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\title{Surfaces of infinite-type are non-Hopfian}
\alttitle{Les surfaces de type infini sont non-Hopfian}

\author{\firstname{Sumanta} \lastname{Das}\IsCorresp}
\address{Department of Mathematics, Indian Institute of Science, Bangalore 560012, India}
\email{sumantadas@iisc.ac.in}

\author{\firstname{Siddhartha} \lastname{Gadgil}}
\address[1]{Department of Mathematics, Indian Institute of Science, Bangalore 560012, India}
\email{gadgil@iisc.ac.in}

\subjclass{57K20, 55S37}

\begin{abstract}
We show that finite-type surfaces are characterized by a topological analogue of the Hopf property. Namely, an oriented surface $\Sigma$ is of finite-type if and only if every proper map $f\colon\thinspace \Sigma \to \Sigma$ of degree one is homotopic to a homeomorphism.
\end{abstract}

\begin{altabstract}
Nous montrons que les surfaces de type fini sont caractérisées par un analogue topologique de la propriété de Hopf. A savoir, une surface orientée $\Sigma$ est de type fini si et seulement si toute application propre $f\colon\thinspace \Sigma \to \Sigma$ de degré un est homotope à un homéomorphisme.
\end{altabstract}

\begin{document}
\maketitle

\section{Introduction}

All surfaces will be assumed to be connected and orientable throughout this note. We will say a surface is of \emph{finite-type} if its fundamental group is finitely generated; otherwise, we will say it is of \emph{infinite-type}.

Recall that a group $G$ is said to be \emph{Hopfian} if every surjective homomorphism $\varphi\co G\twoheadrightarrow G$ is an isomorphism. It is well known that a finitely generated free group is Hopfian, for instance, as a consequence of Grushko's theorem. On the other hand, a free group generated by an infinite set $S$ is not Hopfian as a surjective function $f\co S \to S$ that is not injective extends to a surjective homeomorphism on the free group generated by $S$ which is not injective.

In this note, we show that there is an analogous characterization for orientable surfaces of \emph{finite-type}. The natural topological analog of a surjective homomorphism is a proper map of degree one, and that of an isomorphism is a homotopy equivalence.

One-half of this characterization is classical, namely that any proper map of degree one from a surface of finite-type to itself is a homotopy equivalence. For instance, a theorem of Olum (see~\cite[Corollary~3.4]{MR192475}) says that every proper map of degree one between two oriented manifolds of the same dimension is $\pi_1$-surjective. Now, the fundamental group of any surface is residually finite (see~\cite{MR295352}). Also, any finitely generated residually finite group is Hopfian. Thus, every degree one self map of a finite-type surface is a weak homotopy equivalence, hence a homotopy equivalence by Whitehead's theorem.

Our main result is that infinite-type surfaces are not Hopfian.

\begin{theo}\label{main}
Let $\Sigma$ be any infinite-type surface. Then there exists a proper map $f\co \Sigma\to \Sigma$ of degree one such that $\pi_1(f)\co \pi_1(\Sigma)\to \pi_1(\Sigma)$ is not injective. In particular, $f$ is not a homotopy equivalence.
\end{theo}


\section{Background}

A \emph{surface} is a connected, orientable two-dimensional manifold without boundary and a \emph{bordered surfaces} is a connected, orientable two-dimensional manifold wit non-empty boundary. A (possibly bordered) subsurface $\Sigma'$ of a surface $\Sigma$ is an embedded submanifold of codimension zero.
%Each components of $\partial\Sigma'$ is an embedded curves $C\subset \Sigma$ with a tubular neighbourhood $C \times [-1, 1]$ so that $\Sigma'\cap (C \times [-1, 1]) = C \times [-1, 0]$.


Let $\Sigma$ be a non-compact surface. A \emph{boundary component} of $\Sigma$ is a nested sequence $P_1\supseteq P_2\supseteq \cdots$ of open, connected subsets of $\Sigma$ such that the followings hold:
\begin{itemize}
\item the closure (in $\Sigma$) of each $P_n$ is non-compact,
\item the boundary of each $P_n$ is compact, and
\item for any subset $A$ with compact closure (in $\Sigma$), we have $P_n\cap A=\varnothing$ for all large $n$.
\end{itemize}

We say that two boundary components $P_1\supseteq P_2\supseteq \cdots$ and $P_1'\supseteq P_2'\supseteq \cdots$ of $\Sigma$ are \emph{equivalent} if for any positive integer $n$ there are positive integers $k_n,\ell_n$ such that $P_{k_n}\subseteq P'_n$ and $P'_{\ell_n}\subseteq P_n$. For a boundary component $\mathscr P=P_1\supseteq P_2\supseteq \cdots$, we let $[\mathscr P]$ to denote the equivalence class of $\mathscr P$.

