Every homotopy equivalence of closed surface is homotopic to a homeomorphism

Question

I am reading the first proof of $\text{Theorem 8.9:}$ If $g ≥ 2$, then any homotopy equivalence $S_g → S_g$ is homotopic to a homeomorphism from the book A Primer on Mapping Class Groups.

The proof ends with the following fact:

Let $R,R'$ be homeomorphic to $S_{0,3}$ and $\phi:R\to R'$ be a continuous map such that $\phi^{-1}(\partial R')=\partial R$ with $\phi\big|\partial R\to \partial R'$ a homeomorphism. Then there is a homotopy $H:R\times [0,1]\to R'$ such that $H(-,0)=\phi$, $H(-,1)=\text{homeomorphism}$, and $H(z,t)=\phi(z)$ for all $(z,t)\in \partial R\times [0,1]$.

To prove this, the authors assume that $\phi$ is a smooth map. I think they are using the Whitney Approximation Theorem:

Suppose $N$ is a smooth manifold with or without boundary, $M$ is a smooth manifold (without boundary), and $F: N \to M$ is a continuous map. Then $F$ is homotopic to a smooth map. If $F$ is already smooth on a closed subset $A \subseteq N$, then the homotopy can be taken to be relative to $A$.

But, to apply Whitney Approximation Theorem with $A=\partial R$ and $F=\phi$ one has to ensure that $\phi\big|\partial R\to \partial R'$ is smooth also. So, here is my question.

$\textbf{My Question:}$ Let $\Sigma$ be a compact surface with non-empty boundary and $f:\Sigma\to \Sigma$ be a continuous map such that $f^{-1}(\partial \Sigma)=\partial \Sigma$ with $f\big|\partial \Sigma\to \partial \Sigma$ a homeomorphism. Is it possible to give smooth structure(s) on $\Sigma$ such that $f\big|\partial \Sigma\to \partial \Sigma$ is a diffeomorphism?

Note that every topological surface has a smooth structure unique up to diffeomorphism.

Answer 1

The question you are asking is hard and irrelevant for the purpose of understand the proof in the book you are reading.

1. The posed question is equivalent to:

Is it true that every homeomorphism $f: S^1\to S^1$ is topologically conjugate to a diffeomorphism $g: S^1\to S^1$?

Apparently, the answer depends on the degree of smoothness of $g$: A conjugation exists if we merely require $g$ to be $C^1$ but, in general, does not exist if we require $g$ to be $C^2$.

2. What you really need is much simpler:

Lemma. Every homeomorphism $f$ of the circle is isotopic to a $C^\infty$-diffeomorphism.

Proof. Step 1. Composing $f$ with a rotation $R_\theta$ we can assume that $f$ fixes a point $z$ in $S^1$: Check that $f$ is isotopic to $R_\theta\circ f$.

Step 2. Identify $S^1$ with the unit interval $I=[0,1]$ such that $z$ corresponds to the end-points of $I$; then $f$ defines a homeomorphism $h: I\to I$ which preserves the boundary set. I will assume that $h$ fixes both boundary points (and leave you to work out the case when $h$ swaps the boundary points).

Step 3. Define the linear homotopy $$ h_t(x)= tx + (1-t)h(x), x\in [0,1]. $$ Then $h_0=h, h_1=id$. Check that each $h_t$ is a 1-1, hence, a homeomorphism.

Step 4. Observe that $h_1$ also defines the identity map of the circle, hence, a diffeomorphism. qed