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This question again might be silly, like the last post(deleted). Let me know I will delete it.

Problem: Let $\Sigma$ be a surface without boundary and $f:\Sigma\to \Sigma$ be a proper homotopy equivalence. Suppose for all closed curves $\alpha,\beta:\Bbb S^1\to\Sigma$ we have $i\big([f\circ\alpha],[f\circ\beta]\big)=i\big([\alpha],[\beta]\big)$, i.e. $f$ preserves all geometric intersection numbers. Is it true that $f$ is properly homotopic to a homeomorphism?

Note that if $\Sigma$ is a closed surface, then any homotopy equivalence is homotopic to a homeomorphism. So the problem is clear without the extra assumption: "geometric intersection number is preserved."

Also, note that any homeomorphism of $\Sigma$ preserves the geometric intersection number.

I am not sure about the term "proper," i.e., the problem might be well-posed if one replaces proper homotopy equivalence with ordinary homotopy equivalence and proper homotopy with ordinary homotopy. I used the term proper keeping in mind the open surface.

Even in the closed surface case, is it possible to prove the above problem a priori not assuming "every homotopy equivalence is homotopic to a homeomorphism"?

Any help will be appreciated. Thanks in advance.

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    $\begingroup$ Is your assumption "proper, and homotopy equivalence"? Or is it "has homotopy inverse and all maps (including the homotopy) are proper"? In the first case the answer is "no" using the squaring map on the plane. In the second case, assuming that $\Sigma$ has finite type, the answer is yes, and it is a version of the Dehn-Nielsen-Baer theorem. If $\Sigma$ has infinite topological type then I am not sure... $\endgroup$
    – Sam Nead
    Commented Jan 11, 2021 at 13:00
  • $\begingroup$ All maps are proper, including homotopy. $\endgroup$
    – Sumanta
    Commented Jan 11, 2021 at 13:01
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    $\begingroup$ Then, in the case where $\Sigma$ is finite type, the answer is "yes", and you don't need any assumption on geometric intersection numbers. See Theorem 8.8 of the "Primer on mapping class groups". $\endgroup$
    – Sam Nead
    Commented Jan 11, 2021 at 13:08

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The answer is "no". $\newcommand{\CC}{\mathbb{C}}$

Consider the map $f \colon \CC \to \CC$ given by $f(z) = z^2$. All geometric intersection numbers in $\CC$ are zero, so the extra assumption holds automatically. Note that $f$ is proper, and is a homotopy equivalence.

However, there is no proper homotopy of $f$ to a homeomorphism.

EDIT (after the comments below and above):

As I said, above I am assuming that $f$ is proper, and is a homotopy equivalence. But the original poster has clarified that what is wanted is that $f$ is proper, is a homotopy equivalence, has a homotopy inverse $g$ (which is proper), and the compositions $f \circ g$ and $g \circ f$ are properly homotopic to the identity.

In this case, if $\Sigma$ has finite topological type, then $f$ is indeed properly homotopic to a homeomorphism. This is part of the Dehn-Nielsen-Baer theorem - see Theorem 8.8 of the "Primer on mapping class groups". The assumption on geometric intersection numbers is not needed.

[When $\Sigma$ has infinite topological type, I don't know the answer. But I guess that $f$ is again properly homotopic to a homeomorphism, perhaps via some exhaustion argument?]

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    $\begingroup$ But this map $f$ is not a proper homotopy equivalence (I think maybe you read the problem as asking for a proper map, which is also a homotopy equivalence). Indeed $f$ induces multiplication by $2$ on the degree $2$ Borel-Moore homology of $\mathbb C$, which is a proper homotopy invariant. $\endgroup$
    – Dan Petersen
    Commented Jan 11, 2021 at 12:19
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    $\begingroup$ Yes, that is how I read the question - see the fourth sentence of my answer. If the hypothesis on $f$ is strengthened to be "properly homotopic to the identity" then the answer is becomes "yes, vacuously". Perhaps I am missing something... $\endgroup$
    – Sam Nead
    Commented Jan 11, 2021 at 12:25
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    $\begingroup$ No, I don't think you're missing anything; $f$ is homotopic to a homeomorphism regardless of the condition on intersection numbers. $\endgroup$
    – Dan Petersen
    Commented Jan 11, 2021 at 12:31
  • $\begingroup$ Sorry, is not $f$ induces multiplication by $2$ on $H_\textbf{c}^2(\Bbb C;\Bbb Z)$? I think every proper homotopy equivalence is of degree-$\pm 1$ w.r.t. compactly supported integral second-cohomology. $\endgroup$
    – Sumanta
    Commented Jan 11, 2021 at 12:40
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    $\begingroup$ @SumantaDas "If two maps are proper, they are automatically properly homotopic due to convex combination." Preposterous, take the identity and complex conjugation. $\endgroup$
    – mme
    Commented Jan 11, 2021 at 13:35

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