}}$ such that:

  * •

$S=\bigcup_{i\in\mathbf{N}}K_{i}$,

  * •

$K_{i}\subset K_{j}$ if $i<j$.

  * •

$K_{i}$ has genus at least $3$ for all $i\in\mathbf{N}$.

  * •

$K_{j}\setminus K_{i}$ admits at least one curve nonisotopic to any boundary
curve of $K_{j}$ for $i<j$.

  * •

Every boundary component of $K_{i}$ that is not a boundary component of $S$ is
an essential separating curve of $S$.

For each $i\in\mathbf{N}$ let us write $\partial K_{i}$ for the boundary of
$K_{i}$, $\partial_{S}K_{i}$ for all curves in $\partial K_{i}$ that are part
of the boundary of $S$ and $\partial_{i}K_{i}$ for $\partial
K_{i}\setminus\partial_{S}K_{i}$. Given such a family
$\\{K_{i}\\}_{i\in\mathbb{N}}$ of compact subsurfaces we can find
$\\{\Gamma_{i}\\}_{i\in\mathbb{N}}$ a collection of finite subsets of
$\mathcal{V}(\mathcal{C}(S))$ such that:

  * •

Every boundary component of $K_{i}$ that is not a boundary component of $S$,
is in $\Gamma_{j}$ for $i<j$ and is disjoint from every other curve in
$\cup_{i\in\mathbf{N}}\Gamma_{i}$.

  * •

$\Gamma_{0}$ fills $K_{0}$ and $\Gamma_{j}\setminus\Gamma_{j-1}$ fills
$K_{j}\setminus K_{j-1}$ for all $j>0$. In addition
$\Gamma_{i}\subset\Gamma_{j}$ for $i<j$.

  * •

If we cut $K_{j}\setminus K_{i}$ along $\Gamma_{j}\setminus\Gamma_{i}$ we
obtain either discs or annuli with one boundary component in $\partial K_{k}$,
for $i<j$ and some $k$ with $i\leq k\leq j$.

  * •

For all $\gamma\in(\Gamma_{j}\backslash\Gamma_{i})$ and
$\gamma^{\prime}\in\Gamma_{i}$, we have that $i(\gamma,\gamma^{\prime})=0$.
Moreover, if we define for each $i\in\mathbf{N}$

$\Gamma_{i}^{\prime}=\Gamma_{i}\cup\partial_{i}K_{i}$ (7)

then, for all $\gamma\in(\Gamma_{j}\backslash\Gamma_{i}^{\prime})$ and
$\gamma^{\prime}\in\Gamma_{i}^{\prime}$ we have $i(\gamma,\gamma^{\prime})=0$.

  * •

Both $\Gamma_{i}$ and $\Gamma_{i}^{\prime}$ are Alexander systems in $S$.

Figure 2 shows an example of $\\{K_{i}\\}_{i\in\mathbf{N}}$ and its
corresponding $\\{\Gamma_{j}\\}_{j\in\mathbf{N}}$.

$K_{0}$$K_{1}$ Figure 2: Example for $K_{0},K_{1},\ldots$ and
$\Gamma_{0},\Gamma_{1},\ldots$.

######  Lemma 4.2.

There exist homotopies $H_{i}:S\times[0,1]\rightarrow S$ such that:

  1. 1.

$H_{i}|_{S\times\\{0\\}}$ is the identity for $i\in\mathbf{N}$.

  2. 2.

$H_{i}|_{K_{i}\times\\{1\\}}=h|_{K_{i}}$ for $i\in\mathbf{N}$.

  3. 3.

$H_{i}|_{K_{i}\times[0,1]}=H_{j}|_{K_{i}\times[0,1]}$ for $i<j$.

The proof of this lemma is rather technical, the main difficulty being to
prove that $H_{i}|_{K_{i}\times[0,1]}=H_{j}|_{K_{i}\times[0,1]}$ for $i<j$. We
leave it for later. We will use the lemma to finish the proof of theorem 5.
For every $x\in S$ there exist $i\in\mathbf{N}$ such that $x\in K_{i}$ and
$x\notin\partial_{i}K_{i}$. Define $H:S\times[0,1]\rightarrow S$ as
$H(x,t)=H_{i}(x,t)$. From (3) in the preceding lemma we deduce that $H$ is
well-defined. The function $H$ is clearly continuous, $H|_{S\times\\{0\\}}$ is
the identity and $H|_{S\times\\{1\\}}=h$. Thus $H$ is an homotopy from the
identity to $h$. This, modulo the proof of lemma 4.2, finishes the proof of
theorem 5.

Proof of lemma 4.2. The idea of the proof is a variant of the Alexander method
(see [Farb] for details on this method). By hypothesis, for every
$\gamma\in\mathcal{C}(S)$ , the curves $\gamma$ and $h(\gamma)$ are isotopic.
Using lemma 4.1 we can assure the existence, for each $i\in\mathbf{N}$, of an
isotopy $\tilde{H}_{i}:S\times[0,1]\rightarrow S$, that takes $\gamma$ to
$h(\gamma)$ for all $\gamma\in\Gamma_{i}^{\prime}$ simultaneously. Moreover,
since for all $\gamma\in(\Gamma_{j}\backslash\Gamma_{i}^{\prime})$ and
$\gamma^{\prime}\in\Gamma_{i}^{\prime}$ we have $i(\gamma,\gamma^{\prime})=0$,
we can ask

$\displaystyle\tilde{H}_{i|\hskip
2.84526pt_{\Gamma_{i}^{\prime}\times[0,1]}}=\tilde{H}_{j|\hskip
2.84526pt_{\Gamma_{i}^{\prime}\times[0,1]}},\hskip 22.76219pt\text{for
$i<j$}.$ (8)

In other words, the homotopies can be chosen so that $\tilde{H}_{i}$ moves the
curves in $\Gamma_{i}^{\prime}$ at exactly the same time as $\tilde{H}_{j}$
moves the curves in $\Gamma_{i}^{\prime}$ for $i<j$. Let us define
$f_{i}:={H}_{i|S\times\\{1\\}}$. Remark that $h^{-1}\circ f_{i}$ fixes all the
points in $\Gamma_{i}^{\prime}$. On the other hand, $h$ has to be orientation-
preserving, since otherwise for every compact subsurface
$S^{\prime}\hookrightarrow S$ we could find an homeomorphism that reverses
orientation and at the same time acts trivially on $\mathcal{C}(S^{\prime})$,
which is not possible if $S^{\prime}$ has genus bigger than 3 and at least one
boundary component. Hence $h^{-1}\circ f_{i}$ is orientation-preserving and,
by the same argument used by Farb and Margalit (see proof proposition 2.8, p.
62-63, [Farb]), we have that $h^{-1}\circ f_{i}$ sends each connected region
in $S\setminus\Gamma_{i}^{\prime}$ to itself. By hypotheses $\Gamma_{0}$ fills
$K_{0}$ and $\Gamma_{j}\setminus\Gamma_{j-1}$ fills $K_{j}\setminus K_{j-1}$
for all $j>1$. Hence:

$S\setminus\Gamma_{i}^{\prime}=\left(\bigsqcup_{k=1}^{n_{i}}A_{k}\right)\sqcup\left(\bigsqcup_{k=1}^{m_{i}}D_{k}\right)\sqcup
S_{i}$ (9)

where each $D_{k}$ is homeomorphic to a disc, each $A_{k}$ is homeomorphic to
an annulus and $S_{i}=S\setminus K_{i}$ is an infinite genus surface.
Furthermore:

  1. 1.

