# Topological rigidity of some non-compact surfaces

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.

• 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. Sep 28 '20 at 17:57

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).

• (+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? Sep 29 '20 at 3:51
• That's a great question but I don't know the answer at all. Sep 29 '20 at 13:35
• No problem, only the finite-type surface is enough for me. Just for curiosity, I asked. Thanks again you helped me a lot. Sep 29 '20 at 13:54