The \emph{space of ends} $\Ends (\Sigma)$ of $\Sigma$ is the topological space having equivalence class of boundary components of $\Sigma$ as elements, i.e., as a set $\Ends(\Sigma)\coloneqq\big\{[\mathscr P]\big| \mathscr P\text{ is a boundary component }\big\};$ with the following topology: For any set $X$ with compact boundary, at first, define
\[
X^\dag\coloneqq \big\{[\mathscr P=P_1\supseteq P_2\supseteq\cdots]\big| X\supseteq P_n\supseteq P_{n+1}\supseteq\cdots\text{ for some large }n\big\}.
\]

Now, take the set of all such $X^\dag$ as a basis for the topology of $\Ends(\Sigma)$. The topological space $\Ends(\Sigma)$ is compact, separable, totally disconnected, and metrizable, i.e., homeomorphic to a non-empty closed subset of the Cantor set.

For a boundary component $[\mathscr P]$ with $\mathscr P=P_1\supseteq P_2\supseteq\cdots$, we say $[\mathscr P]$ is \emph{planar} if $P_n$ are homeomorphic to open subsets $\mathbb R^2$ for all large $n$. Define $\Ends_\text{np}(\Sigma)\coloneqq\big\{[\mathscr P]: [\mathscr P]\text{ is }\text{ not }\text{ planar}\big\}$. Thus, $\Ends_\text{np}(\Sigma)$ is a closed subset of $\Ends(\Sigma)$. Also, define the \emph{genus} of $\Sigma$ as $g(\Sigma)\coloneqq\sup g(\Sbf)$, where $\Sbf$ is a compact bordered subsurface of $\Sigma$.

\begin{theo}[{Kerékjártó's classification theorem~\cite[Theorem~1]{MR143186}}]
Let $\Sigma_1,\Sigma_2$ be two non-compact surfaces. Then $\Sigma_1$ is homeomorphic to $\Sigma_2$ if and only if $g(\Sigma_1)=g(\Sigma_2)$, and there is a homeomorphism $\Phi\co \Ends(\Sigma_1)\to \Ends(\Sigma_2)$ with $\Phi\big(\Ends_{\mathrm{np}}(\Sigma_1)\big)=\Ends_{\mathrm{np}}(\Sigma_2)$.\label{richard2}
\end{theo}

Let $\Sigma$ be a non-compact surface, and let $\mathscr E_\text{np}(\Sigma)\subseteq \mathscr E(\Sigma)$ be two closed, totally-disconnected subsets of $\mathbb S^2$ such that the pair $\text{Ends}_\text{np}(\Sigma)\subseteq \text{Ends}(\Sigma)$ is homeomorphic to the pair $\mathscr E_\text{np}(\Sigma)\subseteq \mathscr E(\Sigma)$. Consider a pairwise disjoint collection $\{D_i\subseteq \mathbb S^2\setminus \mathscr E(\Sigma): i\in \mathscr A\}$ of closed disks, where $0\leq |\mathscr A|\leq g(\Sigma)$, such that the following holds: For $p\in \mathbb S^2$, any open neighborhood (in $\mathbb S^2$) of $p$ contains infinitely many $D_i$ if and only if $p\in \mathscr E_{\mathrm{np}}(\Sigma)$. \cite[Theorem~2]{MR143186} describes constructing such a collection of disks.

Now, let $M\coloneqq (\mathbb S^2\setminus \mathscr E(\Sigma))\setminus\bigsqcup_{i\in \mathscr A}\text{int}(D_i)$ and $N\coloneqq \bigsqcup_{i\in \mathscr A} S_{1,1}$, where $S_{1,1}$ is the genus one compact bordered surface with one boundary component. Define a non-compact surface $\Sigma_\text{handle}$ as follows: $\Sigma_\text{handle}\coloneqq M\bigsqcup_{\partial M\equiv \partial N} N$. Then we have the following theorem.

\begin{theo}[{Richards' representation theorem~\cite[Theorems~2 and 3]{MR143186}}]
The surface $\Sigma_{\mathrm{handle}}$ is homeomorphic to $\Sigma$. \label{richard1}
\end{theo}


