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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 $$\deg( f)=\begin{cases}+1&\text{ if }f\text{ is orientation-preserving at }x_0,\\-1&\text{ if }f \text{ is orientation-reserving at }x_0. \end{cases}$$


This is already crossposted here and has no answer yet.

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  • $\begingroup$ Where does this question come from? $\endgroup$
    – Sam Nead
    Apr 26 '21 at 6:22
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    $\begingroup$ Do you care about properness everywhere, or only at “infinity” of the plane? If you don’t mind a bit of nonproperness near the origin, you could crush a peripheral annulus (say about the origin) to a point (say $(1/2,0)$). $\endgroup$
    – Sam Nead
    Apr 26 '21 at 6:34
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    $\begingroup$ Everywhere, 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. $\endgroup$
    – Someone
    Apr 26 '21 at 6:37
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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 $\pi_1$.

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  • $\begingroup$ I can not accept both nice answers(similar). So, I decided to do this: for one case, one upvote but not ✔, and for another one, ✔ but not upvote. Sorry. $\endgroup$
    – Someone
    Apr 26 '21 at 6:58
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What about something like $f(x,y)=(g(x),y)$ where

$$ g(x)=\begin{cases} x\qquad &x\in(-\infty,1/2]\\ 1-x & x\in [1/2,3/2]\\ x-2 & x\in [3/2,\infty) \end{cases} $$

$f$ does not induce the identity mapping on $H_1$, but I do believe it is a proper map of degree $1$.

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  • $\begingroup$ I can not accept both nice answers(similar). So, I decided to do this: for one case, one upvote but not ✔, and for another one, ✔ but not upvote. Sorry. $\endgroup$
    – Someone
    Apr 26 '21 at 6:58

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