The boundary of each disc $D_{k}$ is formed by segments contained in
$\Gamma_{i}$.

  2. 2.

The boundary of each annulus $A_{k}$ is either contained in
$\Gamma_{i}^{\prime}$ or one of its connected components is also a connected
component of the boundary of $S$.

From Alexander’s lemma, we deduce that $h^{-1}\circ f_{i}$ restricted to
$D_{k}$ is isotopic to $Id_{|D_{k}}$. When $A_{k}$ shares a boundary component
with $S$, the restriction of $h^{-1}\circ f_{i}$ to $A_{k}$ is isotopic to the
identity, for we are allowed to perform isotopies on $A_{k}$ that do not fix
the boundary of $S$ pointwise. Finally, when $A_{k}$ shares no boundary
component with the boundary of $S$ the restriction of $h^{-1}\circ f_{i}$ to
$A_{k}$ is also isotopic to the identity for else this restriction will be a
non-trivial Dehn twist and we could then find a curve
$\gamma\in\mathcal{C}(S)$ intersecting the interior of $A_{k}$ which is not
fixed by $h^{-1}$. From this three facts we conclude that $h^{-1}\circ f_{i}$
is isotopic to the identity in $K_{i}$ and hence $f_{i}$ is isotopic to $h$ in
$K_{i}$. The composition of these two isotopies form the desired isotopy
$H_{i}$.∎

###  4.2 Rigidity.

In this section we give the proof of theorem 2 and corollary 1. This requires
some auxiliary facts and lemmas, that we state and prove in the following
paragraphs.
Through this section $S_{1}$ and $S_{2}$ will denote (connected) infinite
genus surfaces with a finite number of boundary components and
$\phi:\mathcal{C}(S_{1})\rightarrow\mathcal{C}(S_{2})$ an isomorphism. We
remark that the image via $\phi$ of any pants decomposition of $S_{1}$ is a
pants decomposition of $S_{2}$. Moreover, if $P$ is a pants decomposition of
$S_{1}$, then $\alpha,\beta\in P$ are adjacent with respect to $P$ if and only
if $\phi(\alpha)$ and $\phi(\beta)$ are adjacent with respect to $\phi(P)$.
The sufficiency of this statement can be found in [shackleton] and the
necessity follows from the fact that we are dealing with an isomorphism of the
curve complex. Therefore
$\phi:\mathcal{C}(S_{1})\rightarrow\mathcal{C}(S_{2})$ induces a map

$\varphi:\mathcal{A}(P)\rightarrow\mathcal{A}(\phi(P))$ (10)

as follows: $\alpha\mapsto\varphi(\alpha):=\phi(\alpha)$. Moreover, $\varphi$
is an isomorphism. For this reason cut points of $\mathcal{A}(P)$ go to cut
points under $\phi$ and this isomorphism sends:

  1. 1.

Non-outer separating curves to non-outer separating curves.

  2. 2.

Nonseparating curves to nonseparating curves.

  3. 3.

Outer curves to outer curves.

The proof of (1) and (2) can be found in [shackleton], where as (3) follows
from (1), (2) and the fact that $\phi$ is an isomorphism. The following lemmas
can be deduced from the work of Irmak (see [Irmak]), but since we use them
several times later, we present elementary and simple proofs.

######  Lemma 4.3.

Let $S_{1}$ and $S_{2}$ be infinite genus surfaces and let
$\phi:\mathcal{C}(S_{1})\rightarrow\mathcal{C}(S_{2})$ be an isomorphism. If
$\alpha$, $\beta$ and $\gamma$ are curves that bound a pair of pants on
$S_{1}$, then their images bound a pair of pants on $S_{2}$.

###### Proof.

If $\alpha\neq\beta=\gamma$, then $\beta$ cannot be an outer curve and hence
its image is not an outer curve. Also, in any pants decomposition $P$ the
curve $\beta$ will have degree one as a vertex of $\mathcal{A}(P)$. Hence
$\phi(\beta)$ will also have degree one as vertex of $\mathcal{A}(\phi(P))$,
given that (10) is an isomorphism. Then, the only option left is for $\beta$
to be the boundary of a pair of pants twice, as in the option of the left in
figure 3. Therefore $\phi(\alpha)$ and $\phi(\beta)=\phi(\gamma)$ bound a pair
of pants on $S_{2}$.

$v$$u$$v$$u$ Figure 3: The two options for $\deg(v)=1$.

It is impossible to bound a pair of pants using two separating curves and one
nonseparating curve. Hence, if $\alpha\neq\beta\neq\gamma\neq\alpha$, we only
have the following cases according to the number of separating curves:

  1. 1.

Three separating curves. In this case, $\phi(\alpha)$, $\phi(\beta)$ and
$\phi(\gamma)$ are three different separating curves, since separating curves
go to separating curves as mentioned before. If these curves did not bound a
pair of pants on $S_{2}$ we would have a pair of pants bounded by
$\phi(\alpha)$ and $\phi(\beta)$ but not bounded by $\phi(\gamma)$, another
pair of pants bounded by $\phi(\beta)$ and $\phi(\gamma)$ but not bounded by
$\phi(\alpha)$ and another pair of pants bounded by $\phi(\gamma)$ and
$\phi(\alpha)$ but not bounded by $\phi(\beta)$, as in figure 4. But then none
of these curves would be separating, leading us to a contradiction. Hence,
$\phi(\alpha)$, $\phi(\beta)$ and $\phi(\gamma)$ bound a pair of pants on
$S_{2}$.

  2. 2.

One separating curve. Let $\alpha$ and $\gamma$ be nonseparating curves and
let $\beta$ be a separating curve. Then $\phi(\alpha)$ and $\phi(\gamma)$ are
nonseparating curves and $\phi(\beta)$ is a separating curve, given the
properties of $\phi$ mentioned before. If these curves did not bound a pair of
pants on $S_{2}$, we would have a pair of pants bounded by $\phi(\alpha)$ and
$\phi(\beta)$ but not bounded by $\phi(\gamma)$, another pair of pants bounded
by $\phi(\beta)$ and $\phi(\gamma)$ but not bounded by $\phi(\alpha)$, but
since $\phi(\beta)$ is a separating curve there cannot exist a pair of pants
bounded by both $\phi(\alpha)$ and $\phi(\gamma)$, given that they are on
different connected components of $S_{2}\backslash\\{\phi(\beta)\\}$, which
leads us to a contradiction ($\phi(\alpha)$ and $\phi(\gamma)$ must be
adjacent). Then $\phi(\alpha)$, $\phi(\beta)$ and $\phi(\gamma)$ bound a pair
of pants on $S_{2}$.

  3. 3.