\section{Proof of Theorem~\ref{main}}

Let $M$ and $N$ be two non-compact, oriented, connected, boundaryless smooth $n$-manifolds. Then the singular cohomology groups with compact support $H^n_{\cbf}(M;\mathbb Z)$ and $H^n_{\cbf}(N;\mathbb Z)$ are infinite cyclic with preferred generators $[M]$ and $[N]$. If $f\colon M\to N$ is a proper map then the degree of $f$ is the unique integer $\deg(f)$ defined as follows: $H^n_{\cbf}(f)([N])=\deg(f)\cdot [M]$. Note that $\deg$ is proper-homotopy invariant and multiplicative. See~\cite[Section~1]{MR192475} for more details.

We will use the following well-known characterization of degree.

\begin{lemm}[{\cite[Lemma~2.1b.]{MR192475}}]
Let $f\co M\to N$ be a proper map between two non-compact, oriented, connected, boundaryless smooth $n$-manifolds. Let $D$ be a smoothly embedded closed disk in $N$ and suppose $f^{-1}(D)$ is a smoothly embedded closed disk in $M$ such that $f$ maps $f^{-1}(D)$ homeomorphically onto $D$. Then $\deg (f)=+1$ or $-1$ according as $f\vert f^{-1}(D)\to D$ is orientation-preserving or orientation-reversing.\label{degreeonemapchecking}
\end{lemm}

We will prove Theorem~\ref{main} by considering the following three cases:
\begin{enumerate}
\item $\Sigma$ has infinite genus.
\item $\Sigma$ has finite genus and the set of isolated points $\mathscr I(\Sigma)$ of $\mathscr E(\Sigma)$ is finite.
\item $\Sigma$ has finite genus and the set of isolated points $\mathscr I(\Sigma)$ of $\mathscr E(\Sigma)$ is infinite.
\end{enumerate}

\begin{rema}\label{infinitetypefinitegenusmeansnumberofendsisinfinite}
If $\Sigma$ is an infinite-type surface of a finite genus, then $\mathscr E(\Sigma)$ is an infinite set.
\end{rema}

Our first result proves Theorem~\ref{main} in the case with infinite genus.

\begin{theo}
Let $\Sigma$ be a surface of the infinite genus. Then there exists a degree one map $f\co \Sigma\to \Sigma$ which is not $\pi_1$-injective. \label{infinitegenus}
\end{theo}

\begin{proof}
Since $\Sigma$ has infinite genus, there exists a compact bordered subsurface $\mathcal S\subset \Sigma$ such that $\mathcal S$ has genus one and one boundary component. Define $\Sigma'\coloneqq \Sigma/\mathcal S$ be the quotient of $\Sigma$ with $\mathcal S$ pinched to a point and let $q\co\Sigma \to \Sigma'$ be the quotient map. Thus, $\Sigma'$ is also an infinite genus surface. Further, there are compact sets in $K\subset \Sigma$ and $K' \subset\Sigma'$ whose complements are homeomorphic, so the pair $(\mathscr E(\Sigma), \mathscr E_\text{np}(\Sigma))$ is homeomorphic to the pair $(\mathscr E(\Sigma'), \mathscr E_\text{np}(\Sigma'))$. Hence, by Theorem~\ref{richard2}, there is a homeomorphism $\varphi\co\Sigma'\to\Sigma$.

Let $f \co \Sigma \to \Sigma$ be the composition $f = \varphi\circ q$. By Lemma~\ref{degreeonemapchecking}, the quotient map $q\co\Sigma\to \Sigma'$ is of degree $\pm 1$. Thus, $\deg(f)=\pm 1$ as homeomorphisms have degree $\pm 1$. Notice that $f$ sends $\partial \mathcal S$ to a point. But $\partial \mathcal S$ does not bound any disk in $\Sigma$, i.e., $\partial \mathcal S$ represents a primitive element of $\pi_1(\Sigma)$, see~\cite[Theorem~1.7. and Theorem~4.2.]{MR214087}. Hence, $f$ is not $\pi_1$-injective. If $\deg(f) = 1$, then we are done. Otherwise, we replace $f$ by $f \circ f$ to get a map that has degree one and is not injective on~$\pi_1$.
\end{proof}

For the remaining two cases, we use a map from the sphere to the sphere, which has degree $\pm 1$ but with some disks identified. We will replace these disks with appropriate surfaces to get $\Sigma$.

