# 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

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

• Where does this question come from? Apr 26 '21 at 6:22
• 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 $(1/2,0)$). Apr 26 '21 at 6:34
• 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. Apr 26 '21 at 6:37

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 $$\pi_1$$.

• 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. Apr 26 '21 at 6:58

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 $$1$$.

• 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. Apr 26 '21 at 6:58