Three nonseparating curves. Given that $\alpha$, $\beta$ and $\gamma$ are
nonseparating curves, we can always find a pants decomposition $P$ such that
all their neighbours in $\mathcal{A}(P)$ are nonseparating, $\alpha$ and
$\beta$ have degree three in $\mathcal{A}(P)$, $\gamma$ has degree four in
$\mathcal{A}(P)$ and $\alpha$ and $\gamma$ only have one common neighbour
$\beta$ in $\mathcal{A}(P)$. For an example consider figure 5. Then,
$\phi(\alpha)$ and $\phi(\beta)$ have degree three, $\phi(\gamma)$ has degree
four, and all their neighbours are nonseparating. If $\phi(\alpha)$,
$\phi(\beta)$ and $\phi(\gamma)$ do not bound a pair of pants on $S_{2}$ then
there exist a pair of pants bounded by $\phi(\alpha)$, $\phi(\beta)$ and
$\delta_{1}\neq\phi(\gamma)$, another pair of pants bounded by $\phi(\beta)$,
$\phi(\gamma)$ and $\delta_{2}\neq\phi(\alpha)$, and another pair of pants
bounded by $\phi(\alpha)$, $\phi(\gamma)$ and $\delta_{3}\neq\phi(\beta)$.
Since $\phi(\beta)$ is the only common neighbour of $\phi(\alpha)$ and
$\phi(\gamma)$, then $\delta_{3}$ is not an essential curve, which means it is
isotopic to a boundary component, but this leads us to a contradiction, since
$\phi(\gamma)$ would then have degree at most $3$.

$\phi(\gamma)$$\phi(\alpha)$$\phi(\beta)$$\delta_{3}$$\delta_{2}$$\delta_{1}$
Figure 4: If $\phi(\alpha)$, $\phi(\beta)$ and $\phi(\gamma)$ do not bound a
pair of pants. $\alpha$$\gamma$$\beta$ Figure 5: Three nonseparating curves
bounding a pair of pants.

∎

######  Remark 4.

If $P$ is a pants decomposition and $\alpha\in P$ is a nonseparating curve of
degree $2$ in $\mathcal{A}(P)$ such that its neighbours are also nonseparating
curves, then $\alpha$ forms part of two peripheral pairs, namely one with each
neighbour (otherwise either of its neighbours or $\alpha$ itself would become
separating).

######  Lemma 4.4.

Let $S_{1}$ and $S_{2}$ be infinite genus surfaces and let
$\phi:\mathcal{C}(S_{1})\rightarrow\mathcal{C}(S_{2})$ be an isomorphism. If
$\alpha$ and $\beta$ form a peripheral pair, then their images form a
peripheral pair. In particular, $S_{1}$ and $S_{2}$ have the same number of
boundary components.

###### Proof.

If $S_{1}$ admits at least $2$ peripheral pairs such that their curves are
pairwise disjoint as in figure 6, then we can always find a pants
decomposition $P$ of $S_{1}$ such that all the neighbours of $\beta$ are
nonseparating, $\deg(\alpha)=3$ and $\deg(\beta)=2$. Then all the neighbours
of $\phi(\beta)$ are nonseparating, and $\phi(\beta)$ has degree $2$, hence it
has to form a peripheral pair with $\phi(\alpha)$ by the previous remark.
If for any two peripheral pairs in $S_{1}$ at least one curve of each pair
intersect each other, we can always find a pants decomposition $P$ of $S_{1}$
such that all the neighbours of $\alpha$ and $\beta$ are nonseparating,
$\deg(\alpha)=\deg(\beta)=3$, and there is only one pair of pants in
$S_{1}\backslash P$ that is bounded by $\alpha$ and $\beta$ at the same time,
namely the one formed by $\alpha$ and $\beta$ being a peripheral pair. Then
$\phi(P)$ is a pants decomposition with all the neighbours of $\phi(\alpha)$
and $\phi(\beta)$ being nonseparating,
$\deg(\phi(\alpha))=\deg(\phi(\beta))=3$ and there exists a pair of pants in
$S_{2}$ bounded by $\phi(\alpha)$, $\phi(\beta)$ and $\delta$. Due to lemma
4.3 applied to $\phi$ and $\phi^{-1}$, $\delta$ cannot be an essential curve
different to both $\phi(\alpha)$ and $\phi(\beta)$, but if
$\delta=\phi(\alpha)$ or $\delta=\phi(\beta)$ then either $\phi(\beta)$ of
$\phi(\alpha)$, respectively, becomes separating. Then $\delta$ is isotopic to
a boundary component and so, $\phi(\alpha)$ and $\phi(\beta)$ form a
peripheral pair.
This result implies that $S_{2}$ has at least as many boundary components as
$S_{1}$, and applying the same result to $\phi^{-1}$ we get that they have the
same number of boundary components, even if this number is infinite. ∎

$\beta$$\alpha$ Figure 6: Example of a convenient pants decomposition.

Proof of theorem 2. Let $P$ be a pants decomposition of $S_{1}$. From the fact
that (10) is an isomorphism and theorem 4 we have
$\mathrm{Ends}(S_{1})\cong\mathrm{Ends}(\mathcal{A}(P))\cong\mathrm{Ends}(\mathcal{A}(\phi(P)))\cong\mathrm{Ends}(S_{2})$.
From the surface classification theorem for infinite surfaces by Richards,
Prishlyak and Mischenko in [Ray] and [PM], it is sufficient to prove that
$S_{1}$ and $S_{2}$ have the same number of boundary components to guarantee
that they are homeomorphic. But this is guaranteed by lemma 4.4.∎

######  Remark 5.

Theorem 2 cannot be extended for infinite genus surfaces with punctures.
Indeed, let $S$ be an infinite genus surface with $n>0$ boundary components
and without planar ends. Let $S^{\prime}$ be the infinite genus surface
obtained from $S$ by glueing one punctured disc to $S$ along a boundary
component. Clearly $S$ and $S^{\prime}$ are not homeomorphic, but
$\mathcal{C}(S)\cong\mathcal{C}(S^{\prime})$.

Proof of corollary 1. The statement is immediate for all arguments given in
the proof of theorem 2 remain valid if we change $\mathcal{C}(S)$ for
$\mathcal{N}(S)$ and take all pants decompositions to be formed just by
nonseparating curves. ∎

###  4.3 Surjectivity.

At the end of this section we give a proof for theorem 1. We begin by proving
the following theorems:

######  Theorem 6.

Let $S$ be an infinite genus surface. Then
$\mathrm{Aut}(\mathcal{G}(S))\cong\mathrm{Aut}(\mathcal{N}(S))$.

######  Theorem 7.

Let $S$ be an infinite genus surface such that
$\mathrm{Ends}(S)=\mathrm{Ends}^{*}(S)$. The natural map:

$\Psi_{\mathcal{G}(S)}:{\rm MCG}^{*}(S)\to{\rm Aut}(\mathcal{G}(S))$ (11)

is surjective.

This two results imply that the natural map:

$\Psi_{\mathcal{N}(S)}:{\rm MCG}^{*}(S)\to{\rm Aut}(\mathcal{N}(S))$ (12)

is surjective. Using the surjectivity of this map, we can deduce the
following:

######  Theorem 8.

Let $S$ be an infinite genus surface such that
$\mathrm{Ends}(S)=\mathrm{Ends}^{*}(S)$. The natural map:

$\Psi_{\mathcal{C}(S)}:{\rm MCG}^{*}(S)\to{\rm Aut}(\mathcal{C}(S))$ (13)

is surjective.

####  4.3.1 Proof of theorems 6 and 7.

The proofs of theorems 6 and 7 require some auxiliary lemmas given in [Irmak]
and [Sch] but adapted to the context of infinite type surfaces. When the
proofs of these lemmas can be easily deduced from the cited works we just
state them without a proof. When this is not the case elementary and simple
proofs are provided. We recall first the different components that a curve
might have.