\begin{lemm}
There exist pairwise disjoint closed disks $\mathcal D_0, \mathcal D_1\subseteq \mathbb S^2$ and a map $f\co \mathbb S^2\to \mathbb S^2$ such that the following hold:
\begin{itemize}
\item $f^{-1}(\mathcal D_0)=\mathcal D_0$ and $f\vert_{\mathcal D_0}\co \mathcal{D}_0\to \mathcal D_0$ is the identity map.
\item $f^{-1}(\mathcal D_1)$ is the union of pairwise-disjoint closed disks $\mathcal D_{1,1}, \mathcal D_{1,2}$, and $\mathcal D_{1,3}$ in $\mathbb S^2$; and $f\vert_{\mathcal{D}_{1k}}\co \mathcal{D}_{1k}\to \mathcal D_{1}$ is a homeomorphism for each $k\in \{1,2,3\}$.\label{foldingmapconstruction}
\end{itemize}
Further, there is a loop $\gamma$ in $\mathbb S^2\setminus \Int(\mathcal D_0\cup \mathcal D_{1, 1}\cup \mathcal D_{1,2}\cup \mathcal D_{1,3})$ which is not homotopically trivial in $\mathbb S^2\setminus \Int(\mathcal D_0\cup \mathcal D_{1, 1}\cup \mathcal D_{1,2}\cup \mathcal D_{1,3})$ but such that $f(\gamma)$ is null-homotopic in $\mathbb S^2\setminus\Int (\mathcal D_0\cup \mathcal D_1)$. \label{foldingmap}
\end{lemm}

\begin{proof}
For each $k\in \{0,1,2,3\}$, choose $(a_k, b_k)\in \mathbb R^2$ such that if we define $\mathcal B_k\coloneqq\left\{(x,y)\in \mathbb R^2 : (x-a_k)^2+(y-b_k)^2\leq 1\right\}$, then $\{\mathcal B_0, \mathcal B_1, \mathcal B_2, \mathcal B_3\}$ is a pairwise-disjoint collection of closed disks.

Define $X\coloneqq\mathbb S^2\setminus\bigcup_{i=0}^3\text{int}(\mathcal B_i)$ and $Y\coloneqq\mathbb S^2\setminus\bigcup_{i=0}^1\text{int}(\mathcal B_i)$. Next, define a map $f\colon \partial X\to Y$ as follows:
\begin{itemize}
\item $f\vert_{\partial \mathcal B_k} \co \partial \mathcal B_k\to \partial \mathcal B_k$ is the identity map for each $k\in\{0,1\}$;
\item $f\vert_{\partial \mathcal B_2} \co \partial \mathcal B_2 \to \partial \mathcal B_1$ is defined as $f(x,y)\coloneqq (-x+a_2+a_1, y-b_2+b_1)$ for all $(x,y)\in \partial \mathcal B_2$.

\item $f\vert_{\partial \mathcal B_3} \co \partial \mathcal B_3\to \partial \mathcal B_1$ is defined as $f(x,y)\coloneqq (x-a_3+a_1, y-b_3+b_1)$ for all $(x,y)\in \partial \mathcal B_3$.
\end{itemize}

\begin{figure}[ht]
\[
\adjustbox{trim={0.0\width} {0.0\height} {0.0\width} {0.0\height},clip}{
\newcommand*\svgwidth{1\linewidth}
\input{figures/flippingmap.pdf_tex}
}
\]
\caption{The four-holed sphere $X$ by attaching a $2$-cell.}\label{attaching}
\end{figure}

For each $k\in \{0,1,2\}$, let $\gamma_k\co[0,1]\hookrightarrow X$ be an embedding such that $\text{im}(\gamma_k)\cap \partial X$ consists of $\gamma_k(0)=(a_k+1,b_k)\in \partial\mathcal B_k$ and $\gamma_k(1)=(a_{k+1}-1,b_{k+1})\in \partial\mathcal B_{k+1}$.

Define $\Gamma_0\co [0,1]\to Y$ as $\Gamma_0(t)\coloneqq\gamma_0(t)$ for all $t\in [0,1]$. Let $\Gamma_1, \Gamma_2\colon [0,1]\to Y$ be the constant loops based at the points $(a_1+1,b_1)\in \partial Y$ and $(a_1-1,b_1)\in \partial Y$, respectively.

