# Big mapping class groups and the co-Hopfian property

Javier Aramayona  Instituto de Ciencias Matemáticas, ICMAT (CSIC-UAM-UC3M-UCM)
<EMAIL_ADDRESS>,  Christopher J. Leininger  Rice University
<EMAIL_ADDRESS>and  Alan McLeay  University of Luxembourg
<EMAIL_ADDRESS>
###### Abstract.

We study injective homomorphisms between big mapping class groups of infinite-
type surfaces. First, we construct (uncountably many) examples of surfaces
without boundary whose (pure) mapping class groups are not co-Hopfian; these
are the first such examples of injective endomorphisms of mapping class groups
that fail to be surjective.

We then prove that, subject to some topological conditions on the domain
surface, any continuous injective homomorphism between (arbitrary) big mapping
class groups that sends Dehn twists to Dehn twists is induced by a subsurface
embedding.

Finally, we explore the extent to which, in stark contrast to the finite-type
case, superinjective maps between curve graphs impose no topological
restrictions on the underlying surfaces.

J.A. was supported by grant PGC2018-101179-B-I00. He acknowledges financial
support from the Spanish Ministry of Science and Innovation, through the
“Severo Ochoa Programme for Centres of Excellence in R&D” (CEX2019-000904-S).
C.L. was partially supported by NSF grants DMS-1811518 and DMS-2106419. A.M
was supported by the Luxembourg National Research Fund OPEN grant
O19/13865598.

##  1\. Introduction and main results

Throughout this article, all surfaces will be assumed to be connected,
orientable and second-countable, unless otherwise specified. A surface $S$ has
finite type if its fundamental group is finitely generated, and has infinite
type otherwise.

The mapping class group $\operatorname{Map}(S)$ is the group of orientation-
preserving homeomorphisms of $S$, up to homotopy; if $\partial
S\neq\emptyset$, then we require that all homeomorphisms and homotopies fix
$\partial S$ pointwise. In the case when $S$ has finite type,
$\operatorname{Map}(S)$ is well-known to be finitely presented. If, on the
contrary, $S$ has infinite type, then $\operatorname{Map}(S)$ becomes an
uncountable, totally disconnected, non-locally compact topological group with
respect to the quotient topology stemming from the compact-open topology on
the homeomorphism group of $S$. We refer the reader to Section 2 for expanded
definitions, and to the recent survey [5] for a detailed treatment of mapping
class groups of infinite-type surfaces, now commonly known as big mapping
class groups.

###  1.1. Algebraic rigidity for mapping class groups

A seminal result of Ivanov [23] states that if $S$ is a (sufficiently
complicated) finite type surface $S$, then every automorphism of
$\operatorname{Map}(S)$ is a conjugation by an element of the extended mapping
class group $\operatorname{Map}^{\pm}(S)$, namely the group of all
homeomorphisms of $S$ up to isotopy. In other words, every isomorphism
$\operatorname{Map}(S)\to\operatorname{Map}(S)$ is induced by a homeomorphism
$S\to S$. The analog for infinite-type surfaces was recently obtained by
Bavard-Dowdall-Rafi [8]; see Theorem Theorem below.

The proofs in the finite and infinite type settings proceed along similar
lines, based on the work of Ivanov [23]. First, one proves an algebraic
characterization of Dehn twists and uses it to deduce that any isomorphism
$\phi:\operatorname{Map}(S)\to\operatorname{Map}(S)$ must send (powers of)
Dehn twists to (powers of) Dehn twists. In particular, $\phi$ induces a
simplicial automorphism $\phi_{*}:\mathcal{C}(S)\to\mathcal{C}(S)$ of the
curve complex. At this point, the argument boils down to showing that every
simplicial automorphism of $\mathcal{C}(S)$ is induced by a homeomorphism of
$S$ [23]. In fact, this approach applies, with a finite-list of exceptional
surfaces, to any isomorphism between finite-index subgroups of
$\operatorname{Map}(S)$ or the pure mapping class group
$\operatorname{PMap}(S)$, the subgroup of $\operatorname{Map}(S)$ whose
elements fix every end of $S$ [23, 9, 10, 21, 22, 20, 31]

###  1.2. The co-Hopfian property

The combination of results of Ivanov-McCarthy [24] and Bell-Margalit [10]
imply that, with the possible exception of the twice-punctured torus, if $S$
has finite type then every injective homomorphism
$\operatorname{Map}(S)\to\operatorname{Map}(S)$ is induced by a homeomorphism
of $S$. In particular, the mapping class group of a (sufficiently complicated)
finite-type surface is co-Hopfian: every injective endomorphism is surjective.

For infinite-type surfaces, Question 4.5 of the AIM Problem List on Surfaces
of Infinite Type [1] (see also [5, Question 5.2]) asks:

######  Question 1.1.

Are big mapping class groups co-Hopfian?

######  Remark.

The answer to the above question is immediately “no” for surfaces with non-
empty boundary. For instance, if a surface $S$ has non-empty boundary and its
space of ends is homeomorphic to a Cantor set, then there exists a proper
$\pi_{1}$-injective continuous map $S\to S$ that is not surjective. In turn,
this map induces an injective homomorphism
$\operatorname{Map}(S)\to\operatorname{Map}(S)$ that is not surjective;
compare with Theorem 2 below.

Our first result states that the answer to Question 1.1 is also negative for
surfaces without boundary. Recall that the Loch Ness Monster is the unique (up
to homeomorphism) connected orientable surface of infinite genus and exactly
one end. We will prove:

######  Theorem 1.

Let $S$ be either the once-punctured Loch Ness Monster or a torus minus the
union of a Cantor set and an isolated point. Then there exists a homomorphism
$\phi:\operatorname{Map}(S)\to\operatorname{Map}(S)$ such that:

  1. (1)

$\phi$ is continuous and injective, but not surjective,

  2. (2)

There is a Dehn twist $t$ such that $\phi(t)$ is not supported on any finite-
type subsurface of $S$. In the particular case when $S$ is the Loch Ness
Monster, no power of $\phi(t)$ is supported on a finite-type subsurface of
$S$.

  3. (3)

There exists a partial pseudo-Anosov $f\in\operatorname{Map}(S)$ such that
$\phi(f)$ is a multitwist.

######  Remark.

Part (2) of the above theorem yields a negative answer to Problem 4.75 of the
AIM Problem List on Surfaces of Infinite Type [1].

The construction behind Theorem 1 is inspired by the construction of the non-
geometric injective homomorphism of the first two authors with Souto [2,
Theorem 2], building on a well-known construction of homomorphisms from covers
(see e.g. [12, 24, 34]). Namely, we construct a covering map $S^{\prime}\to S$
which induces an injective homomorphism
$\operatorname{Map}(S)\to\operatorname{Map}(S^{\prime})$, in such a way that
the surface $S^{\prime\prime}$ obtained by filling in all but one of the
punctures of $S^{\prime}$ is homeomorphic to $S$, and yet the homomorphism
$\operatorname{Map}(S)\to\operatorname{Map}(S^{\prime\prime})$ remains
injective. Once this has been done, the resulting homomorphism is easily seen
to not be surjective in light of Bavard-Dowdall-Rafi’s result [8] mentioned
above, since $\phi$ sends some finitely supported elements to elements which
are not finitely supported.

Motivated by this construction, we also observe the following.

######  Theorem 1.2.

Let $S$ denote the closed surface of genus $g\geq 1$ minus the union of a
Cantor set and an isolated point. Then there exists a continuous injective
homomorphism
$\operatorname{Map}(S)\to\operatorname{Map}(\mathbb{R}^{2}\smallsetminus C)$,
where $C$ denotes a Cantor set.

As far as we know, these are the first examples of injective homomorphisms
between mapping class groups where the genus decreases from domain to
codomain.

###  1.3. Not co-Hopfian pure mapping class groups

If we restrict Question 1.1 to pure mapping class groups, we will see that
there is a much richer palette of examples of injective, but not surjective,
homomorphisms. We say that a surface $S$ is _self-doubling_ if there exists a
multicurve $\mathcal{B}$ such that;

  1. (1)

$S\setminus\mathcal{B}$ has two connected components $L$ and $R$,

  2. (2)

$L$ and $R$ are both homeomorphic to $S$, and

  3. (3)

there exists orientation-reversing homeomorphism $\iota:S\to S$ of order two
such that $\iota(L)=R$ and $\iota(b)=b$ for each $b\in\mathcal{B}$.

######  Theorem 2.

For every self-doubling surface $S$ there exists a continuous injective
homomorphism $\operatorname{PMap}(S)\to\operatorname{PMap}(S)$ that is not
surjective.

Examples. We now list some concrete examples of surfaces to which Theorem 2
applies.

  1. (1)

A first example of a self-doubling surface $S$ is the sphere minus the union
of a Cantor set and the north and south pole: any essential curve $b$ that
separates the poles splits $S$ into two subsurfaces $L$ and $R$ with interiors
homeomorphic to $S$.

  2. (2)

Alternatively, suppose $S$ has infinite genus such that every planar end is
isolated, and for every non-planar end there is a sequence of planar ends
converging to it. (We remark that there are uncountably many surfaces with
this property.) Then $S$ is also self-doubling; see Theorem 4.2.

We give an equivalent definition of self-doubling surfaces, along with more
examples, in Section 4.

###  1.4. Twist-preserving homomorphisms

In light of the discussion above, we now focus on homomorphisms that send Dehn
twists to Dehn twists; we call these twist-preserving homomorphisms. In this
section we will allow surfaces to have a non-empty boundary, though we assume
it is compact (and is thus a finite union of circles). We will prove the
following result; recall that a map between topological spaces is proper if
the preimage of every compact set is compact:

######  Theorem 3.

Let $S$ and $S^{\prime}$ be surfaces of infinite type, where $S$ has positive
genus. Assume further that either the boundary of $S$ is empty, or else $S$
has at most one end accummulated by genus. If
$\phi:\operatorname{PMap}(S)\to\operatorname{PMap}(S^{\prime})$ is a
continuous injective twist-preserving homomorphism, then there is a proper
$\pi_{1}$-injective embedding $S\to S^{\prime}$ that induces $\phi$.