######  Definition 4.5 (Curve components).

Let $\alpha$ and $\beta$ be nonseparating curves such that
$i(\alpha,\beta)\geq 2$. Let $\beta_{1}$ be a connected component of $\beta$
in $S_{\alpha}$. If the surface resulting from cutting $S_{\alpha}$ along
$\beta_{1}$ is connected, then $\beta_{1}$ is called a nonseparating component
of $\beta$ (with respect to $\alpha$). Otherwise, $\beta_{1}$ is called a
separating component of $\beta$ (with respect ot $\alpha$). If $\beta_{1}$
connects the two different boundary components of $S_{\alpha}$ induced by
$\alpha$, then $\beta_{1}$ is called a two-sided component. Otherwise it is
called one-sided.

######  Lemma 4.6.

[Sch] Let $S$ be an infinite genus surface and
$\alpha,\beta\in\mathcal{V}(\mathcal{N}(S))$ such that $i(\alpha,\beta)\geq
2$. If $\beta$ has a nonseparating component $\beta_{1}$ with respect to
$\alpha$, then there exists
$\gamma,\gamma^{\prime}\in\mathcal{V}(\mathcal{N}(S))\backslash\\{\alpha,\beta\\}$
such that $N(\alpha,\beta)\subset(N(\gamma)\cup N(\gamma^{\prime}))$.
Moreover, if $\beta_{1}$ is one-sided, then $\alpha,\gamma,\gamma^{\prime}$
are mutually disjoint; if $\beta_{1}$ is two-sided, then
$\\{\alpha,\gamma,\gamma^{\prime}\\}$ is a triple with

$i(\alpha,\beta)=i(\beta,\gamma)+i(\beta,\gamma^{\prime})$

and $\min\\{i(\beta,\gamma),i(\beta,\gamma^{\prime})\\}>0$.

######  Lemma 4.7.

_[_Ibid._]_ Let $S_{1}$ and $S_{2}$ be infinite genus surfaces and let
$\phi:\mathcal{G}(S_{1})\rightarrow\mathcal{G}(S_{2})$ be an isomorphism. Then
for any disjoint curves $\alpha$ and $\beta$, their images under $\phi$ will
also be disjoint.

Proof theorem 6. Let $\phi\in\mathrm{Aut}(\mathcal{N}(S))$. Since any
automorphism of $\mathcal{N}(S)$ (and $\mathcal{G}(S)$ respectively) is
uniquely determined by the function on its vertices and
$\mathcal{V}(\mathcal{N}(S))=\mathcal{V}(\mathcal{G}(S))$, then $\phi$ induces
a bijection $\phi^{*}:\mathcal{G}(S)\rightarrow\mathcal{G}(S)$. From the work
of Irmak [Irmak] on the characterization of two curves that intersect once,
one can deduce that if $S_{1}$ and $S_{2}$ are infinite genus surfaces, and
$\phi_{1}:\mathcal{C}(S_{1})\rightarrow\mathcal{C}(S_{2})$ and
$\phi_{2}:\mathcal{N}(S_{1})\rightarrow\mathcal{N}(S_{2})$ are isomorphisms,
then for any curves $\alpha_{1}$ and $\alpha_{2}$ such that
$i(\alpha_{1},\alpha_{2})=1$ we have that
$i(\phi_{1}(\alpha_{1}),\phi_{1}(\alpha_{2}))=i(\phi_{2}(\alpha_{1}),\phi_{2}(\alpha_{2}))=1$.
This fact applied to $\phi$ and $\phi^{-1}$ implies that $\phi^{*}$ must
preserve adjacency and non-adjacency. Hence we can define the function

$\Phi:\mathrm{Aut}(\mathcal{N}(S))\rightarrow\mathrm{Aut}(\mathcal{G}(S))$
(14)

as $\phi\mapsto\phi^{*}$. This function is clearly an injective group
homomorphism.
In the same way, for any automorphism of $\mathcal{G}(S)$ we can induce a
bijection from $\mathcal{N}(S)$ to itself, and due to lemma 4.7 this bijection
will become an automorphism of $\mathcal{N}(S)$. Therefore $\Phi$ is an
isomorphism. ∎

######  Remark 6.

From the proof of theorem 6 and the proof of corollary 1 we conclude that the
statements of lemmas 4.3 and 4.4 remain valid if we change $\mathcal{C}(S)$
for $\mathcal{G}(S)$.

The following four lemmas are used in the proof of theorem 7. Let us recall
first the notion of triple of curves.

######  Definition 4.8 (Triples of curves).

Let $\alpha$, $\beta$ and $\gamma$ be nonseparating curves of $S$. We will say
$\\{\alpha,\beta,\gamma\\}$ is a triple if
$i(\alpha,\beta)=i(\alpha,\gamma)=i(\beta,\gamma)=1$ and there exists a
subsurface of $S$ which contains $\alpha$, $\beta$ and $\gamma$, and is
homeomorphic to a torus with one boundary component.

######  Lemma 4.9.

Let $S$ be infinite genus surface and
$\alpha,\beta\in\mathcal{V}(\mathcal{N}(S))$ be such that $i(\alpha,\beta)\geq
2$. If $\beta$ does not have two-sided components with respect to $\alpha$,
then there exists
$\gamma,\gamma^{\prime}\in\mathcal{V}(\mathcal{N}(S))\backslash\\{\alpha,\beta\\}$
such that $\\{\alpha,\gamma,\gamma^{\prime}\\}$ is a triple with

$i(\alpha,\beta)=i(\beta,\gamma)+i(\beta,\gamma^{\prime})$

and $\min\\{i(\beta,\gamma),i(\beta,\gamma^{\prime})\\}>0$.

###### Proof.

Let $\alpha_{1}$ and $\alpha_{2}$ be the boundary components on $S_{\alpha}$
induced by $\alpha$. Since $\beta$ does not have two-sided components then it
only has one-sided components and therefore we can choose a curve $\gamma$
that intersects $\alpha$ once, does not intersect any one sided component of
$\beta$ based on $\alpha_{1}$ and intersects $\beta$ in such a way that
$0<i(\gamma,\beta)\leq\frac{1}{2}i(\alpha,\beta)$. This can be done by drawing
$\gamma$ disjoint from every one-sided component of $\beta$ based on
$\alpha_{1}$, we keep on going “following” a convenient one-sided component of
$\beta$ based on $\alpha_{2}$ until before we reach $\alpha_{2}$, then we
intersect $\alpha_{2}$ in the corresponding point seeking the desired
inequality. See figure 7 for examples.

$\beta$$\gamma$$\beta$$\beta$$\gamma$$\beta$ Figure 7: Examples of $\beta$ and
$\gamma$ in $S_{\alpha}$

Then let $N$ be a regular neighbourhood of $\alpha$ and $\gamma$; since
$i(\alpha,\gamma)=1$ then $N$ is homeomorphic to a torus with one boundary
component. Let $\gamma^{\prime}$ be the image of $\gamma$ under a Dehn twist
along $\alpha$ on $N$. See figure 8 for the corresponding diagram.

$\alpha$$\gamma$$\alpha$$\gamma^{\prime}$$\beta$ Figure 8: Diagram of $N$.