Next, define $X^{(1)}\coloneqq \partial X\cup \text{im}(\gamma_0)\cup \text{im}(\gamma_1)\cup\text{im}(\gamma_2)$. Extend $f\colon \partial X\to Y$ to a map $X^{(1)}\to Y$, which we again denote by $f\co X^{(1)}\to Y$, by mapping $\gamma_0$ onto $\Gamma_0$ by the identity, and, for each $k=1,2$, mapping $\gamma_k$ to the constant loop $\Gamma_k$.

Let $\theta_0$ (resp. $\theta_3$) be the simple loop that traverses $\partial \mathcal B_0$ (resp. $\partial \mathcal B_3$) in the counter-clockwise direction starting from $(a_0+1,b_0)$ (resp. $(a_3-1,b_3)$).

Let $\theta_{1,l}$ (resp. $\theta_{1,u}$) be the simple arc that traverses $\partial \mathcal B_1\cap \{y\leq b_1\}$ (resp. $\partial \mathcal B_1\cap \{y\geq b_1\}$) counter-clockwise direction. Similarly, define $\theta_{2, l}$ and $\theta_{2,u}$.

Now, $X\cong X^{(1)}\cup_\varphi \mathbb D^2$, (see Figure~\ref{attaching}) where the attaching map $\varphi\colon \mathbb S^1\to X^{(1)}$ can be described as
\[
\varphi\coloneqq \theta_0*\gamma_0*\theta_{1,l}*\gamma_1*\theta_{2,l}*\gamma_2*\theta_3*\overline \gamma_2*\theta_{2,u}*\overline \gamma_1*\theta_{1,u}*\overline{\gamma_0}.
\]

Notice that $f(\gamma_1)=\Gamma_1$ and $f(\gamma_2)=\Gamma_2$ are constant loops. Also, as in Figure~\ref{map-extension}, $\overline{f\circ\theta_{1,l}}=f\circ \theta_{2,l}$ and $\overline{f\circ\theta_{1,u}}=f\circ \theta_{2,u}$. Thus, $f\circ\varphi$ is homotopic to $(f\circ\theta_0)*\Gamma_0*(f\circ\theta_3)*\overline{\Gamma_0}$.

\begin{figure}[ht]
\[
\adjustbox{trim={0.0\width} {0.0\height} {0.0\width} {0.0\height},clip}{
\newcommand*\svgwidth{1\linewidth}
\input{figures/flippingmap2.pdf_tex}
}
\]
\caption{The map on the $X^{(1)}$}\label{map-extension}
\end{figure}

If $r\co Y\cong \mathbb S^1\times [0,1]\to \mathbb S^1$ is the projection then $r\circ f\circ\theta_0$ and $r\circ f\circ\theta_3$ traverse $\mathbb S^1$ in opposite directions. Since $r$ is a strong deformation retract, $(f\circ\theta_0)*\Gamma_0*(f\circ\theta_3)*\overline{\Gamma_0}$, and hence $f\circ\varphi$ is null-homotopic. Now, the null-homotopic map $f\circ\varphi \colon \mathbb S^1\to Y$ can be extended to a map $\mathbb D^2\to Y$. Thus $f\co X^{(1)}\to Y$ can be extended to a map $X\cong X^{(1)}\cup_\varphi\mathbb D^2\to Y$, which will be again denoted by $f\co X\to Y$.

Note that every homeomorphism $\mathbb S^1\to \mathbb S^1$ can be extended to a homeomorphism $\mathbb D^2\to \mathbb D^2$ naturally. Thus, we can extend $f\co X\to Y$ to a map $\mathbb S^2\to \mathbb S^2$, which will be again denoted by $f\co \mathbb S^2\to \mathbb S^2$. Let $\mathcal D_0$ (resp. $\mathcal D_1$) be any closed disk, which is contained in $\text{int}(\mathcal B_0)$ (resp. $\text{int}(\mathcal B_1)$).

Finally, observe that if $\gamma=\theta_{1u}*\theta_{1l}*\gamma_1*\theta_{2l}*\theta_{2u}*\overline{\gamma_1}$, then $\gamma$ is a loop in $\mathbb S^2\setminus \Int(\mathcal D_0\cup \mathcal D_{1, 1}\cup \mathcal D_{1,2}\cup \mathcal D_{1,3})$ which is not homotopically trivial in $\mathbb S^2\setminus \Int(\mathcal D_0\cup \mathcal D_{1, 1}\cup \mathcal D_{1,2}\cup \mathcal D_{1,3})$, but such that $f(\gamma)$ is null-homotopic in $\mathbb S^2\setminus \Int(\mathcal D_0\cup \mathcal D_1)$, as claimed.
\end{proof}