######  Remark.

Theorem 3 no longer holds if $\partial S\neq\emptyset$ and $S$ has more than
one end accumulated by genus; see the final remark of Section 5. Also, in the
same remark we will see that the result we will prove in fact applies to a
larger class of homomorphisms than injective ones.

Continuing with our discussion, observe that if $\partial S=\emptyset$, any
proper $\pi_{1}$-injective embedding $S\to S^{\prime}$ is homotopic to a
homeomorphism. Hence we obtain:

######  Corollary 1.3.

Let $S$ and $S^{\prime}$ be surfaces of infinite type, where $S$ has positive
genus and no boundary. If
$\phi:\operatorname{PMap}(S)\to\operatorname{PMap}(S^{\prime})$ is a
continuous injective twist-preserving homomorphism, then it is induced by a
homeomorphism $S\to S^{\prime}$; in particular, it is surjective.

To prove Theorem 3, one first observes that $\phi$ induces a simplicial map
$\mathcal{C}(S)\to\mathcal{C}(S^{\prime})$ that preserves intersection number
one. Once this has been shown, the result follows considering exhaustions and
a result of Souto and the first named author [4], plus a continuity argument.
We conjecture that, although continuity is needed in our proof, it is in fact
not necessary for Theorem 3 to hold. In this direction, a result of Mann [26]
states that the mapping class group of certain surfaces have the automatic
continuity property. Specifically, this automatic continuity holds when the
surface is a closed surface punctured along the union of a Cantor set and a
finite set; for these surfaces we obtain:

######  Corollary 1.4.

Let $S$ and $S^{\prime}$ be surfaces of infinite type, where $S$ is obtained
from a closed surface of positive genus by removing the union of a Cantor set
and a finite set. If
$\phi:\operatorname{PMap}(S)\to\operatorname{PMap}(S^{\prime})$ is an
injective twist-preserving homomorphism, then there is a homeomorphism $S\to
S^{\prime}$ that induces $\phi$.

###  1.5. Injections between curve graphs

In the finite-type setting, an important step for establishing the co-Hopfian
property for finite-index subgroups of mapping class groups is that
superinjective self-maps of the curve graph are induced by homeomorphisms; see
[21], for instance. We recall that a simplicial map between curve graphs is
superinjective if it maps pairs of curves with non-zero intersection number to
pairs of curves with the same property.

While it is known that every automorphism of the curve graph of an infinite-
type surface is induced by a homeomorphism, [17, 8], it is easy to see that
this is no longer true if one drops the requirement that the map be
surjective; see [19, 5] for concrete examples. Our final result highlights the
extent of this failure:

######  Theorem 4.

Let $S$ and $S^{\prime}$ be infinite-type surfaces, where $S^{\prime}$ has
infinite genus. Then:

  1. (1)

There exists a superinjective simplicial map $\mathcal{C}(S)\to\mathcal{C}(S)$
that is not surjective.

  2. (2)

There exists a superinjective simplicial map
$\mathcal{C}(S)\to\mathcal{C}(S^{\prime})$.

In the case of infinite-genus surfaces, a proof of part (1) Theorem 4 was
previously obtained by Hernández-Valdez [19] using different methods.

######  Remark.

If $S$ and $S^{\prime}$ are arbitrary infinite-type surfaces, then there is
always an injective simplicial map $\mathcal{C}(S)\to\mathcal{C}(S^{\prime})$.
To see this, simply observe that the curve graph of any infinite-type surface
contains the complete graph on countably many vertices, and that the curve
graph of any surface has a countable set of vertices.

Theorem 4 should be compared with Theorem 3, which implies that every
superinjective map between curve graphs that comes from an injective
homomorphism between the corresponding mapping class groups is induced by an
embedding of the underlying surfaces.

Plan of the paper. In Section 2 we will give all the necessary background
needed in the article. Sections 3.2 and 3.3 are devoted to the proof of
Theorem 1 in the cases, respectively, of the one-holed torus minus a Cantor
set and the once-punctured Loch Ness Monster. In Section 4 we deal with
Theorem 2. Section 5 is devoted to the proof Theorem 3. Finally, in Section 6
we establish Theorem 4.

Acknowledgements. We are grateful to the organizers of the AIM workshop
“Surfaces of Infinite Type” for the discussions that are the origins of this
paper. Thanks also to AIM for its hospitality and financial support. We would
like to thank Nick Vlamis and Henry Wilton for enlightening conversations
about Lemma 6.2, and Vlamis for the reference [14]. We also thank Israel
Morales for bringing to our attention the refererence [6], and Ty Ghaswala for
comments on an earlier version.

##  2\. Preliminaries

In this section we will briefly introduce all the background material needed
for our results. We refer the reader to the survey paper [5] for a detailed
account on these topics.

###  2.1. Surfaces

In what follows, by a surface we will mean a orientable second-countable
topological surface, with a finite number (possibly zero) of boundary
components, all of them assumed to be compact. If the fundamental group of a
surface $S$ is finitely generated, we will say that $S$ has finite type;
otherwise it has infinite type. The space of ends of $S$ is

$\operatorname{Ends}(S):=\lim_{\leftarrow}\pi_{0}(S\smallsetminus K),$

where $K$ denotes a compact subset of $S$. The space of ends becomes a
topological space with respect to the final topology obtained from endowing
each set defining the inverse system with the discrete topology. It is well-
known that $\operatorname{Ends}(S)$ is a closed subset of a Cantor set. We say
that an end is planar if it has a planar neighborhood; otherwise we will say
that the end is non-planar or that it is accumulated by genus. An isolated
planar end is usually called a puncture; as is customary, we will treat
punctures both as ends and as marked points on the surface, switching between
the two viewpoints without further mention. Finally, we will denote by
$\operatorname{Ends}_{\infty}(S)$ the subspace of non-planar ends, which is
closed in $\operatorname{Ends}(S)$.

A classical result [29] asserts that any surface $S$ is uniquely determined by
the quadruple $(g,b,\operatorname{Ends}(S),\operatorname{Ends}_{\infty}(S))$,
where $g\in\mathbb{N}\cup\\{\infty\\}$ is the genus, $b\in\mathbb{N}$ is the
number of boundary components. More concretely, two surfaces are homeomorphic
if and only if their genera and number of boundary components are equal, and
there is a homeomorphism between the spaces of ends that restricts to a
homeomorphism between the spaces of non-planar ends.

###  2.2. Curves and multicurves

A simple closed curve on $S$ is said to be inessential if it bounds a disk, a
once-punctured disk, or an annulus whose other boundary is a boundary
component of $S$; otherwise we say the curve is essential. By a curve we mean
the isotopy class of an essential simple closed curve on $S$. A multicurve is
a set of curves on the surface that have pairwise disjoint representatives.
Given a multicurve $Q$, we will write $S\smallsetminus Q$ to mean the
(possibly disconnected) surface obtained from $S$ by cutting along (pairwise
disjoint representatives of) each element of $Q$.

###  2.3. Mapping class group

Let $\operatorname{Homeo}^{+}(S)$ be the group of all orientation-preserving
self-homeomorphisms of $S$, equipped with the compact-open topology. We will
denote by $\operatorname{Homeo}_{0}(S)$ the connected component of the
identity in $\operatorname{Homeo}^{+}(S)$, noting it is a normal subgroup. The
mapping class group of $S$ is

$\operatorname{Map}(S):=\operatorname{Homeo}^{+}(S)/\operatorname{Homeo}_{0}(S).$

As is customary, if $\partial S\neq\emptyset$, in this definition we
implicitly require that all homeomorphisms fix $\partial S$ pointwise.

The mapping class group is naturally a topological group with respect to the
quotient topology coming from the compact-open topology on
$\operatorname{Homeo}^{+}(S)$. It is a classical fact that
$\operatorname{Map}(S)$ is a finitely presented group if $S$ has finite type,
while it is easy to see that it is uncountable whenever $S$ has infinite type.
Moreover, in this latter case, $\operatorname{Map}(S)$ is totally disconnected
and not locally compact; see [5] for more details.

One of the motivating results for us is the following theorem of Bavard-
Dowdall-Rafi [8] mentioned in the introduction.

######  Theorem ([8]).

For any infinite-type surface $S$, any isomorphism
$\operatorname{Map}(S)\to\operatorname{Map}(S)$ is induced by a homeomorphism
$S\to S$. In particular, any such isomorphism is continuous.

The pure mapping class group $\operatorname{PMap}(S)$ is the subgroup of
$\operatorname{Map}(S)$ whose elements fix every end of $S$. If $S$ has finite
type, it is well-known that $\operatorname{PMap}(S)$ is generated by Dehn
twists along a finite set of curves (see [16, Section 4.4]). When $S$ has
infinite type, this is no longer true in general; instead, we have the
following result; see [28] for the definition of a handle shift:

######  Theorem 2.1 ([28]).

$\operatorname{PMap}(S)$ is topologically generated by Dehn twists if and only
if $S$ has at most one end accumulated by genus. Otherwise, it is
topologically generated by Dehn twists and handle shifts.

Here, being topologically generated by Dehn twists means that the subgroup
generated by Dehn twists (i.e. the compactly-supported mapping class group,
see below) is dense in $\operatorname{PMap}(S)$.

####  2.3.1. Some important subgroups

Following [8], we will say that an element of $\operatorname{Map}(S)$ has
finite support if it represented by a homeomorphism which is the identity
outside some finite-type subsurface of $S$. We will write
$\operatorname{Map}_{f}(S)$ for the subgroup consisting of elements with
finite support. Although not directly relevant to our arguments, we record the
following beautiful result of Bavard-Dowdall-Rafi [8], which gives an
algebraic characterization of elements with finite support. In particular, it
serves to shed light on the potential differences between isomorphisms and
injective homomorphisms between big mapping class groups.

######  Proposition 2.2 ([8]).

An element of $\operatorname{Map}(S)$ has finite support if and only if its
conjugacy class in $\operatorname{Map}(S)$ is countable.