Thus $\\{\alpha,\gamma,\gamma^{\prime}\\}$ form a triple and by construction
$i(\beta,\gamma^{\prime})=i(\alpha,\beta)-i(\beta,\gamma)$, with both curves
intersecting $\beta$ at least once. ∎

######  Lemma 4.10.

Let $S$ be an infinite genus surface and let
$\phi:\mathcal{G}(S)\rightarrow\mathcal{G}(S)$ be an automorphism. Then
$i(\alpha,\beta)=i(\phi(\alpha),\phi(\beta))$ for all
$\alpha,\beta\in\mathcal{V}(\mathcal{G}(S))$.

###### Proof.

Let $\alpha,\beta\in\mathcal{V}(\mathcal{G}(S))$. If $i(\alpha,\beta)=0$, then
due to lemma 4.7 we have that $i(\phi(\alpha),\phi(\beta))=0$. If
$i(\alpha,\beta)=1$, then due to $\phi$ being an automorphism
$i(\phi(\alpha),\phi(\beta))=1$. For $i(\alpha,\beta)\geq 2$, we will proceed
by induction on the geometric intersection number.
Let us suppose the geometric intersection number is preserved under
automorphisms for curves which intersect at most $k$ times for a $k\geq 1$.
Let $i(\alpha,\beta)=k+1$. Due to lemmas 4.6 and 4.9, we know there exists
$\gamma,\gamma^{\prime}\in\mathcal{V}(\mathcal{G}(S))\backslash\\{\alpha,\beta\\}$
such that $\\{\alpha,\gamma,\gamma^{\prime}\\}$ is a triple,
$i(\alpha,\beta)=i(\beta,\gamma)+i(\beta,\gamma^{\prime})$ and
$\min\\{i(\beta,\gamma),i(\beta,\gamma^{\prime})\\}>0$.
Since $i(\beta,\gamma),i(\beta,\gamma^{\prime})<k+1$, then
$i(\beta,\gamma)=i(\phi(\beta),\phi(\gamma))$ and
$i(\beta,\gamma^{\prime})=i(\phi(\beta),\phi(\gamma^{\prime}))$. From the work
of Schmutz [_Ibid_.] one can deduce that if $S$ is an infinite genus surface
and $\phi:\mathcal{G}(S)\rightarrow\mathcal{G}(S)$ an automorphism, then for
every triple $\\{\alpha,\beta,\gamma\\}$ we have that
$\\{\phi(\alpha),\phi(\beta),\phi(\gamma)\\}$ form a triple. Therefore
$\\{\phi(\alpha),\phi(\gamma),\phi(\gamma^{\prime})\\}$ form a triple. Using a
diagram of the torus with one boundary component which contains this triple
(see figure 8), we can see that each time $\phi(\beta)$ intersects
$\phi(\alpha)$ then either $\phi(\beta)$ intersects $\phi(\gamma)$ or
$\phi(\beta)$ intersects $\phi(\gamma^{\prime})$. Therefore
$i(\phi(\beta),\phi(\gamma))+i(\phi(\beta),\phi(\gamma^{\prime}))\geq
i(\phi(\alpha),\phi(\beta))$. Thus $i(\alpha,\beta)\geq
i(\phi(\alpha),\phi(\beta))$. Applying the same argument on $\phi^{-1}$ we
obtained the symmetric inequality, therefore
$i(\alpha,\beta)=i(\phi(\alpha),\phi(\beta))$. ∎

######  Lemma 4.11.

Let $S$ be an infinite genus surface and let
$\phi:\mathcal{G}(S)\rightarrow\mathcal{G}(S)$ be an automorphism. If $P$ is a
pants decomposition of $S$, then there exist an homeomorphism
$h\in\mathrm{MCG}^{*}(S)$ such that $h(\alpha)=\phi(\alpha)$ for all
$\alpha\in P$.

###### Proof.

From remark 6, we know that $\phi(P)$ is a pants decomposition and the
boundaries of pair of pants in $S_{\phi(P)}$ induced by curves of $\phi(P)$
are boundaries of pair of pants in $S_{P}$ induced by curves of $P$. Then we
can define an homeomorphism of $S$ by parts using homeomorphisms from the
connected components of $S_{P}$ to the corresponding connected components of
$S_{\phi(P)}$; this homeomorphism by construction will agree with $\phi$ for
every element in $P$. ∎

######  Remark 7.

It is clear, using theorem 6, that this lemma remains valid if we substitute
$\mathcal{G}(S)$ by $\mathcal{N}(S)$.

######  Lemma 4.12.

[Sch] Let $S^{\prime}$ be a surface of genus zero and four boundary
components. Let $\alpha,\beta\in\mathcal{V}(\mathcal{C}(S^{\prime}))$ with
$i(\alpha,\beta)=2$.

  1. 1.

Let $\gamma\in\mathcal{V}(\mathcal{C}(S^{\prime}))$ such that
$i(\alpha,\gamma)=2$. Then there exists $h\in\mathrm{MCG}^{*}(S^{\prime})$
such that $h(\alpha)=\alpha$ and $h(\beta)=\gamma$.

  2. 2.

There are exactly two curves
$\gamma_{1},\gamma_{2}\in\mathcal{V}(\mathcal{C}(S^{\prime}))$ such that
$i(\alpha,\gamma_{i})=i(\beta,\gamma_{i})=2$ for $i=1,2$. Moreover, there
exists $h\in\mathrm{MCG}^{*}(S^{\prime})$ such that $h(\alpha)=\alpha$,
$h(\beta)=\beta$, and $h(\gamma_{1})=\gamma_{2}$.

######  Remark 8.

The homeomorphism of part (1) in the preceding lemma is just a Dehn twist
about $\alpha$, where as the homeomorphism from part (2) is an orientation-
reversing involution that leaves invariant each connected component in the
boundary of $S_{0,4}$.

The proof theorem 7 uses the notion of Dehn-Thurston coordinates. Therefore we
recall it and discuss it briefly in the context of infinite surfaces in the
following paragraphs.

######  Definition 4.13 (Dehn-Thurston coordinates).

A Dehn-Thurston coordinates system of curves is a set $D$ of curves that
parametrize every curve $\alpha\in\mathcal{V}(\mathcal{C}(S))$ using the
geometric intersection number, i.e. for
$\alpha,\beta\in\mathcal{V}(\mathcal{C}(S))$ if
$i(\alpha,\gamma)=i(\beta,\gamma)$ for all $\gamma\in D$, then $\alpha=\beta$.

