
> 
> 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 **not** homotopic 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
$$\deg( f)=\begin{cases}+1&\text{ if }f\text{ is orientation-preserving at }x\_0,\\-1&\text{ if }f \text{ is orientation-reserving at }x\_0. \end{cases}$$




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This is already crossposted [here](https://math.stackexchange.com/q/4112179/873110) and has no answer yet.



Degree one self-map of $\Bbb R^2\big\backslash \big\{(n,0):n\in \Bbb Z\big\}$ not homotopic to any self-homotopy equivalence
We can produce such a map by *folding*.


We define $f(x,y)$ to be $(x,y)$ if $x < 0$, to be $(-x,y)$ if $0 \leq x \leq 1$, and to be $(x-2,y)$ if $x > 1$.


This map is proper and degree one, but is not injective at the level of fundamental groups.


Edit: Beaten by 45 seconds! I’ll leave this here as (perhaps) it is a (tiny) bit easier to see that my map is not injective on $\pi\_1$.