The compactly-supported mapping class group $\operatorname{Map}_{c}(S)$ is the
subgroup of $\operatorname{Map}(S)$ whose elements are represented by
homeomorphisms which are the identity outside some compact subsurface of $S$.
Observe that $\operatorname{Map}_{c}(S)$ is a subgroup of
$\operatorname{Map}_{f}(S)$, proper when $S$ has more than one puncture, and
of $\operatorname{PMap}(S)$. On the other hand, $\operatorname{Map}_{f}(S)$ is
a subgroup of $\operatorname{PMap}(S)$ if and only if $S$ has at most one
puncture. Before ending this section, we record the following immediate
observation for future use:

######  Lemma 2.3.

For every infinite-type surface $S$, we have

$\operatorname{Map}_{c}(S)=\lim_{\longrightarrow}\operatorname{Map}(X),$

where the direct limit is taken over all compact subsurfaces $X\subset S$,
directed with respect to inclusion.

###  2.4. Curve graph

The curve graph $\mathcal{C}(S)$ is the simplicial graph whose vertex set is
the set of curves on $S$, and where two vertices are deemed to be adjacent if
the corresponding curves may be realized disjointly on $S$. In what follows we
will not distinguish between vertices of the curve graph and the curves
representing them.

Observe that $\operatorname{Map}(S)$ acts on $\mathcal{C}(S)$ by simplicial
automorphisms. A classical fact, discovered initially by Ivanov [23] is that
every simplicial automorphism of $\mathcal{C}(S)$ is induced by a
homeomorphism of $S$, provided $S$ is not the twice-punctured torus. The
analog for infinite-type surfaces has been obtained independently by
Hernández-Morales-Valdez [18] and Bavard-Dowdall-Rafi [8]. A crucial step in
their proofs is the following so-called Alexander method for infinite-type
surfaces [18], which we record for future use:

######  Theorem 2.4.

Let $S$ be a connected orientable infinite-type surface with empty boundary.
The natural action of $\operatorname{Map}(S)$ on $\mathcal{C}(S)$ has trivial
kernel; in other words, if $f\in\operatorname{Map}(S)$ induces the identity
transformation on $\mathcal{C}(S)$, then it is the identity in
$\operatorname{Map}(S)$.

##  3\. Covers and forgetting

In this section we prove Theorem 1 and Theorem 1.2. These theorems are about
homomorphisms between mapping class groups of surfaces with a single puncture,
and the proofs exploit their relationship with automorphism groups and the
Birman exact sequence. We start with some generalities, then focus on the
specific cases of the theorems.

###  3.1. Covers and automorphism groups

In what follows, suppose $S$ is an orientable, connected surface without
punctures such that $\pi_{1}(S)$ is non-abelian. Given $z\in S$ any point, let
$S^{z}=S\smallsetminus\\{z\\}$ denote the surface obtained from $S$ by
removing $z$ (thus producing a puncture we call the $z$–puncture). Any
homeomorphism $f$ of $S^{z}$ induces a homeomorphism of $S$ that fixes the
point $z$, which by an abuse of notation we also denote by $f$. This defines a
homomorphism

$\operatorname{Homeo}^{+}(S^{z})\to\operatorname{Homeo}^{+}(S).$

This homomorphism sends $\operatorname{Homeo}_{0}(S^{z})$ into
$\operatorname{Homeo}_{0}(S)$ and so induces a homomorphism

$\operatorname{Map}(S^{z})\to\operatorname{Map}(S).$

Given a loop $\gamma$ representing an element of $\pi_{1}(S,z)$, one
associates a homeomorphism $f_{\gamma}\colon S^{z}\to S^{z}$ by “point pushing
along $\gamma$”. More precisely, $f_{\gamma}\colon S\to S$ is a homeomorphism
isotopic to the identity by an isotopy $f_{t}$ with $f_{0}=f_{\gamma}$,
$f_{1}=\text{id}$, and $f_{t}(z)=\gamma(t)$, for all $t\in[0,1]$. When $S$ has
finite type, Birman proved that $[\gamma]\mapsto[f_{\gamma}]$ defines an
injective homomorphism $\pi_{1}(S,z)\to\operatorname{Map}(S^{z})$ in such a
way that the sequence

(1)
$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(S,z)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Map}(S^{z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Map}(S)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1}$

is exact; see [11]. The infinite type case is proved by Dickmann-Domat in the
appendix to Domat’s paper [14].

######  Proposition 3.1.

For any connected, orientable surface $S$ without punctures and non-abelian
fundamental group, there is an injection of
$\pi_{1}(S,z)\to\operatorname{Map}(S^{z})$ given by
$[\gamma]\mapsto[f_{\gamma}]$ making (1) exact. ∎

For the remainder of this section, we use this theorem to identify
$\pi_{1}(S,z)$ with its image in $\operatorname{Map}(S^{z})$.

Remark. The theorem also holds when $S$ has punctures, replacing
$\operatorname{Map}(S^{z})$ with the subgroup preserving the $z$–puncture.

Given $f\in\operatorname{Homeo}^{+}(S^{z})$, after extending over $z$ we have
an induced automorphism $f_{*}\in\operatorname{Aut}(\pi_{1}(S,z))$, which
descends to a homomorphism
$\iota\colon\operatorname{Map}(S^{z})\to\operatorname{Aut}(\pi_{1}(S,z))$
given by $\iota([f])=f_{*}$. This further descends to a homomorphism
$\operatorname{Map}(S)\to\operatorname{Out}(\pi_{1}(S))$ which is injective by
Theorem 2.4. The inclusion of $\pi_{1}(S,z)$ into $\operatorname{Map}(S^{z})$,
composed with $\iota$ is precisely the isomorphism onto the group of inner
automorphisms, and thus we get a homomorphism of short exact sequences:

$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(S,z)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Map}(S^{z})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota}$$\textstyle{\operatorname{Map}(S)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1\,\,}$$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\pi_{1}(S,z)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Aut}(\pi_{1}(S,z))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Out}(\pi_{1}(S,z))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1.}$

The first vertical map is the identity and the last is injective, from which
it follows that the middle is also injective.

A regular covering space $p\colon\widetilde{S}\to S$ with the property that
every homeomorphism of $S$ lifts to a homeomorphism of $\widetilde{S}$ will be
called a geometrically characteristic cover. If
$p_{*}(\pi_{1}(\widetilde{S},\widetilde{z}))$ is characteristic, general map
lifting implies that the cover is geometrically characteristic, but this is
not a necessary condition in general; see the examples below.

Now suppose $p\colon\widetilde{S}\to S$ is geometrically characteristic, let
$z\in S$ be a point, and fix any $\widetilde{z}\in p^{-1}(z)$. Being regular
and geometrically characteristic implies that for any
$f\in\operatorname{Homeo}^{+}(S^{z})$, after extending over $z$, there is a
unique lift $\widetilde{f}\colon\widetilde{S}\to\widetilde{S}$ that fixes
$\widetilde{z}$, and thus determines a homeomorphism of the same name
$\widetilde{f}\in\operatorname{Homeo}^{+}(\widetilde{S}^{\widetilde{z}})$. The
general fact we need is the following (compare with [2, Theorem 2]).

######  Proposition 3.2.

If $p\colon\widetilde{S}\to S$ is a geometrically characteristic covering
space, $\pi_{1}(S)$ is non-abelian, and $\pi_{1}(\widetilde{S})$ is non-
trivial, then the assignment $f\mapsto\widetilde{f}$ described above descends
to a continuous, injective homomorphism

$\phi\colon\operatorname{Map}(S^{z})\to\operatorname{Map}(\widetilde{S}^{\widetilde{z}}).$

Moreover, via the inclusions from the Birman exact sequence we have $\phi\circ
p_{*}=\text{id}|_{\pi_{1}(\widetilde{S},\widetilde{z})}$.

###### Proof.

Continuity of
$\operatorname{Homeo}^{+}(S^{z})\to\operatorname{Homeo}^{+}(\widetilde{S}^{\widetilde{z}})$
is clear. Since $f\mapsto\widetilde{f}$ maps $\operatorname{Homeo}_{0}(S^{z})$
to $\operatorname{Homeo}_{0}(\widetilde{S},\widetilde{z})$ (by lifting
isotopies), we get a well-defined, continuous homomorphism
$\phi\colon\operatorname{Map}(S^{z})\to\operatorname{Map}(\widetilde{S}^{\widetilde{z}})$.
Since the homomorphisms
$\iota\colon\operatorname{Map}(S^{z})\to\operatorname{Aut}(\pi_{1}(S,z))$ and
$\tilde{\iota}\colon\operatorname{Map}(\widetilde{S}^{\widetilde{z}})\to\operatorname{Aut}(\pi_{1}(\widetilde{S},\widetilde{z}))$
are injective, to prove the first part of the proposition it suffices to prove
injectivity of the homomorphism
$\phi_{*}\colon\iota(\operatorname{Map}(S^{z}))\to\widetilde{\iota}(\operatorname{Map}(\widetilde{S}^{\widetilde{z}}))$
defined by $\phi_{*}\circ\iota=\widetilde{\iota}\circ\phi$, or more explicitly

$\iota([f])=f_{*}\stackrel{{\scriptstyle\phi_{*}}}{{\longmapsto}}\tilde{f}_{*}=\tilde{\iota}([\tilde{f}])=\tilde{\iota}(\phi([f])).$

Consequently, we need only show that the kernel of this homomorphism is
trivial.

To this end, suppose that $f\in\operatorname{Homeo}^{+}(S^{z})$ is any element
such that $\tilde{f}_{*}$ is the identity. Let

$G=p_{*}(\pi_{1}(\widetilde{S},\widetilde{z}))\triangleleft\pi_{1}(S,z),$

so that $p_{*}$ is an isomorphism from $\pi_{1}(\widetilde{S},\widetilde{z})$
onto the normal subgroup $G$ (normality follows from regularity of $p$). Since
$\widetilde{f}$ is a lift of $f$ that preserves $\widetilde{z}$, we have that

$p_{*}\circ\tilde{f}_{*}=f_{*}\circ p_{*}.$

In particular, this means that $f_{*}$ preserves $G$ and $p_{*}$ conjugates
$\widetilde{f}_{*}$ to $f_{*}|_{G}$. Since $\tilde{f}_{*}$ is the identity,
this implies that $f_{*}|_{G}$ is the identity. We claim that $f_{*}$ is the
identity. To prove this, first observe that for all $x\in G$ and
$a\in\pi_{1}(S,z)$, we have $axa^{-1}\in G$, and thus

$axa^{-1}=f_{*}(axa^{-1})=f_{*}(a)f_{*}(x)f_{*}(a)^{-1}=f_{*}(a)xf_{*}(a)^{-1}.$

This implies that for all $a\in\pi_{1}(S,z)$ and all $x\in G$,
$a^{-1}f_{*}(a)$ commutes with $x$.