For compact surface, it is well known that Dehn-Thurston coordinate systems
exist, see [HP]. For noncompact surfaces such a system of curves can be
realized in the following way. Let $\\{\alpha_{i}\\}_{i\in\mathbf{N}}$ be a
pants decomposition, $\\{\beta_{i}\\}_{i\in\mathbf{N}}$ be curves such that
$i(\alpha_{i},\beta_{i})=2$ and $i(\alpha_{i},\beta_{j})=0$ for $i\neq j$, and
$\\{\gamma_{i}\\}_{i\in\mathbf{N}}$ be curves such that
$i(\alpha_{i},\gamma_{i})=i(\beta_{i},\gamma_{i})=2$ and
$i(\alpha_{i},\gamma_{j})=0$ for $i\neq j$. Then the set of curves $D$ formed
by the union of elements in $\\{\alpha_{i}\\}_{i\in\mathbf{N}}$,
$\\{\beta_{i}\\}_{i\in\mathbf{N}}$ and $\\{\gamma_{i}\\}_{i\in\mathbf{N}}$ is
a Dehn-Thurston coordinate system. Indeed, any curve $\delta$ in $S$ will only
intersect finitely many curves in $D$, hence we can take any compact
subsurface $S^{\prime}$, such that it contains $\delta$ and there is a
(finite) subset $D^{\prime}$ of $D$ that is a Dehn-Thurston coordinate system
of $S^{\prime}$. Any other curve in $S$ with the same Dehn-Thurston
coordinates as $\delta$ on the system $D$, would have to be isotopic to a
curve contained in $S^{\prime}$ and thus would have the same Dehn-Thurston
coordinates as $\delta$ on the system $D^{\prime}$, therefore it would be
isotopic to $\delta$. We must remark that, when $S$ is an infinite genus
surface such that $\mathrm{Ends}(S)=\mathrm{Ends}^{*}(S)$, we can also
construct the Dehn-Thurston coordinate system $D$ with families
$\\{\alpha_{i}\\}_{i\in\mathbf{N}}$, $\\{\beta_{i}\\}_{i\in\mathbf{N}}$ and
$\\{\gamma_{i}\\}_{i\in\mathbf{N}}$ formed exclusively by nonseparating
curves.

Proof theorem 7. Given that $\mathrm{Ends}(S)=\mathrm{Ends}^{*}(S)$ we can
construct $P=\\{\alpha_{i}\\}_{i\in\mathbf{N}}$ a pants decomposition of $S$
formed by nonseparating curves. Let $\phi:\mathcal{G}(S)\to\mathcal{G}(S)$ an
automorphism. Due to lemma 4.11 there exists an homeomorphism
$h_{1}:S\rightarrow S$ such that $h_{1}(\alpha_{i})=\phi(\alpha_{i})$ for all
$\alpha_{i}\in P$.
Again, since $\mathrm{Ends}(S)=\mathrm{Ends}^{*}(S)$ we can construct
$\\{\beta_{i}\\}_{i\in\mathbf{N}}$ a collection of nonseparating curves such
that $i(\alpha_{i},\beta_{i})=2$ for all $i$ and $i(\alpha_{i},\beta_{j})=0$
for $i\neq j$. We can define an homeomorphism $h_{2}:S\rightarrow S$ such that
$h_{2}(h_{1}(\alpha_{i}))=h_{1}(\alpha_{i})=\phi(\alpha_{i})$ and
$h_{2}(h_{1}(\beta_{i}))=\phi(\beta_{i})$ in the following way. For every
$i\in\mathbf{N}$ the curves $\alpha=h_{1}(\alpha_{i})$,
$\beta=h_{1}(\beta_{i})$ and $\gamma=\phi(\beta_{i})$ satisfy the hypotheses
of part (1) in lemma 4.12 and lie in a subsurface $S_{i}$ homeomorphic to
$S_{0,4}$ that does not contain any element in
$h_{1}(P)\backslash\\{h(\alpha_{i})\\}$, $S_{i}$ contains $h_{1}(\beta_{i})$
and $\phi(\beta_{i})$, and its boundary components are isotopic to the curves
adjacent to $h_{1}(\alpha_{i})$ with respect to $h_{1}(P)$. Let
$h_{2,i}:S_{i}\rightarrow S_{i}$ be the homeomorphism from $(1)$ in lemma
4.12. This homeomorphism is just a Dehn twist about $\alpha$, therefore it
preserves orientation and its support $K_{i}\subset S_{i}$ satisfies that
$K_{i}\cap K_{j}=\emptyset$ for $i\neq j$ for all $i,j\in\mathbf{N}$. Hence
$h_{2}$ can be defined by parts using $\\{h_{2,i}\\}_{i\in\mathbf{N}}$.
Let $\\{\gamma_{i}\\}_{i\in\mathbf{N}}$ be a collection of curves such that
$i(\alpha_{i},\gamma_{i})=i(\beta_{i},\gamma_{i})=2$ and
$i(\alpha_{i},\gamma_{j})=0$ for $i\neq j$. We can define an homeomorphism
$h_{3}:S\rightarrow S$ such that
$h_{3}(h_{2}(h_{1}(\alpha_{i})))=h_{2}(h_{1}(\alpha_{i})=\phi(\alpha_{i})$,
$h_{3}(h_{2}(h_{1}(\beta_{i})))=h_{2}(h_{1}(\beta_{i}))=\phi(\beta_{i})$ and
$h_{3}(h_{2}(h_{1}(\gamma_{i})))=\phi(\gamma_{i})$ in the following way. For
every $i\in\mathbf{N}$, let now $\alpha=h_{1}(h_{2}(\alpha_{i}))$,
$\beta=h_{1}(h_{2}(\beta_{i}))$, $\gamma_{1}=h_{1}(h_{2}(\gamma_{i}))$ and
$\gamma_{2}=\phi(\gamma_{i})$. Analogously to the preceding case, these curves
satisfy the hypotheses of part (2) in lemma 4.12. Let
$h_{3,i}:R_{i}\rightarrow R_{i}$ be the (orientation-reversing) homeomorphism
from part (2) in lemma 4.12, where $R_{i}$ is homeomorphic to $S_{0,4}$ and
contains the curves $\alpha$, $\beta$, $\gamma_{1}$ and $\gamma_{2}$. It is
not difficult to see that if $i\neq j$ and $R_{i}\cap R_{j}\neq\emptyset$,
then $R_{i,j}=R_{i}\cap R_{j}\cong S_{0,3}$. Moreover $h_{3,i}$ and $h_{3,j}$
coincide in $R_{i,j}$ and hence we can define $h_{3}$ by parts using
$\\{h_{3,i}\\}_{i\in\mathbf{N}}$.
Let $h=h_{3}\circ h_{2}\circ h_{1}$. Since
$P^{\prime}=P\cup\\{\beta_{i}\\}\cup\\{\gamma_{i}\\}$ form a Dehn-Thurston
coordinates system of curves, then $h(P^{\prime})$ is a Dehn-Thurston
coordinates system of curves, and by construction
$h(\varepsilon)=\phi(\varepsilon)$ for all $\varepsilon\in P^{\prime}$.
Therefore, due to lemma 4.10, for all $\delta\in\mathcal{V}(\mathcal{G}(S))$
and all $\varepsilon\in P^{\prime}$:

$i(\phi(\delta),\phi(\varepsilon))=i(\delta,\varepsilon)=i(h(\delta),h(\varepsilon))=i(h(\delta),\phi(\varepsilon)),$
(15)

then $\phi(\delta)=h(\delta)$ for all $\delta\in\mathcal{V}(\mathcal{G}(S))$,
which implies $\Psi_{\mathcal{G}(S)}$ is surjective. ∎

######  Corollary 4.

Let $S_{1}$ and $S_{2}$ be infinite genus surfaces, such that
$\mathrm{Ends}(S_{i})=\mathrm{Ends}^{*}(S_{i})$ for $i=1,2$ and let
$\phi:\mathcal{G}(S_{1})\rightarrow\mathcal{G}(S_{2})$ be an isomorphism. Then
$S_{1}$ and $S_{2}$ are homeomorphic and $\phi$ is induced by a mapping class
in ${\rm MCG}^{*}(S_{1})$.

