8
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Where does this question come from? $\endgroup$
    – Sam Nead
    Apr 26, 2021 at 6:22
  • 2
    $\begingroup$ 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, 2021 at 6:34
  • 2
    $\begingroup$ 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, 2021 at 6:37

2 Answers 2

7
$\begingroup$

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 $\pi_1$.

$\endgroup$
1
  • $\begingroup$ 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, 2021 at 6:58
7
$\begingroup$

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 $1$.

$\endgroup$
1
  • $\begingroup$ 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, 2021 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.