Consider the surface $\Sigma=\Bbb R^2\big\backslash \big\{(n,0):n\in \Bbb Z\big\}$. Does there exist a proper map $f\colon \Sigma\to \Sigma$ of degree $1$ and

nothomotopic to any self-homotopy equivalence of $\Sigma$ ?

$\bullet$ Here $H_\mathbf{c}^2(\Sigma;\Bbb Z)=\Bbb Z$, so for any *proper* map $f\colon \Sigma\to \Sigma$ we can talk about the an interger, denoted by $\deg(f)$ so that the induced map $f^*\colon H_\mathbf{c}^2(\Sigma;\Bbb Z)\to H_\mathbf{c}^2(\Sigma;\Bbb Z)$ is multiplication by $\deg(f)$. This integer is invariant under *proper* homotopy.

$\bullet$ Note that $\pi_1(\Sigma)$ is a free group on countably infinitely many generators, and proper $\deg$ $1$ map is $\pi_1$-surjective, but may not be $\pi_1$-injective as a free group on infinitely many generators is not Hopfian.

$\bullet$ Here is a lemma that I am trying to use to construct a degree one map.

**Lemma:** Let $f\colon M\to N$ be a proper map between two non-compact orientable connected manifolds without boundary of same dimension. Let $x_0\in M$ be such that $f^{-1}\big(f(x_0)\big)=\{x_0\}$. Suppose $f$ is a local homeomorphism near $x_0$. Then
$$\deg( f)=\begin{cases}+1&\text{ if }f\text{ is orientation-preserving at }x_0,\\-1&\text{ if }f \text{ is orientation-reserving at }x_0. \end{cases}$$

This is already crossposted here and has no answer yet.

compact inverse image is compact. I am taking this as a definition. Just as a note, all my surfaces are boundaryless, so there is no chance of considering properness in terms of the boundary. $\endgroup$