Since $\pi_{1}(S,z)$ is a non-abelian surface group (either free or a closed
surface group), we know that centralizers of elements are cyclic, and since
$G$ is a non-trivial normal subgroup of $\pi_{1}(S,z)$, we can find two
elements $x,y\in G$ who centralizers intersect trivially. Now for any
$a\in\pi_{1}(S,z)$, $a^{-1}f_{*}(a)$ is in the centralizer of $x$ and $y$, and
thus $a^{-1}f_{*}(a)=1$. Therefore, $f_{*}(a)=a$, and hence $f_{*}$ is the
identity proving that $\phi_{*}$, and hence $\phi$, is injective.

To prove the last statement, given any element $g$ of a group, let $c_{g}$
denote the inner automorphism determined by conjugating by $g$. From the
inclusions in the Birman exact sequence as noted above, given any
$[\gamma]\in\pi_{1}(\widetilde{S},\widetilde{z})$ we have

$\widetilde{\iota}([\gamma])=c_{[\gamma]}\quad\mbox{ and
}\quad\iota(p_{*}([\gamma]))=c_{p_{*}([\gamma])}.$

Since $\phi_{*}\circ\iota=\widetilde{\iota}\circ\phi$, we have

$\widetilde{\iota}(\phi(p_{*}([\gamma])))=\phi_{*}(\iota(p_{*}([\gamma])))=\phi_{*}(c_{p_{*}([\gamma])})=c_{[\gamma]}=\widetilde{\iota}([\gamma]),$

where the second-to-last equality comes from the fact that the isomorphism
described above, $p_{*}\colon\pi_{1}(\widetilde{S},\widetilde{z})\to
G<\pi_{1}(S,z)$, conjugates $c_{[\gamma]}$ to $c_{p_{*}([\gamma])}$, while the
argument above shows that it conjugates $\phi_{*}(c_{p_{*}([\gamma])})$ to
this as well, hence $c_{[\gamma]}=\phi_{*}(c_{p_{*}([\gamma])})$. Therefore,
$\widetilde{\iota}\circ\phi\circ p_{*}=\widetilde{\iota}$ and since
$\widetilde{\iota}$ is injective, it follows that $\phi\circ p_{*}$ is the
identity on $\pi_{1}(\widetilde{S},\widetilde{z})$, as required. ∎

###  3.2. The once-punctured torus minus a Cantor set

Here we prove Theorem 1 in one of the two cases.

###### Proof of Theorem 1 for the once-punctured torus minus a Cantor set.

Let $T$ be a torus and choose, once and for all, a Cantor set $C\subset T$,
set $S=T\smallsetminus C$, fix $z\in S$, and write $S^{z}=S\smallsetminus z$
as above. Then $C\cup\\{z\\}$ is canonically homeomorphic to
$\operatorname{Ends}(S^{z})$ and $T$ is the end-compactification of $S^{z}$.
Consider the surjective homomorphism

$\pi_{1}(S,z)\to\pi_{1}(T,z)\cong\mathbb{Z}\times\mathbb{Z}$

given by “filling in” every element of $C$. Fixing any integer $m\geq 2$ and
reducing modulo $m$, we get a surjective homomorphism

$\rho:\pi_{1}(T,z)\to\mathbb{Z}_{m}\times\mathbb{Z}_{m}.$

Let $p:\widetilde{T}\to T$ be the regular cover corresponding to $\ker(\rho)$,
and fix some $\tilde{z}\in p^{-1}(z)$. Write $\widetilde{C}=p^{-1}(C)$, and
observe that the surface
$\widetilde{S}=\widetilde{T}\smallsetminus\widetilde{C}$ is homeomorphic to
$S$, and
$\widetilde{S}^{\widetilde{z}}=\widetilde{S}\smallsetminus\\{\widetilde{z}\\}$
is homeomorphic to $S^{z}$.

Every homeomorphism of $T$ lifts to $\widetilde{T}$ since the kernel of

$\pi_{1}(T,z)\to\mathbb{Z}_{m}\times\mathbb{Z}_{m}$

is characteristic. Moreover, every homeomorphism of $S$ extends uniquely to a
homeomorphism of $T$ and hence the restricted covering (with the same name)
$p\colon\widetilde{S}\to S$ is geometrically characteristic. Proposition 3.2
therefore provides an injective, continuous homomorphism
$\phi\colon\operatorname{Map}(S^{z})\to\operatorname{Map}(\widetilde{S}^{\widetilde{z}})$.
Since $\widetilde{S}^{\widetilde{z}}$ and $S^{z}$ are homeomorphic surfaces,
by choosing a homeomorphism between them we can view $\phi$ as a continuous
homomorphism
$\phi\colon\operatorname{Map}(S^{z})\to\operatorname{Map}(S^{z})$. We have
already shown that $\phi$ is injective. We now verify the remaining properties
claimed in the statement:

Compact to non-finite support. To prove part (2) of Theorem 1 we will show
that the $\phi$–image of the Dehn twist
$t_{a}\in\operatorname{Map}_{c}(S^{z})$ about a non-separating curve $a$ is
not supported on any finite-type subsurface of $S^{z}$.

In order to achieve this it suffices to find a point of $\tilde{C}$ that is
not fixed by our chosen lift $\widetilde{t}_{a}=\phi(t_{a})$. Let
$\beta\subset S^{z}$ be an arc which begins at $z$ and ends at a point $e\in
C$ so that $\beta$ essentially intersects $a$ exactly once. If $\tilde{\beta}$
is the lift of $\beta$ starting at $\widetilde{z}$ then
$\widetilde{t}_{a}(\tilde{\beta})$ also starts at $\widetilde{z}$, but these
two arcs terminate at distinct points in the preimage of $e$. Therefore
$\phi(t_{a})$ does not have finite support. (Observe, however, that there is a
nontrivial power of $\phi(t_{a})$ that has finite support; this will not
happen in the case of the Loch Ness Monster.)

Not co-Hopfian. Suppose $\phi$ were surjective, and thus an automorphism.
Theorem Theorem implies that $\phi$ is induced by a homeomorphism of $S^{z}$,
and in particular preserves the property of having finite support, which is a
contradiction to the previous paragraph. This proves part (1)

Non-geometric. To prove part (3), and thus complete the proof of Theoreom 1,
we must show that there are partial pseudo-Anosovs in
$\operatorname{Map}(S^{z})$ that map to multitwists in
$\operatorname{Map}(\widetilde{S}^{\widetilde{z}})$. The proof is essentially
the same as that of [2, Theorem 2(1)]; we sketch it for completeness.

As in [2], one can find a loop $\gamma$ representing an element in
$\pi_{1}(\widetilde{S},\widetilde{z})$ which is simple, but for which
$p_{*}([\gamma])$ is not represented by any simple closed curve. The mapping
class associated to $[\gamma]$ via the Birman exact sequence is a multi-twist
about the boundary of a regular neighborhood of $\gamma$, while by a result of
Kra [25], $p_{*}([\gamma])$ is a pseudo-Anosov on $X\smallsetminus\\{z\\}$,
where $X$ is the subsurface filled by a loop representing $p_{*}([\gamma])$
with minimal self-intersection. Since $\phi\circ p_{*}([\gamma])=[\gamma]$ by
Proposition 3.2, it follows that $p_{*}([\gamma])$ is pseudo-Anosov on a
proper subsurface while $\phi(p_{*}([\gamma]))$ is a multi-twist. This
completes the proof of the theorem in this case. ∎

###  3.3. The Loch Ness Monster

In this section we prove Theorem 1 for the once-punctured Loch Ness Monster.
In what follows, $S$ will denote the Loch Ness Monster, that is, the connected
orientable surface of infinite genus and exactly one end. As in the case of a
torus minus a Cantor set, we fix once and for all, a point $z\in S$. We will
again apply Proposition 3.2 for an appropriate cover $p\colon\widetilde{S}\to
S$ to induces an injective homomorphism
$\phi\colon\operatorname{Map}(S^{z})\to\operatorname{Map}(\widetilde{S}^{\widetilde{z}})$.
This time the cover will be of infinite degree which, in turn, will be the key
to the existence of compactly-supported elements with image for which no non-
trivial power has compact support. Consider the homomorphism

$\rho:\pi_{1}(S,z)\to H_{1}(S,\mathbb{Z}_{2})$

obtained by first abelianizing and then reducing modulo 2. Observe that
$H:=\ker(m)$ is a characteristic subgroup of $\pi_{1}(S,z)$. Let
$p:\tilde{S}\to S$ be the cover associated to $H$, which is usually called the
mod-2 homology cover of $S$. We claim:

######  Proposition 3.3.

$\widetilde{S}$ is homeomorphic to $S$.

###### Proof.

First, observe that the Loch Ness Monster $S$ is a characteristic cover of the
closed surface $\Sigma$ of genus $2$; more precisely, it is the covering
surface associated to the commutator subgroup of $\pi_{1}(\Sigma)$. Since
$\widetilde{S}$ is a characteristic cover of $S$, it follows that
$\widetilde{S}$ is also a characteristic cover of $\Sigma$.

Let $H$ be the characteristic subgroup of $\pi_{1}(\Sigma)$ corresponding to
$\pi_{1}(\widetilde{S})$ and $D=\pi_{1}(\Sigma)/H$ the group of deck
transformations of $\tilde{S}\to\Sigma$. Since this action is properly
discontinuous and cocompact, the S̆varc-Milnor Lemma (see e.g. [13]) implies
that $D$ is quasi-isometric to $\widetilde{S}$. By Stallings’ Theorem [33]
$D$, or equivalently $\tilde{S}$, has one end, two ends, or infinitely many
ends. By the classification of infinite-type surfaces [29], it follows that
$\widetilde{S}$ is homeomorphism to either the Loch-Ness Monster $S$; Jacob’s
Ladder $L$; the Cantor tree $T$; or the Blooming Cantor Tree $B$; see [7] for
details. Since $S$ is the only one of these that has one end, we suppose
$\widetilde{S}$ has more than one end and derive a contradiction.

