This question again might be silly, like the last post([deleted](https://mathoverflow.net/questions/380672/a-self-homotopy-equivalence-of-a-closed-surface-preserves-geometric-intersection)). 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.



Homotopy equivalence preserving all geometric intersection numbers
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?]