We now prove Theorem~\ref{main} in the two remaining cases, in both of which we have a finite genus surface. Note that for a finite genus surface, all ends are planar, so in applying Theorem~\ref{richard2}, it suffices to consider the genus and the space of ends.

\begin{theo}
Let $\Sigma$ be a finite genus infinite-type surface such that $\mathscr E(\Sigma)$ has finitely many isolated points. Then there is a degree one map $f\co \Sigma\to \Sigma$ which is not $\pi_1$-injective.\label{finitegenswithfinitelymanyisolatedends}
\end{theo}

\begin{proof}
Let $\mathscr I(\Sigma)$ be the set of all isolated points of $\mathscr E(\Sigma)$, let $k\in\mathbb{N}\cup\{0\}$ be the cardinality of $\mathscr{I}(\Sigma)$, and let $g$ be the genus of $\Sigma$. Then $\mathscr C(\Sigma)\coloneqq\mathscr E(\Sigma)\setminus\mathscr I(\Sigma)$ is a non-empty, perfect, compact, totally-disconnected, metrizable space as it is infinite (by Remark~\ref{infinitetypefinitegenusmeansnumberofendsisinfinite}) and has no isolated points. Thus $\mathscr C(\Sigma)$ is a Cantor space (see~\cite[Theorem~8 of Chapter~12]{MR0488059}).

Let $\mathcal D_0, \mathcal D_1, \mathcal D_{1,1},\mathcal D_{1, 2},\mathcal D_{1, 3}\subseteq \mathbb S^2$, $f\co\mathbb S^2\to \mathbb S^2$, and let $\gamma$ be as in the conclusion of Lemma~\ref{foldingmapconstruction}. Let $C_1\subset \text{int}(\mathcal D_1)$ be a subset homeomorphic to the Cantor set and let $\mathscr I\subset \text{int}(\mathcal D_0)$ be a set consisting of $k$ points (hence homeomorphic to $\mathscr I(\Sigma)$). Let $C_{1, j} = f^{-1}(C_1)\cap \mathcal D_{1, j}$ for $j=1,2,3$. Note that each $C_{1, j}$ is homeomorphic to the Cantor set. See Figure~\ref{cantortree}.

\begin{figure}[ht]
\[
\adjustbox{trim={0.0\width} {0.0\height} {0.13\width} {0.0\height},clip}{
\newcommand*\svgwidth{1\linewidth}
\input{figures/drawing12.pdf_tex}
}
\]
\caption{A non $\pi_1$-injective degree $\pm 1$ map $f\colon \Sigma\to \Sigma$, where $g=3$ and $|\mathscr I|=4$.}\label{cantortree}
\end{figure}

As $f^{-1}(\mathcal D_0)=\mathcal D_0$ and $f\vert_{\mathcal D_0}\co \mathcal{D}_0\to \mathcal D_0$ is the identity map, we can say that $f^{-1}(\mathscr I) = \mathscr{I}$. Let $\Sigma_1$ be the surface obtained from $\mathbb S^2\setminus (\mathscr I\cup C_1)$ by attaching $g$ handles along disjoint disks $\Delta_{k}\subset \text{int}(\mathcal D_0)\setminus \mathscr I$, $1\leq k \leq g$ and let $\Sigma_2$ be the surface obtained from $\mathbb S^2\setminus (\mathscr I\cup C_{1, 1}\cup C_{1, 2}\cup C_{1, 3})$ by attaching $g$ handles along the (same) disks $\Delta_{k}$, $1\leq k \leq g$. Then $f$ induces a proper map, which we also call $f$, from $\Sigma_2$ to $\Sigma_1$. By Lemma~\ref{degreeonemapchecking}, $\deg(f)=\pm 1$.

Further, we claim that $f\co\Sigma_2\to\Sigma_1$ is not injective on $\pi_1$. Namely, the fundamental group of $\Sigma_2$ is the amalgamated free product of four groups, one of which is $\pi_1(\mathbb S^2\setminus \Int(\mathcal D_0\cup\mathcal D_{1, 1}\cup\mathcal D_{1,2}\cup\mathcal D_{1,3}))$. As $\gamma$ is not homotopic to the trivial loop in $\mathbb S^2\setminus \Int(\mathcal D_0\cup\mathcal D_{1, 1}\cup\mathcal D_{1,2}\cup\mathcal D_{1,3})$, and components of an amalgamated free product inject, $\gamma$ is not homotopic to the trivial loop in $\Sigma_2$. However, $f(\gamma)$ is homotopic to the trivial loop in $\mathbb S^2\setminus \Int(\mathcal D_0\cup\mathcal D_1)$ and hence in $\Sigma_1$. Hence, $f$ is not injective on $\pi_1$.