To this end, we appeal to Stallings Theorem again and note that $D$ admits a
non-trivial action on a tree $A$ with finite edge stabilizers and without edge
inversions. From this action construct an equivariant map
$f\colon\widetilde{S}\to A$, which we can assume is transverse to the union of
midpoints $X\subset T$ of every edge, so that $f^{-1}(X)$ is a properly
embedded $1$–submanifold which is $D$–invariant (c.f. [32]). This descends to
a closed $1$–submanifold in $\Sigma$, and via the action of
$\pi_{1}(\Sigma)\to D$ on $A$, any loop in the complement of the
$1$–submanifold fixes a vertex. Observe that there is a non-separating simple
closed curve in a component of this complement (this is true for any
$1$–submanifold in the genus $2$ surface $\Sigma$) which, when viewed as an
element of $\pi_{1}(\Sigma)$, fixes a vertex. Since the cover is
characteristic, every non-separating simple closed curve fixes a vertex of
$A$.

Now choose a generating set $a_{1},\ldots,a_{4}$ so that all $a_{i}$ are
simple as are $a_{i}a_{j}$, for all $i\neq j$. Then all $a_{i}$ and
$a_{i}a_{j}$ act elliptically, in which case Serre’s criterion [30] implies
$\pi_{1}(\Sigma)$ fixes a point of $A$, which is a contradiction. ∎

###### Proof of Theorem 1 for the once-punctured Loch Ness Monster.

Fixing $\widetilde{z}\in p^{-1}(z)$, Proposition 3.2 implies that $p$ induces
an injective homomorphism
$\phi\colon\operatorname{Map}(S^{z})\to\operatorname{Map}(\widetilde{S}^{\widetilde{z}})$.

Since $p$ has infinite degree, there are Dehn twists whose image does not have
compact support, even up to taking powers. For the same reason, $\phi$ is not
surjective. The fact that there exist partial pseudo-Anosovs whose image is a
Dehn twist follows along the exact same lines as in the torus case. Finally,
since $\widetilde{S}$ and $S$ are homeomorphic, so are
$\widetilde{S}^{\widetilde{z}}$ and $S^{z}$, and thus we may view the
homomorphism $\phi$ as an injective, but not surjective, endomorphism of
$\operatorname{Map}(S^{z})$. This finishes the proof of the Theorem in the
case of the Loch Ness Monster. ∎

######  Remark.

After this article was finished, we learned that Proposition 3 had already
been established in [6, Proposition 4.1].

###  3.4. Decreasing genus

Our final application of Proposition 3.2 is the following.

###### Proof of Theorem 1.2.

For $g\geq 1$, let $\Sigma_{g}$ denote the surface of genus $g\geq 1$, let
$C_{g}\subset\Sigma_{g}$ be a Cantor set, and $S=\Sigma_{g}\smallsetminus
C_{g}$. Let $p\colon\widetilde{\Sigma}\to\Sigma_{g}$ be the universal cover.
Choosing a disk $D\subset\Sigma$ that contains $C_{g}$, we note that
$p^{-1}(C_{g})$ is a disjoint union of Cantor sets that only accumulate at
infinity of $\widetilde{\Sigma}$. Consequently, since the one-point
compactification of $\widetilde{\Sigma}$ is the sphere, it follows that

$\widetilde{S}=\widetilde{\Sigma}\smallsetminus p^{-1}(C_{g})$

is homeomorphic to the $2$–sphere minus a Cantor set.

Now observe that $p\colon\widetilde{S}\to S$ is a geometrically characteristic
cover, and so for any basepoint $z\in S$ and choice of $\widetilde{z}\in
p^{-1}(z)$, Proposition 3.2 implies that $p$ induces an injective homomorphism
$\phi\colon\operatorname{Map}(S^{z})\to\operatorname{Map}(\widetilde{S}^{\widetilde{z}})$.
Observe that $S^{z}$ is a surface of genus $g$ minus a Cantor set and an
isolated point, while $\widetilde{S}^{\widetilde{z}}$ is a $2$–sphere minus a
Cantor set and an isolated point, which is also homeomorphic to the plane
$\mathbb{R}^{2}$ minus a Cantor set. This completes the proof. ∎

##  4\. Self-Doubling

In this section we will prove Theorem 2 and provide a large class of self-
doubling surfaces. We will use an argument similar to the example in [24,
Section 2], although adapted to the case of surfaces without boundary.

###  4.1. The construction

Let $\bar{S}$ be an orientable connected surface with non-empty boundary. In
this section, we still require each boundary component of $\bar{S}$ be
compact, although this time we allow the set of boundary components to be
countable. Let $\mathcal{B}=\\{b_{1},b_{2},\dots\\}$ be a (possibly finite)
subset of the set of boundary components, and $S$ be the surface that results
from $\bar{S}$ by gluing disks with marked points $z_{i}$ onto $b_{i}$. This
operation gives rise to a boundary deleting homomorphism
$\operatorname{PMap}(\bar{S})\to\operatorname{PMap}(S)$, which fits in a short
exact sequence

(2) $1\to\Pi_{i}\langle
T_{b_{i}}\rangle\to\operatorname{PMap}(\overline{S})\to\operatorname{PMap}(S)\to
1,$

see [16] for details. Now suppose $d_{\mathcal{B}}S$ is the surface obtained
by gluing two disjoint copies of $\bar{S}$ along the boundary components in
$\mathcal{B}$. This operation induces two natural inclusion maps

$\psi_{1}:\bar{S}\hookrightarrow d_{\mathcal{B}}S\quad\text{ and
}\quad\psi_{2}:\bar{S}\hookrightarrow d_{\mathcal{B}}S$

such that
$\operatorname{int}(\psi_{1}(\bar{S}))\cap\operatorname{int}(\psi_{2}(\bar{S}))=\emptyset$
and $\psi_{1}(b_{i})=\psi_{2}(b_{i})$ is an essential curve. We abuse notation
by writing $b_{i}$ and $\mathcal{B}$ for the images of $b_{i}$ and
$\mathcal{B}$, respectively. Furthermore, there is an induced orientation-
reversing homeomorphism

$\iota:d_{\mathcal{B}}S\to d_{\mathcal{B}}S$

that swaps the images of $\psi_{1}$ and $\psi_{2}$ and, in particular, fixes
the set $\mathcal{B}\subset d_{\mathcal{B}}S$. On the level of mapping class
groups, we have two injective homomorphisms

$\Psi_{1}:\operatorname{PMap}(\bar{S})\hookrightarrow\operatorname{PMap}(d_{\mathcal{B}}S)\quad\text{
and
}\quad\Psi_{2}:\operatorname{PMap}(\bar{S})\hookrightarrow\operatorname{PMap}(d_{\mathcal{B}}S),$

such that for all $f\in\operatorname{PMap}(\overline{S})$, we have
$\Psi_{1}(f)=f_{1}$, $\Psi_{2}(f)=f_{2}$, where $f_{1}=\iota f_{2}\iota^{-1}$;
here we consider $f_{1}$, $f_{2}$, and $\iota$ as elements of
$\operatorname{Map}^{\pm}(d_{\mathcal{B}}S$). In particular, a left twist
about $b_{i}$ is sent to a left twist by $\Psi_{1}$, and to a right twist by
$\Psi_{2}$. Consider now the map

$d:\operatorname{PMap}(\bar{S})\to\operatorname{PMap}(d_{\mathcal{B}}S)$

such that $d(f)=f_{1}f_{2}$. Note that $d$ is indeed a homomorphism as the
images of $\Psi_{1}$ and $\Psi_{2}$ commute. Furthermore, the kernel of $d$ is
equal to $\Pi_{i}\langle T_{b_{i}}\rangle$, so from the short exact sequence
(2) we have an induced injective homomorphism

$\operatorname{PMap}(S)\hookrightarrow\operatorname{PMap}(d_{\mathcal{B}}S).$

### Self-doubling surfaces

With the doubling construction to hand, we now prove Theorem 2. To this end,
we are tasked with finding a surface $\bar{S}$ with boundary components
$\mathcal{B}$ such that the surfaces $S$ and $d_{\mathcal{B}}S$ constructed
above are homeomorphic. Note that this is equivalent to the definition of
self-doubling from the introduction.

###### Proof of Theorem 2.

Suppose the surface $S$ is self-doubling, and let $L$ and $R$ be the
subsurfaces whose interiors are homeomorphic to $S$. We then define
$\mathcal{B}=\partial L$. Following the doubling construction above, we have
the injective homomorphism

$\operatorname{PMap}(S)\cong\frac{\operatorname{PMap}(L)}{\Pi_{i}\langle
T_{b_{i}}\rangle}\hookrightarrow\operatorname{PMap}(S),$

where $b_{i}\in\mathcal{B}$. ∎

In the remainder of this section, we give conditions on the space of ends
giving rise to self-doubling surfaces.

###  4.2. Ends and orbits

A surface $S$ is _stable_ if for any end $e\in\operatorname{Ends}(S)$ there
exists a sequence of nested subsurfaces $U_{1}\supset U_{2}\supset\dots$
defining $e$ such that $U_{i}\cong U_{i+1}$ [15]. Here, we call each $U_{i}$ a
_stable neighborhood_ of $e$.

We say that an end $e\in\operatorname{Ends}(S)$ is of _higher rank_ than
$e^{\prime}\in\operatorname{Ends}(S)$ if each stable neighborhood of $e$
contains an element of the $\operatorname{Map}(S)$-orbit of $e^{\prime}$.
Finally, we define $\mathcal{F}\subset\operatorname{Ends}(S)$ to be the set of
ends whose $\operatorname{Map}(S)$-orbit is finite [15].

We can now prove the first case of Theorem 2.