###### Proof.

Every isomorphism $\phi:\mathcal{G}(S_{1})\rightarrow\mathcal{G}(S_{2})$
induces an isomorphism $\phi:\mathcal{N}(S_{1})\rightarrow\mathcal{N}(S_{2})$.
Indeed, take $u,v$ two curves such that $i(u,v)=0$. Suppose
$i(\phi(u),\phi(v))\geq 2$ and remark that, as in the proof of theorem 7,
lemmas 1, 3 and 5 in [Sch] remain valid in the context of this corollary.
Hence we obtain a contradiction. On the other hand it is clear that
$i(\phi(u),\phi(v))\neq 1$, for $\phi$ is an isomorphism. Hence the only
possibility left is that $i(\phi(u),\phi(v))=0$. By corollary 1 we obtain that
$S_{1}$ is homemorphic to $S_{2}$. The rest of the proof follows from theorems
1 and 7. ∎

####  4.3.2 Proof of theorem 8.

Any $\phi\in\mathrm{Aut}(\mathcal{C}(S))$ sends nonseparating curves to
nonseparating curves, hence
$\phi|_{\mathcal{N}(S)}\in\mathrm{Aut}(\mathcal{N}(S))$ and then due to
theorem 7 there exists $h\in\mathrm{MCG}^{*}(S)$ such that
$\phi|_{\mathcal{N}(S)}(\alpha)=h(\alpha)$ for all
$\alpha\in\mathcal{V}(\mathcal{N}(S))$. Hence we only need to check that
$\phi$ and $h$ coincide in the separating curves of $S$. Let $\alpha$ be a
separating curve of $S$; we consider three cases.

  1. 1.

If both connected components of $S_{\alpha}$ have positive genus, then we can
find a pants decomposition $P$ such that $\alpha\in P$,
$(P\backslash\\{\alpha\\})\subset\mathcal{V}(\mathcal{N}(S))$ and
$\deg(\alpha)=4$ in $\mathcal{A}(P)$; let $\beta_{1}$, $\gamma_{1}$,
$\beta_{2}$ and $\gamma_{2}$ be the neighbours of $\alpha$ in $\mathcal{A}(P)$
such that $\beta_{i}$ and $\gamma_{i}$ are in the same connected component of
$S_{\alpha}$ for $i=1,2$. Let also $\delta_{1}$ and $\delta_{2}$ be
nonseparating curves such that $i(\alpha,\delta_{i})=0$ and
$i(\beta_{i},\delta_{i})=i(\gamma_{i},\delta_{i})=1$ for $i=1,2$. See figure 9
for an example.

$\gamma_{1}$$\gamma_{2}$$\beta_{1}$$\beta_{2}$$\delta_{1}$$\delta_{2}$$\alpha$
Figure 9: Catching $\alpha$ in a $S_{0,4}$.

By construction and lemma 4.3, $\phi(\alpha)$ and $h(\alpha)$ are contained in
the $S_{0,4}$ subsurface bounded by $\phi(\beta_{1})$, $\phi(\gamma_{1})$,
$\phi(\beta_{2})$ and $\phi(\gamma_{2})$ (recall that
$\phi(\beta_{i})=h(\beta_{i})$ and $\phi(\gamma_{i})=h(\gamma_{i})$ for
$i=1,2$ since they are nonseparating curves). Even more, since
$i(\alpha,\delta_{i})=0$ for $i=1,2$ then $\phi(\alpha)$ and $h(\alpha)$ must
be contained in the annulus formed by cutting the aforementioned $S_{0,4}$
subsurface along the arcs of $\phi(\delta_{i})=h(\delta_{i})$ for $i=1,2$;
therefore $\phi(\alpha)=h(\alpha)$.

  2. 2.

If $\alpha$ is an outer curve, then let $P$ be a pants decomposition such that
the peripheral pairs of $P$ bounding the same boundary components as $\alpha$,
are consecutive to one another (similar to the proof of lemma 4.4), and
$\alpha$ intersects only one curve in $P$ (namely $\beta$); let also $\gamma$
be a nonseparating curve that intersects each curve in the peripheral pairs
bounding the same boundary component as $\alpha$ only once while being
disjoint from $\alpha$. Figure 10 illustrates this situation.

$\beta$$\alpha$$\gamma$ Figure 10: Catching $\alpha$ again in a $S_{0,4}$.

Due to $\phi$ being an isomorphism, $\phi(\alpha)$ will intersect
$\phi(\beta)$ and be disjoint of every other curve in $P$. Using that and
lemma 4.4, we know that $\phi(\alpha)$ and $h(\alpha)$ are contained in the
$S_{0,4}$ subsurface bounded by two boundary components of $S$ and the images
of the adjacent curves in $\mathcal{A}(P)$ of $\beta$; even more,
$\phi(\alpha)$ and $h(\alpha)$ must be contained in the pair of pants
resulting from cutting the aforementioned $S_{0,4}$ subsurface that contains
them along the arc of $\phi(\gamma)=h(\gamma)$. Since there is only one curve
in this pair of pants which is an essential curve of $S$, then
$\phi(\alpha)=h(\alpha)$.

  3. 3.

Let $S_{1}$ and $S_{2}$ be the two connected components of $S_{\alpha}$ and
suppose that $S_{1}$ has genus zero and $n^{\prime}\geq 3$ boundary
components. We can find the following: a finite sequence
$\\{\beta_{i}\\}_{i=1}^{n^{\prime}-1}$ composed of outer curves, such that
$i(\beta_{i},\alpha)=0$ for $i=1,\ldots,n^{\prime}-1$,
$i(\beta_{i},\beta_{i+1})=2$ for $i=1,\ldots,n^{\prime}-2$ and
$i(\beta_{i},\beta_{j})=0$ for $j\notin\\{i-1,i+1\\}$; a pants decomposition
$P$ (composed solely of nonseparating curves) of the infinite genus connected
component of $S\backslash\\{\alpha\\}$; and finally, a curve $\gamma$ which
intersects once the curves $\delta_{1}$ and $\delta_{2}$ forming the
peripheral pair that bounds the boundary of $S_{2}$ induced by $\alpha$.
Figure 11 illustrates this situation.

$\alpha$$\gamma$$\delta_{1}$$\delta_{2}$$\beta_{1}$$\beta_{2}$$\beta_{3}$
Figure 11: Catching $\alpha$ in an annulus.

Given that isomorphism of $\mathcal{C}(S)$ send outer curves to outer curves,
part $(2)$ of this proof, the fact that $\phi(\alpha)$ and $h(\alpha)$ must
both be essential curves and they must be different from every element of
$\phi(\\{\beta_{i}\\}_{i=1}^{n^{\prime}})\cup\phi(P)\cup\\{\phi(\gamma)\\}$;
we can conclude that $\phi(\alpha)$ and $h(\alpha)$ must be contained in the
annulus obtained by cutting $S$ along
$\phi(\\{\beta_{i}\\}_{i=1}^{n^{\prime}})\cup\phi(P)\cup\\{\phi(\gamma)\\}$.
The boundary components of this annulus are formed by arcs of
$\phi(\beta_{i})$ for $i=1,\ldots,n^{\prime}-1$, $\phi(\gamma)$,
$\phi(\delta_{1})$ and $\phi(\delta_{2})$. Therefore $\phi(\alpha)=h(\alpha)$.