Both $\Sigma_1$ and $\Sigma_2$ have genus the same as $\Sigma$, and the space of ends homeomorphic to that of $\Sigma$ (as a finite disjoint union of Cantor spaces is a Cantor space by the universality of the Cantor set) with all ends planar. Hence, by Theorem~\ref{richard2} both $\Sigma_1$ and $\Sigma_2$ are homeomorphic to $\Sigma$.

Identifying $\Sigma_1$ and $\Sigma_2$ with $\Sigma$ by homeomorphisms, we get a proper map $f\co\Sigma\to\Sigma$ which is not injective on $\pi_1$. As homeomorphisms have degree $\pm 1$, it follows that $\deg(f)=\pm 1$. Replacing $f$ by $f\circ f$ if necessary, we obtain a proper map of degree one that is not injective on $\pi_1$.
\end{proof}

\begin{theo}
Let $\Sigma$ be a finite genus surface such that $\mathscr E(\Sigma)$ has infinitely many isolated points. Then there is a degree one map $f\co \Sigma\to \Sigma$ which is not $\pi_1$-injective.\label{finitegenswithINfinitelymanyisolatedends}
\end{theo}

\begin{proof}
Let $\mathscr I(\Sigma)$ be the set of all isolated points of $\mathscr E(\Sigma)$ and let $g$ be the genus of $\Sigma$. Let $\mathcal D_0, \mathcal D_1, \mathcal D_{1,1},\mathcal D_{1, 2},\mathcal D_{1, 3}\subseteq \mathbb S^2$, $f\co\mathbb S^2\to \mathbb S^2$, and let $\gamma$ be as in the conclusion of Lemma~\ref{foldingmapconstruction}. Let $\mathscr E$ be a subset of $\text{int}(\mathcal D_0)$ such that $\mathscr E$ is homeomorphic to $\mathscr E(\Sigma)$. Also, let $p_1\in \text{int}(\mathcal D_1)$ and $p_{1, i}\in\text{int}(\mathcal D_{1, i})$, $i=1, 2, 3$ be points such that $f(p_{1,i})=p_1$ for each $i=1,2,3$. See Figure~\ref{flutedpattern}.
\begin{figure}[ht]
\[
\adjustbox{trim={0.0\width} {0.0\height} {0.07\width} {0.0\height},clip}{
\newcommand*\svgwidth{1\linewidth}
\input{figures/drawing13.pdf_tex}
}
\]
\caption{A non $\pi_1$-injective degree $\pm 1$ map $f\colon \Sigma\to \Sigma$, where $g=5$ and $\mathscr I$ is an infinite set.}\label{flutedpattern}
\end{figure}

Recall that $f^{-1}(\mathcal D_0)=\mathcal D_0$ and $f\vert_{\mathcal D_0}\co \mathcal{D}_0\to \mathcal D_0$ is the identity map. Thus $f^{-1}(\mathscr E)=\mathscr E$. Now, let $\Sigma_1$ be the surface obtained from $\mathbb S^2\setminus (\mathscr E\cup\{p_1\})$ by attaching $g$ handles along disjoint disks $\Delta_{k}\subset \text{int}(\mathcal D_0)\setminus \mathscr I$, $1\leq k \leq g$ and let $\Sigma_2$ be the surface obtained from $\mathbb S^2\setminus (\mathscr E\cup \{p_{1,1}, p_{1, 2}, p_{1, 3}\})$ by attaching $g$ handles along the same disks $\Delta_{k}$, $1\leq k \leq g$. Then $f$ induces a proper map, which we also call $f$, from $\Sigma_2$ to $\Sigma_1$. By Lemma~\ref{degreeonemapchecking}, $\deg(f)=\pm 1$.