######  Theorem 4.1.

Let $S$ be a stable surface with infinitely many punctures, and suppose the
genus of $S$ is either zero or infinite. If $\mathcal{F}=\emptyset$ then $S$
is self-doubling.

###### Proof.

Let $\overline{S}$ be the surface obtained by removing an open disk
surrounding one puncture $z$. Let $d_{b}S$ be the doubled surface of
$\overline{S}$ along the sole boundary component $b$. The induced map on the
space of ends gives us the homeomorphism

$\displaystyle(\operatorname{Ends}(d_{b}S),\operatorname{Ends}_{\infty}(d_{b}S))$
$\displaystyle=(\operatorname{Ends}(\overline{S}),\operatorname{Ends}_{\infty}(\overline{S}))\cup(\operatorname{Ends}(\overline{S}),\operatorname{Ends}_{\infty}(\overline{S}))$
$\displaystyle\cong(\operatorname{Ends}(S),\operatorname{Ends}_{\infty}(S))\cup(\operatorname{Ends}(S),\operatorname{Ends}_{\infty}(S)).$

Now, suppose $e\in\operatorname{Ends}(S)$ has highest rank and denote the
$\operatorname{Map}(S)$-orbit of $e$ by $O(e)$. Since $O(e)$ is infinite,
there must be an accumulation point of its elements in
$\operatorname{Ends}(S)$. However, as each element of $O(e)$ is of highest
rank this accumulation point must also belong to $O(e)$; it follows that
$O(e)$ is a Cantor set, see [27, Proposition 4.7] for more details.

Since $S$ is stable, there exists a maximal finite set of highest rank ends
$e_{1},e_{2},\dots,e_{n}$ such that $O(e_{i})\neq O(e_{j})$. Indeed, if there
exist infinitely many highest rank ends with different orbits then any
accumulation point of such ends would not have a stable neighborhood. Write
$E(e_{i})$ for the end space of a stable neighborhood of $e_{i}$ and
$E_{\infty}(e_{i})=E(e_{i})\cap\operatorname{Ends}_{\infty}(S)$. Since each
$E(e_{i})$ is a Cantor set we have that

$(\operatorname{Ends}(S),\operatorname{Ends}_{\infty}(S))\cong\bigcup_{i}(E(e_{i}),E_{\infty}(e_{i}))\cong\bigcup_{i}(E(e_{i}),E_{\infty}(e_{i}))\cup(E(e_{i}),E_{\infty}(e_{i})).$

We can now conclude that

$\displaystyle(\operatorname{Ends}(d_{b}S),\operatorname{Ends}_{\infty}(d_{b}S))$
$\displaystyle\cong(\operatorname{Ends}(S),\operatorname{Ends}_{\infty}(S))\cup(\operatorname{Ends}(S),\operatorname{Ends}_{\infty}(S))$
$\displaystyle\cong\bigcup_{i}(E(e_{i}),E_{\infty}(e_{i}))\cup(E(e_{i}),E_{\infty}(e_{i}))$
$\displaystyle\cong(\operatorname{Ends}(S),\operatorname{Ends}_{\infty}(S)),$

so $S$ and $d_{\mathcal{B}}S$ are homeomorphic. ∎

### Doubling along infinitely many boundary components

We define an end to be _truly non-planar_ if it is non-planar and has a
neighborhood that does not contain a puncture.

Figure 1. The figure shows two copies of the surface $S$. The double obtained
using half the punctures is also homeomorphic to $S$. There is a natural
orientation reversing homeomorphism $\iota$ which fixes the boundary
components of $\overline{S}$.

######  Theorem 4.2.

Suppose $S$ is a stable surface of infinite genus with infinitely many
punctures. If $\mathcal{F}$ contains no planar nor truly non-planar ends then
$S$ is self-doubling.

###### Proof.

If $\mathcal{F}$ is empty then we the result follows from Theorem 4.1. As in
the previous subsection, since $S$ is stable we have that $|\mathcal{F}|=n$ is
finite as otherwise there would be an accumulation point without a stable
neighborhood [15]. Therefore we have a partition of $\operatorname{Ends}(S)$

$\operatorname{Ends}(S)=P_{0}\cup P_{1}\cup\dots\cup P_{n}$

such that $P_{0}$ contains no elements of $\mathcal{F}$, and all other $P_{k}$
contain a single element of $\mathcal{F}$. Moreover, we may choose each
$P_{k}$ to be the end space of a stable neighborhood of the corresponding
element of $\mathcal{F}$.

Our strategy is to choose an infinite collection of punctures in each set
$P_{k}$, where $k\geq 1$, remove disk neighborhoods of these punctures
producing a surface $\overline{S}$ with boundary $\mathcal{B}$, then prove
that the resulting surface $d_{\mathcal{B}}S$ is homeomorphic to $S$. Note
that since no element of $\mathcal{F}$ is truly non-planar, each $P_{k}$ does
indeed contain infinitely many punctures, when $k\geq 1$.

Suppose $k\geq 1$ and denote by $e$ the unique end in $P_{k}\cap\mathcal{F}$.
Let $U_{0}\supset U_{1}\supset\dots$ be a nested sequence of homeomorphic
subsurfaces with connected boundary such that
$\operatorname{Ends}(U_{0})=P_{k}$. Define $X_{i}=U_{i}\setminus U_{i+1}$.
Without loss of generality we may assume that $X_{i}\cong X_{i+1}$ and that
$X_{i}$ contains at least two punctures (it may contain infinitely many
punctures). Now, remove an open neighborhood of one puncture in each $X_{i}$.
Repeating this process for each $P_{k}$ (where $k\geq 1$) results in a surface
$\overline{S}$ and a set of boundary components $\mathcal{B}$. We define
$d_{\mathcal{B}}S$ as usual, and write $d_{\mathcal{B}}(U_{i})\subset
d_{\mathcal{B}}S$ for the subsurface obtained by doubling $U_{i}$.

Note that $d_{\mathcal{B}}(U_{0})\supset d_{\mathcal{B}}(U_{1})\supset\dots$
is a nested sequence of homeomorphic subsurfaces, in particular, it defines a
unique end of $d_{\mathcal{B}}S$. We will now show that the end spaces of
$d_{\mathcal{B}}(U_{0})$ and $U_{0}$ are homeomorphic. Indeed, if each $X_{i}$
has infinitely many punctures, then by construction we have that

$\big{(}\operatorname{Ends}(d_{\mathcal{B}}(X_{i})),\operatorname{Ends}_{\infty}(d_{\mathcal{B}}(X_{i}))\big{)}\cong\big{(}\operatorname{Ends}(X_{i}\cup
X_{i+1}),\operatorname{Ends}_{\infty}(X_{i}\cup X_{i+1})\big{)}.$

However, since $U_{0}\supset U_{2}\supset U_{4}\supset\dots$ also defines the
end $e$, it follows that the end spaces of $d_{\mathcal{B}}(U_{0})$ and
$U_{0}$ are homeomorphic. This shows that $d_{\mathcal{B}}(U_{0})$ is
homeomorpic to $U_{0}$ with an open disk removed, that is,
$d_{\mathcal{B}}(U_{0})$ has two boundary components.

If each $X_{i}$ contains finitely many punctures then the argument above
follows similarly as both $d_{\mathcal{B}}(U_{0})$ and $U_{0}$ contain
infinitely many punctures with a single accumulation point. Finally, since
$P_{0}$ contains no elements of $\mathcal{F}$, the argument from Theorem 4.1
implies that $P_{0}\cong P_{0}\cup P_{0}$. It follows that

$\displaystyle(\operatorname{Ends}(d_{\mathcal{B}}S),\operatorname{Ends}_{\infty}(d_{\mathcal{B}}S))$
$\displaystyle\cong(P_{0}\cup P_{0})\cup P_{1}\cup\dots P_{n}$
$\displaystyle\cong P_{0}\cup P_{1}\cup\dots P_{n}$
$\displaystyle\cong(\operatorname{Ends}(S),\operatorname{Ends}_{\infty}(S))$

Since $d_{\mathcal{B}}S$ has infinite genus and no boundary components, it is
homeomorphic to $S$. ∎

##  5\. Twist-pair preserving homomorphisms

The purpose of this section is to prove Theorem 3. The key ingredient of our
arguments is that a twist-pair preserving homomorphism preserves certain
combinatorial configurations of curves that we term spanning chains, and which
we now define; the reader should keep Figure 2 in mind:

1pt $\alpha_{1}$ at 29 105 $\alpha_{2}$ at 65 104 $\alpha_{3}$ at 105 101
$\alpha_{4}$ at 145 105 $\alpha_{5}$ at 180 100 $\alpha_{6}$ at 205 100
$\gamma_{1}$ at 225 115 $\gamma_{2}$ at 280 104 $\gamma_{3}$ at 285 74
$\gamma_{4}$ at 286 49 $\gamma_{5}$ at 283 20 $\gamma_{6}$ at 215 25 $\beta$
at 140 10

Figure 2. A spanning chain for the surface $S_{3,5}$.

######  Definition.

Let $S_{g,p}$ be the connected orientable surface of genus $g$ and for which
the number of punctures plus boundary components is $p$. A spanning chain is a
set

$\\{\beta,\alpha_{1},\alpha_{2},\ldots,\alpha_{2g-1},\alpha_{2g},\gamma_{1},\ldots,\gamma_{p+1}\\}$

of non-separating curves on $S_{g,p}$ such that:

  * •

$i(\beta,\alpha_{4})=1$ and $i(\beta,\alpha_{j})=i(\beta,\gamma_{i})=0$ for
all $i$ and all $j\neq 3$.

  * •

$i(\alpha_{i},\alpha_{i+1})=1$, and $i(\alpha_{i},\alpha_{j})=0$ otherwise.

  * •

$i(\gamma_{i},\alpha_{2g})=1$, for all $i$, and $i(\gamma_{i},\alpha_{j})=0$
otherwise.

  * •

$i(\gamma_{i},\gamma_{j})=0$ for all $i,j$.

