Every homotopy equivalence of closed surface is homotopic to a homeomorphism 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.
 A: The question you are asking is hard and irrelevant for the purpose of understand the proof in the book you are reading.

*

*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 $C^2$.


*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