Further, we claim that $f\co\Sigma_2\to\Sigma_1$ is not injective on $\pi_1$. Namely, the fundamental group of $\Sigma_2$ is the amalgamated free product of four groups, one of which is $\pi_1(\mathbb S^2\setminus \Int(\mathcal D_0\cup \mathcal D_{1, 1}\cup\mathcal D_{1,2}\cup\mathcal D_{1,3}))$. As $\gamma$ is not homotopic to the trivial loop in $\mathbb S^2\setminus \Int(\mathcal D_0\cup\mathcal D_{1, 1}\cup\mathcal D_{1,2}\cup\mathcal D_{1,3})$, and components of an amalgamated free product inject, $\gamma$ is not homotopic to the trivial loop in $\Sigma_2$. However, $f(\gamma)$ is homotopic to the trivial loop in $\mathbb S^2\setminus \Int(\mathcal D_0\cup\mathcal D_1)$ and hence in $\Sigma_1$. Hence, $f$ is not injective on $\pi_1$.

Both $\Sigma_1$ and $\Sigma_2$ have genus the same as $\Sigma$ and, by Lemma~\ref{add-finite} below, $\E(\Sigma_1)$ and $\E(\Sigma_2)$ are homeomorphic to $\E(\Sigma)$. Further, all ends of $\Sigma$, $\Sigma_1$ and $\Sigma_2$ are planar. Hence, by Theorem~\ref{richard2} both $\Sigma_1$ and $\Sigma_2$ are homeomorphic to $\Sigma$.

Identifying $\Sigma_1$ and $\Sigma_2$ with $\Sigma$ by homeomorphisms, we get a proper map $f\co\Sigma\to\Sigma$ which is not injective on $\pi_1$. As homeomorphisms have degree $\pm 1$, it follows that $\deg(f)=\pm 1$. Replacing $f$ by $f\circ f$ if necessary, we obtain a proper map of degree one that is not injective on $\pi_1$.
\end{proof}

\goodbreak
\begin{lemm}
Let $\mathscr E$ be a closed totally disconnected subset of $\mathbb S^2$. Let $\mathscr I$ be the set of all isolated points of $\mathscr E$. Assume $\mathscr I$ is infinite. If $\mathscr F$ is a finite subset of $\mathbb S^2\setminus\mathscr E$, then $\mathscr E\cup \mathscr F$ is homeomorphic to~$\mathscr E.$ \label{add-finite}
\end{lemm}

\begin{proof}
Let $\mathscr A\coloneqq\{a_1,a_2,\dots\}$ be a subset of $\mathscr I$ such that $a_n\to \ell\in \mathscr E$ ($\mathscr{A}$ exists as $\mathscr E$ is compact and infinite). Define $\mathscr B\coloneqq\mathscr A\cup \mathscr F$. Write $\mathscr B$ as $\mathscr B= \{b_1,b_2,\dots\}$. Then the map $g\co \mathscr E\cup \mathscr F\to\mathscr E$ defined by
\[
g(z)\coloneqq
\begin{cases}
z&\text{if }z\in (\mathscr E\cup \mathscr F)\setminus \mathscr B,\\
a_n&\text{if }z=b_n\in \mathscr B,
\end{cases}
\]
is a homeomorphism. To prove this, note that $g$ is a bijection from a compact space to a Hausdorff space, so it suffices to show that $g$ is continuous. But observe that $g$ restricted to the closed set $(\mathscr E\cup \mathscr F)\setminus \mathscr B$ is the identity, so $g$ is continuous on $(\mathscr E\cup \mathscr F)\setminus \mathscr B$. Also $g$ restricted to the closed set $\mathscr B\cup \{\ell\}$ is continuous as $b_n\to \ell$ and $g(b_n)=a_n\to \ell = g(\ell)$, and all other points of $\mathscr B\cup \{\ell\}$ are isolated. Thus $g$ is continuous, as required.
\end{proof}

\begin{rema}
In the paper~\cite{MR4353971}, the authors have proved that for every infinite-type surface $\Sigma$, there exists a subsurface homeomorphic to $\Sigma$ such that the inclusion map is not homotopic to a homeomorphism. As our surfaces are connected, this type of inclusion map can’t be proper because of the following two facts:
\begin{itemize}
\item Any injective map between two boundaryless topological manifolds of the same dimension is an open map. This follows from the invariance of domain.
\item Any proper map between two topological manifolds is a closed map, as manifolds are compactly generated spaces, see~\cite{MR254818}.
\end{itemize}
Also, notice that all our results are related to proper maps.
\end{rema}


\subsection*{Acknowledgements}

The first author is supported by a fellowship from the National Board for Higher Mathematics. We would like to thank Hugo Parlier for telling us about the excellent result of~\cite{MR4353971}. We are grateful to the anonymous referee for his careful reading of the paper and his comments and suggestions, which helped considerably in improving the manuscript.

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