####  4.3.3 Proof of theorem 1.

Theorems 5 and 8 imply that $\Psi_{\mathcal{C}(S)}$ is an isomorphism. From
theorem 7 we know that the natural map:

$\Psi_{\mathcal{N}(S)}:{\rm MCG}^{*}(S)\to{\rm Aut}(\mathcal{N}(S))$ (16)

is surjective. Let us suppose $h_{1},h_{2}\in\mathrm{MCG}^{*}(S)$ are such
that $h_{1}\neq h_{2}$ and
$\Psi_{\mathcal{N}(S)}(h_{1})=\Psi_{\mathcal{N}(S)}(h_{2})$. Then since
$\Psi_{\mathcal{C}(S)}$ is injective we have that
$\Psi_{\mathcal{C}(S)}(h_{1})\neq\Psi_{\mathcal{C}(S)}(h_{2})$ even though
their restrictions to $\mathcal{N}(S)$ are the same. This implies that
$\Psi_{\mathcal{C}(S)}(h_{1})$ and $\Psi_{\mathcal{C}(S)}(h_{2})$ differ in
some separating curves. But given that the restrictions of
$\Psi_{\mathcal{C}(S)}(h_{1})$ and $\Psi_{\mathcal{C}(S)}(h_{2})$ to
$\mathcal{N}(S)$ are the same, we can use the same technique as in the proof
of theorem 8, for catching the separating curves in an annulus (or a pair of
pants), which means
$\Psi_{\mathcal{C}(S)}(h_{1})(\alpha)=\Psi_{\mathcal{C}(S)}(h_{2})(\alpha)$
for every separating curve $\alpha$. Thus we have reached a contradiction and
therefore $\Psi_{\mathcal{N}(S)}$ is injective, hence it is an isomorphism. We
finish the proof by remarking that
$\Psi_{\mathcal{G}(S)}=\Phi\circ\Psi_{\mathcal{N}(S)}$, where $\Phi$ is the
isomorphism between $\mathrm{Aut}(\mathcal{N}(S))$ and
$\mathrm{Aut}(\mathcal{G}(S))$ defined in (14). ∎

######  Remark 9.

Using theorem 1 we can deduce that, for an infinite genus surface $S$ such
that $\mathrm{Ends}(S)=\mathrm{Ends}^{*}(S)$, every automorphism $\varphi$ of
${\rm MCG}^{*}(S)$ sending Dehn twists to Dehn twist must be an inner
automorphism. The proof of this fact is taken verbatim from the proof of
theorem 2, in [Ivanov]. However, it is still unknown if, as in the compact
case, every automorphism of ${\rm MCG}^{*}(S)$ sends Dehn twists to Dehn
twists.

##  5 Counterexamples

In this section we show that theorem 2 is not valid if the morphism between
curve complexes is not an isomorphism. For that, let us first recall the
notion of _superinjective map_.

######  Definition 5.1 (Superinjectivity).

A simplicial map $f:\mathcal{C}(S_{1})\rightarrow\mathcal{C}(S_{2})$ is called
superinjective if for any two vertices $\alpha$ and $\beta$ in
$\mathcal{C}(S_{1})$ such that $i(\alpha,\beta)\neq 0$ we have that
$i(f(\alpha),f(\beta))\neq 0$.

Every superinjective map is injective. For compact surfaces, we have the
following theorem concerning superinjective maps.

######  Theorem 9.

[Irmak] Let $S$ be a closed, connected, orientable surface of genus at least
3. A simplicial map, $f:\mathcal{C}(S)\to\mathcal{C}(S)$, is superinjective if
and only if $f$ is induced by an homeomorphism of $S$.

The following lemma shows that this result is not true for a large class of
surfaces of infinite genus and, in this sense, theorem 2 is optimal.

######  Lemma 5.2.

Let $S$ be a surface such that $\mathrm{Ends}^{*}(S)\neq\emptyset$. Then there
exist a simplicial superinjective map $f:\mathcal{C}(S)\to\mathcal{C}(S)$
which is not surjective.

###### Proof.

This proof makes reference to figure 12. Let
$\alpha\in\mathcal{V}(\mathcal{C}(S))$ be a separating curve. Without loss of
generality we can think that $\alpha$ is contained in a subsurface $S_{i}$ in
$[S_{1}\supseteq S_{2}\supseteq\ldots]\in\mathrm{Ends}^{*}(S)$ where $i$ is
large enough.

$\alpha$$\beta$$\beta^{\prime}$$\alpha$$\gamma$$f(\beta)$$f(\beta^{\prime})$$f$
Figure 12: A superinjective but not surjective simplicial map.

We describe $f$ topologically. Let $S_{1}$ and $S_{2}$ be the two connected
components of $S_{\alpha}$. Cut $S$ along $\alpha$ and then glue in a copy of
$S_{1,2}$. This operation produces a new surface $S^{\prime}=S_{1}\cup
S_{2}\cup S_{1,2}$. Remark that $S$ is homeomorphic to $S^{\prime}$ and that
there is a natural inclusion map $f_{i}:S_{i}\hookrightarrow S^{\prime}$, for
$i=1,2$. If $\beta\in\mathcal{V}(\mathcal{C}(S_{i}))$, then we define
$f(\beta)=f_{i}(\beta)$ for $i=1,2$. On the other hand, if $\beta^{\prime}$
intersects the curve $\alpha$ we define $f(\beta^{\prime})$ as depicted in
figure 12. Clearly, $f$ is superinjecive but no essential curve properly
contained in the copy of $S_{1,2}$ that we introduced is in the image of $f$.
Hence $f$ is not surjective and, in particular, $f$ cannot be induced by a
class in ${\rm MCG}^{*}(S)$. ∎

We think that this result can be optimized in the following way.

######  Conjecture 1.

Let S be a surface such that $\mathrm{Ends}^{*}(S)\neq\emptyset$ and
$\\{\alpha_{1},\ldots,\alpha_{n}\\}\subset\mathcal{C}(S)$ be simplex. Then
there exists a simplicial superinjective map
$f:\mathcal{C}(S)\to\mathcal{C}(S)$ whose image does not intersect
$\\{\alpha_{1},\ldots,\alpha_{n}\\}$.

The following result shows that the statement of theorem 2 is not valid for
superinjective maps.

$\alpha$$\beta$$\beta^{\prime}$$\alpha$$f(\beta^{\prime})$$f(\beta)$$f$ Figure
13: A superinjecive map between two nonhomeomorphic surfaces.

######  Lemma 5.3.

There exist uncountably many examples of pairs of nonhomeomorphic infinite
genus surfaces $S_{1}$ and $S_{2}$ for which there exists a superinjective map
$f:\mathcal{C}(S_{1})\to\mathcal{C}(S_{2})$.

###### Proof.

The arguments are similar to those of the proof of lema 5.2. Let $S_{1}$ be
the Loch Ness monster and $\alpha\in\mathcal{C}(S_{1})$ be a separating curve.
Let $S$ be your favorite infinite genus surface and suppose that $S$ has at
least two boundary components. We describe $f$ topologically. Cut $S_{1}$
along $\alpha$ and then glue in a copy of $S$ as indicated in figure 13. This
produces $S_{2}$. The rest of the proof is analogous to the proof of lemma
5.2.
∎

## References