Problem $1$: Let $M$ be the non-compact manifold obtained from $\Bbb R^2$ removing $n$-distinct points of $\Bbb R^2$. Suppose $f:M\to M$ is a homotopy-equivalence, i.e. there is a map $g:M\to M$ such that both $f\circ g$ and $g\circ f$ are homotopic to $\text{Id}_M$. Is it true that $f:M\to M$ is homotopic to a homeomorphism of $\psi:M\to M$?
Motivation: A closed topological manifold $X$ is called topological rigid if any homotopy equivalence $F : Y → X$ with some manifold $Y$ as source and $X$ as target is homotopic to a homeomorphism. It is well-known that, any homotopy equivalence of closed surfaces deforms to a homeomorphism. Also, there are rigidity theorems, like Mostow's rigidity theorem, Bieberbach's Theorem, etc, but these manily deal with closed-manifolds, and in some cases dimensions higher than $2$.
Thoughts: Here I am considering most elementary non-compact surface, namely punctured plane $\Bbb R^2-0$. Note that, any two self-maps of $\Bbb R^2$ are homotopic as $\Bbb R^2$ is convex, so $\Bbb R^2$ excluded. Now, any homeomorphism is a proper map, so I have to find an invariant of proper map that is fixed or fully stable under ordinary homotopy. The only fact I know is, the set of regular values of a proper map is open and dense. But, my guess is : it is not a fully stable property.
My second thought is to use compactly supported cohomology, we can also consider de-Rham type cohomology as we have enough smooth maps for approximation. Note that $H^2_{\text{c}}(\Bbb R^2-0)=\Bbb R$ and, we can consider the degree of a map between compactly supported chomology groups induced by a proper map, and by checking degrees of two proper maps we can say they are properly homotopic or not. But the homotopy equivalence may not necessarily homotopic to a proper homotopy equivalence. And, this thought gives me another question written below.
Problem $2$: Is every proper self-homotopy equivalence of punctured plane properly homotopic to a self-homeomorphism of punctured plane? What about if I replace the term "punctured plane" by $M$?
My third thought is to construct an explicit homotopy equivalence of punctured plane not homotopic to a homeomorphism. Here I am tring to construct a homotopy-equivalence $f:\Bbb R^2-0\longrightarrow \Bbb R^2-0$ with $f(z)=z$ for $1<|z|<2$ and $f$ is "bad-enough" near $0$ or $\infty$ so that it is far from being homotopic to a proper map. Maybe annuls fixing property is not necessary, I am considering just because to induce an self-isomorphsim of $\pi_1(\Bbb R^2-0)=\Bbb Z$.
As you can see, I could not complete my thoughts, so, any help, comment, reference will be highly appreciated. Thanks in advance.
from Hot Weekly Questions - Mathematics Stack Exchange
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