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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 $\pi_1(\Bbb R^2-0)=\Bbb Z$.

Any help, comment, reference will be highly appreciated. Thanks in advance.

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    $\begingroup$ The homotopy automorphisms of this space correspond to homotopy automorphisms of a wedge of circles, one for each puncture. Try looking at the one associated to $a \rightarrow ab^2$ and $b \rightarrow b$. Use the winding number. $\endgroup$ Sep 28 '20 at 17:57
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The Dehn-Nielsen-Baer-Epstein theorem gives you a necessary and sufficient condition for a homotopy equivalence $f : M \to M$ to be homotopic to a homeomorphism. Here's the statement.

In the rank $n$ free group $\pi_1 M$, let $g_1,...,g_n$ be the free basis represented by loops going around the respective punctures that are pairwise disjoint except for having a common base point. By arranging these loops appropriately, the element $g_{n+1}=g_1...g_n$ represents a loop bounding a disc that contains each of the given loops, i.e. a "loop going around infinity". Let $\mathcal D = \{D_1,...,D_{2n+2}\}$ denote the set of conjugacy classes of $g_1^{\pm 1},...,g_{n+1}^{\pm 1}$ in the group $\pi_1 M$, so $D_1 = [g_1]$, $D_2 = [g_1^{-1}]$, etc.

Any homotopy equivalence $f : M \to M$ induces a permutation of the set of conjugacy classes of $\pi_1 M$. The Dehn-Nielsen-Baer-Epstein theorem says that $f$ is homotopic to a homeomorphism if and only if the induced isomorphism $f_* : \pi_1 M \to \pi_1 M$ induces a permutation of the set $\mathcal D $.

So, in the case of a 2-punctured plane $M$ for example, there is a homotopy equivalence that induces the free group automorphism defined by $g_1 \mapsto g_2$ and $g_2 \mapsto g_2 g_1$ (the existence of this homotopy equivalence follows from the easy fact that the 2-punctured sphere is an Eilenberg-Maclane space). And we have $g_3 = g_1g_2 \mapsto g_2^2 g_1$. You can immediately see that $\mathcal D = \{[g_1],[g_1^{-1}],[g_2],[g_2^{-1}],[g_3],[g_3^{-1}]\}$ is not preserved. So this homotopy equivalence is not homotopic to a homeomorphism.

Finally, it isn't too hard to see that a proper homotopy equivalence must indeed permute $\mathcal D$ and so is indeed homotopic to a homeomorphism, by application of the Dehn-Nielsen-Baer-Epstein theorem (in fact it is properly homotopic).

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  • $\begingroup$ (+1)Thanks for your answer. I will read the Dehn-Nielsen-Baer-Epstein theorem. But before that, I have a question. Can I apply this theorem in case the fundamental group of a surface is not finitely generated, let us say our surface is $\Bbb R^2\backslash\text{Cantor Set}$. If not, could you suggest me some other theorem or reference that tells any (proper)homotopy equivalence of infinite-type surface is (properly)homotopic to a homeomorphism? $\endgroup$
    – Atnamus
    Sep 29 '20 at 3:51
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    $\begingroup$ That's a great question but I don't know the answer at all. $\endgroup$
    – Lee Mosher
    Sep 29 '20 at 13:35
  • $\begingroup$ No problem, only the finite-type surface is enough for me. Just for curiosity, I asked. Thanks again you helped me a lot. $\endgroup$
    – Atnamus
    Sep 29 '20 at 13:54

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