Observe that any surface filled by a collection of distinct curves satisfying
the conditions in the definition of a spanning chain above must have genus $g$
since a regular neighborhood of the union of the curves can be reconstructed
from the data, up to homeomorphism (not necessarily preserving the names of
the curves); see again Figure 2. We will need the following well-known fact;
see [16, Section 4.4.4.] for instance:

######  Proposition 5.1.

Suppose $g\geq 1$. If $C$ is a spanning chain on $S_{g,p}$, then the Dehn
twists along the elements of $C$ generate $\operatorname{PMap}(S_{g,p})$.

We will also make use of the braid relation between Dehn twists; see e. g.
[16, Section 3.5.1]:

######  Lemma 5.2.

Let $\alpha,\beta$ be curves on a surface. Then

$t_{\alpha}t_{\beta}t_{\alpha}=t_{\beta}t_{\alpha}t_{\beta}$

if and only if $i(a,b)=1$.

We are now ready to prove Theorem 3:

###### Proof of Theorem 3.

Let $\phi:\operatorname{Map}(S)\to\operatorname{Map}(S^{\prime})$ be a
continuous injective twist-preserving homomorphism. In particular, $\phi$
induces a simplicial map

$\phi_{*}:\mathcal{C}(S)\to\mathcal{C}(S^{\prime})$

by the rule

$\phi_{*}(\alpha)=\beta\iff\phi(t_{\alpha})=t_{\beta}.$

Moreover, $\phi_{*}$ is superinjective, meaning that $i(\alpha,\beta)=0$ if
and only if $i(\phi_{*}(\alpha),\phi_{*}(\beta))\neq 0$. Finally, Lemma 5.2
implies that $i(\phi_{*}(\alpha),\phi_{*}(\beta))=1$ if and only if
$i(\alpha,\beta)=1$.

Now, let $Z_{1}\subset Z_{2}\subset\ldots$ be an exhaustion of $S$ by
connected, properly embedded, $\pi_{1}$–injective, finite type subsurfaces for
which the inclusion of $Z_{i}$ into $S$ induces an injection
$\operatorname{PMap}(Z_{i})\to\operatorname{PMap}(S)$. Without loss of
generality, we may assume each component of $\partial Z_{i}$ is contained in
the interior of $Z_{i+1}$ or is a component of $\partial S$. Note that since
each $Z_{i}$ is properly embedded and $\operatorname{PMap}(Z_{i})$ injects
into $\operatorname{PMap}(S)$, any puncture of $S$ is a puncture of $Z_{i}$
for some $i$. Since the genus of $S$ is positive and $S$ has infinite type,
without loss of generality we may assume that the same is true for the genus
of $Z_{i}$ for all $i$, and that $Z_{i}$ is not a torus with one
puncture/boundary component for any $i$.

For each $i$, choose a spanning chain $C_{i}$ of $Z_{i}$. By the discussion
above, $\phi_{*}(C_{i})$ is a set of curves on $S^{\prime}$ of the same
cardinality as $C_{i}$, and whose elements have the same intersection pattern
as the curves in $C_{i}$. Let $Z_{i}^{\prime}$ be the surface obtained from
the regular neighborhood of $\phi_{*}(C_{i})$ by adding any complementary
once-punctured disks or peripheral annuli which a boundary component of the
neighborhood may bound. By construction $Z_{i}^{\prime}$ is filled by
$\phi_{*}(C_{i})$, and thus $Z_{i}$ and $Z_{i}^{\prime}$ have the same genus.
Moreover,
$\operatorname{PMap}(Z_{i}^{\prime})<\operatorname{PMap}(S^{\prime})$, and
Proposition 5.1 implies that the homomorphism $\phi$ restricts to an injective
homomorphism

$\phi_{i}:\operatorname{PMap}(Z_{i})\to\operatorname{PMap}(Z_{i}^{\prime}).$

Since the genera of $Z_{i}$ and $Z_{i}^{\prime}$ are equal and the
homomorphism $\phi_{i}$ is injective, [4, Theorem 1.1] implies that $\phi_{i}$
is induced by a (unique) proper $\pi_{1}$-injective embedding

$h_{i}:Z_{i}\to Z_{i}^{\prime},$

which is in fact a homeomorphism since $Z_{i}$ and $Z_{i}^{\prime}$ are each
filled by a spanning chain of the same cardinality and combinatorial type.

######  Remark.

Although the main result of [4] is stated for surfaces of genus at least four
(see the remark below the statement of Theorem 1.1 in [4]), it is also true
for twist-preserving injective homomorphisms between pure mapping class groups
of surfaces of the same positive genus. The reader can check that (after a
suitable reduction of the target surface), the standing assumption of [4,
Section 10] holds, and that every argument goes through without modification
from then on.

We also note that the main result of [4] deals with arbitrary non-trivial
homomorphisms between pure mapping class groups, and under suitable genus
bounds, any such homomorphism is induced by an embedding between the
underlying surfaces. In the case of an injective homomorphism, the reader will
quickly verify that the definition of embedding of [4] simply means “proper
topological embedding”.

Fix a complete hyperbolic metric with geodesic boundary on $S^{\prime}$. We
view $h_{i}$ as a proper embedding from $Z_{i}$ to $S^{\prime}$, and note that
$h_{i}$ and $\phi_{*}$ agree on every curve in $Z_{i}$. Since $Z_{i}$ is more
complicated than a torus with one boundary/puncture, $h_{i}$ is uniquely
determined, up to isotopy, by this agreement with $\phi_{*}$ on curves. As the
punctures of $Z_{i}^{\prime}$ were punctures of $S^{\prime}$ (by
construction), it follows that $h_{i}$ maps punctures of $Z_{i}$ to punctures
of $S^{\prime}$. Now $Z_{i}\subset Z_{i+1}$, and it follows from the
uniqueness statement that $h_{i}$ and $h_{i+1}$ agree on $Z_{i}$, up to
isotopy. Starting with $h_{1}$, and adjusting $h_{2}$ by isotopy if necessary,
we may assume that $h_{1}$ and $h_{2}$ agree on $Z_{1}$. Continuing
inductively and adjusting $h_{i+1}$ to agree with $h_{i}$ on $Z_{i}$, we get a
well-defined injective continuous function

$h\colon S\to S^{\prime},$

that agrees with $h_{i}$ on $Z_{i}$ for all $i$. Without loss of generality,
by arranging it to be the case at every step, we may assume that $h(\partial
Z_{i})=h_{i}(\partial Z_{i})$ is a union of closed geodesics in the fixed
hyperbolic metric. Since any curve in $S$ is contained in $Z_{i}$ for some
$i$, it follows that $h(\delta)=\phi_{*}(\delta)$ for every curve $\delta$ on
$S$.

We now claim that $h$ is proper. To see this, suppose for contradiction that
this were not the case. Then there exists a compact set $K$ of $S^{\prime}$
such that $h^{-1}(K)$ is not compact. We may enlarge $K$ if necessary to a
finite type, connected, properly embedded subsurface of $W\subset S^{\prime}$
with geodesic boundary (in our fixed hyperbolic metric). Then $h^{-1}(W)$ is a
non-empty, non-compact, closed subset of $S$. Now we observe that
$h^{-1}(W)\cap\partial Z_{i}\neq\emptyset$ for all $i$ sufficiently large:
otherwise for some $i$, $W$ would be entirely contained in
$h(Z_{i})=h_{i}(Z_{i})$, and hence so would $K$, implying that
$h^{-1}(K)=h_{i}^{-1}(K)$ is compact (by properness of $h_{i}$), a
contradiction. Therefore, we can find components $\alpha_{i}\subset\partial
Z_{i}$ so that for all $i$ sufficiently large, $h(\alpha_{i})$ transversely
and essentially intersects $W$. In particular, the sequence
$\\{t_{\alpha_{i}}\\}$ of mapping classes converges to the identity in
$\operatorname{Map}(S)$. In contrast, the image sequence $h(\alpha_{i})$ is a
sequence of pairwise distinct curves, all of which intersect $W$. As a
consequence, the sequence $\\{t_{h(\alpha_{i})}\\}$ cannot converge to the
identity in $\operatorname{Map}(S^{\prime})$. This contradicts the fact that
$\phi$ is continuous, as

$\phi(t_{\alpha_{i}})=t_{\phi_{*}(\alpha_{i})}=t_{h(\alpha_{i})}.$

Since $h$ is a proper embedding it induces
$h_{\sharp}\colon\operatorname{PMap}(S)\to\operatorname{PMap}(S^{\prime})$, an
injective homomorphism.

By hypothesis, $S$ either has empty boundary or else has at most one end
accumulated by genus. In the former case, it follows that $h$ is in fact a
homeomorphism, and thus $h_{\sharp}$ and $\phi$ are equal by Theorem 2.4. In
the latter case, we know that $h$ and $\phi_{*}$ agree on every isotopy class
of simple closed curve, and hence $h_{\sharp}$ and $\phi$ agree on every Dehn
twist. Since Dehn twists topologically generate $\operatorname{PMap}(S)$ [28]
and $\phi$ is continuous, it follows that $\phi$ and $h_{\sharp}$ are equal,
as required. ∎

######  Remark.

(1) As mentioned in the introduction, Theorem 3 no longer holds if $\partial
S\neq\emptyset$ and $S$ has more than one end accumulated by genus. Indeed,
suppose that $S$ has at least two ends accumulated by genus, in which case
there exists a surjective homomorphism
$\rho:\operatorname{PMap}(S)\to\mathbb{Z}$ by [3]. Since $\partial
S\neq\emptyset$, we may find a surface $S^{\prime}$ for which there exists a
$\pi_{1}$-injective embedding $\iota:S\to S^{\prime}$ and such that the
mapping class group of $S^{\prime}\setminus S$ is infinite.

Consider the injective homomorphism
$\phi:\operatorname{PMap}(S)\to\operatorname{PMap}(S^{\prime})$ induced by
this embedding, noting that its image is supported on
$\operatorname{PMap}(\iota(S))$. Now choose an infinite-order element
$g\in\operatorname{PMap}(S^{\prime})$ supported on
$S^{\prime}\setminus\iota(S)$. Using the homomorphism $\rho$ we construct a
homomorphism
$\rho_{g}:\operatorname{PMap}(S)\to\operatorname{PMap}(S^{\prime})$ whose
image is the cyclic group generated by $g$; in particular, the image of
$\rho_{g}$ is supported on $\operatorname{PMap}(S^{\prime}\setminus\iota(S))$

As the images of $\phi$ and $\rho_{g}$ commute, we may consider the
homomorphism
$(\phi,\rho_{g}):\operatorname{PMap}(S)\to\operatorname{PMap}(S^{\prime})$,
whose image is contained in

$\operatorname{PMap}(\iota(S))\times\operatorname{PMap}(S^{\prime}\setminus\iota(S))<\operatorname{PMap}(S^{\prime}).$

Observe that this homomorphism is continuous, injective (since $\phi$ is) and
twist-preserving (since the $\rho$-image of every Dehn twist is trivial, by
[3]). However, it is not induced by any subsurface embedding $S\to
S^{\prime}$, as required.

(2) As may have become apparent in the proof of Theorem 3, the requirement
that the homomorphism $\phi$ be injective may be relaxed in various ways. For
instance, say that a twist-preserving homomorphism preserves twist pairs if
two Dehn twists commute if and only if their images commute. With this
definition, a minor modification of the proof above yields:

Let $S$ and $S^{\prime}$ be surfaces of infinite type, and assume $S$ has
positive genus. Assume further that either the boundary of $S$ is empty, or
else $S$ has at most one end accummulated by genus. If
$\phi:\operatorname{PMap}(S)\to\operatorname{PMap}(S^{\prime})$ a continuous
twist-pair preserving homomorphism, then there is a proper $\pi_{1}$-injective
embedding $h:S\to S^{\prime}$ that induces $\phi$.

Indeed, the only subtlety when mimicking the proof above occurs in justifying
why the restriction homomorphism
$\phi_{i}:\operatorname{PMap}(Z_{i})\to\operatorname{PMap}(Z_{i}^{\prime})$ is
injective. To this end, if a non-central element
$f\in\operatorname{PMap}(Z_{i})$ is in the kernel of $\phi_{i}$, we may use
its Nielsen-Thurston normal form to find a curve $\alpha\subset Z_{i}$ such
that $i(f(\alpha),\alpha)>0$. However, since $f\in\ker(\phi_{i})$ we have that
$i(\phi_{*}(f(\alpha)),\phi_{*}(\alpha))=0$, which contradicts that $\phi_{*}$
is superinjective.

If on the other hand, there is $f\in\ker(\phi_{i})$ which is central, then we
have an induced injective homomorphism between the pure mapping class groups
modulo (the relevant part of) their centers, at which point the main result of
[4] applies.

##  6\. Superinjective maps of curve graphs

Finally, in this section we give a proof of Theorem 4. We separate the proof
into two parts.

###### Proof of part (1) of Theorem 4.

Equip $S$ with a fixed hyperbolic metric, and realize every vertex of
$\mathcal{C}(S)$ by its unique geodesic representative in its homotopy class.

Suppose first that $S$ has infinite genus. By choosing two points $p,q$ in the
complement of the union of all simple closed geodesics on $S$, we obtain a
(non-surjective) superinjective map
$\mathcal{C}(S)\to\mathcal{C}(S\smallsetminus\\{p,q\\})$. Now, observe that
$\mathcal{C}(S\smallsetminus\\{p,q\\})\cong\mathcal{C}(S^{\prime})$, where
$S^{\prime}$ is obtained by removing two open discs (with disjoint closures)
from $S$. Finally, we may glue a cylinder to $S^{\prime}$ along its boundary
components, obtaining a surface $S^{\prime\prime}\cong S$ and the inclusion
$S^{\prime}\to S^{\prime\prime}$ induces a superinjective map
$\mathcal{C}(S^{\prime})\to\mathcal{C}(S^{\prime\prime})\cong\mathcal{C}(S)$.
The composition of these superinjective maps/isomorphisms

$\mathcal{C}(S)\to\mathcal{C}(S\smallsetminus\\{p,q\\})\cong\mathcal{C}(S^{\prime})\to\mathcal{C}(S^{\prime\prime})\cong\mathcal{C}(S)$

is a superinjective map which is not surjective (since the first map is non-
surjective).

Suppose now that $S$ has finite genus. Then either $S$ has infinitely many
punctures or its space of ends is a Cantor set union a finite set. In the
former case we may puncture $S$ at a point $p$ in the complement of the union
of simple closed geodesics, producing a non-surjective, superinjective map

$\mathcal{C}(S)\to\mathcal{C}(S\smallsetminus\\{p\\})\cong\mathcal{C}(S),$

where the isomorphism comes from fact that $S$ and $S\smallsetminus\\{p\\}$
are homeomorphic. In the latter case, we may again remove a point $p$ missing
all simple closed geodesics to get a non-surjective, superinjective map
$\mathcal{C}(S)\to\mathcal{C}(S\smallsetminus\\{p\\})\cong\mathcal{C}(S^{\prime})$,
where $S^{\prime}$ is obtained from $S$ by removing an open disk, then glue in
a disk minus a Cantor set to $S^{\prime}$ produce a surface
$S^{\prime\prime}\cong S$. This gives a superinjective map
$\mathcal{C}(S^{\prime})\to\mathcal{C}(S^{\prime\prime})\cong\mathcal{C}(S)$,
and composing with the maps above gives the required non-surjective,
superinjective map in this case. ∎

We are now going to prove part (2) of Theorem 4. The proof is divided into two
lemmas. In what follows, we denote by $M$ the Loch Ness Monster surface, which
we again recall is the unique, up to homeomorphism, infinite-genus surface
with exactly one end.

######  Lemma 6.1.

Let $S$ be an arbitrary infinite-type surface. Then there exists a
superinjective map $\mathcal{C}(S)\to\mathcal{C}(M)$.

###### Proof.

We construct the desired map via a series of intermediate steps. First, let
$S_{1}$ be the surface obtained from $S$ by first replacing each puncture by a
boundary component, and then gluing, to each of these new boundary components,
a torus with one boundary component. Observe that if $S_{1}$ has any planar
ends, then they must belong to a Cantor set. We have a superinjective map

(3) $\mathcal{C}(S)\to\mathcal{C}(S_{1}).$

Fix a principal exhaustion $P_{1}\subset P_{2}\subset\dots$ of $S_{1}$, namely
an exhaustion of $S_{1}$ by compact subsurfaces such that every component of
$\partial P_{i}$ is separating in $S_{1}$. Let
$X^{n}_{1},X^{n}_{2},\dots,X^{n}_{k_{n}}$ be the connected components of
$P_{n}\setminus P_{n-1}$. We puncture $S_{1}$ along $y^{n}_{i}\in X^{n}_{i}$,
and then replace each of these punctures by a boundary component $b^{n}_{i}$,
obtaining a new surface $S_{2}$ and a superinjective map

(4) $\mathcal{C}(S_{1})\to\mathcal{C}(S_{2})$

The surface $S_{2}$ is naturally equipped with a principal exhaustion
$Q_{1}\subset Q_{2}\subset\dots$ coming from the obvious subsurface embedding
$S_{2}\to S_{1}$; abusing notation, we denote the connected components of
$P_{n}\setminus P_{n-1}$ in $S_{2}$ by
$Z^{n}_{1},Z^{n}_{2},\dots,Z^{n}_{k_{n}}$.

Now, we glue a sphere with $k_{n}$ boundary components to the union of the
$Z_{i}^{n}$, thus obtaining a connected surface $Y_{n}$ which contains
$Q_{n}$, see Figure 3. Moreover, $Y_{n}\subset Y_{n+1}$, so we may consider
$S_{3}=\bigcup Y_{n}$. Since $S_{2}=\bigcup Q_{n}$, we have that $S_{3}$
contains a subsurface homeomorphic to $S_{2}$, and in particular there is a
superinjective map

(5) $\mathcal{C}(S_{2})\to\mathcal{C}(S_{3})$

It remains to show that $S_{3}$ is homeomorphic to $M$. By construction,
$S_{3}$ has infinite genus and no planar ends. Since the complement of any
finite-type subsurface of $S_{3}$ contains only one component of infinite-
type, we have that $S$ has a single end which is not planar. Therefore,
$S_{3}$ is homeomorphic to $M$, as desired. At this point, the composition of
the maps in (3),(4) and (5) gives the superinjective map

$\mathcal{C}(S)\to\mathcal{C}(M).$

This finishes the proof of the lemma. ∎

Figure 3. One of the steps towards obtaining the surface $S_{3}$

Next, we prove that every surface $S^{\prime}$ of infinite genus contains a
subsurface homeomorphic to $M$.

######  Lemma 6.2.

If $S$ is a surface with infinite genus then it contains a subsurface $S\cong
M$. In particular, there is a superinjective map
$\mathcal{C}(M)\to\mathcal{C}(S)$.

###### Proof.

Let $N$ be a neighborhood of a non-planar end $e$. Let
$c_{1},c_{2},\dots\subset N$ be a set of pairwise disjoint separating curves,
each of which bound bounding a one-holed torus. Let $a_{i}$ be an arc
connecting $c_{i-1}$ and $c_{i}$ such that $a_{i}\cap a_{j}=\emptyset$.

Now, the boundary of a neighborhood of $\bigcup a_{i}\cup c_{i}$ contains a
separating arc $\alpha$. We define $\Sigma$ to be the component of
$S\setminus\alpha$ containing every $c_{i}$ (here, we take the complement
$S\setminus\alpha$ to be without boundary). It follows that $\Sigma$ has
infinite genus. Moreover, the complement of any compact subsurface contains a
single non-compact component, so $\Sigma$ has a single end and is therefore
homeomorphic to $M$. ∎

Finally, we put together the above pieces in order to prove Theorem 4:

###### Proof of Theorem 4.

Let $S$ and $S^{\prime}$ be surfaces of infinite type, and assume $S^{\prime}$
has infinite genus. By Lemma 6.1, there is a superinjective map

$\mathcal{C}(S)\to\mathcal{C}(M).$

In turn, Lemma 6.2 tells us there exists a superinjective map

$\mathcal{C}(M)\to\mathcal{C}(S^{\prime}),$

from which the result follows. ∎

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