I'm re-reading a paper of Bers and for the second time, and I am yet again confused about the claim in the title, which Bers declares to be easy to prove.
For context, I'll lay out some terminology. We call a pair $(S, \alpha)$ a marked Riemann surface when $S$ is a Riemann surface and $\alpha$ is an equivalence class of generators of $\pi_1S$, each of which $\{a_i\}_{i=1}^{2g}$ satisfies the standard one relation $1= \prod_{j=1}^{2g} a_{2j-1} a_{2j} a_{2j-1}^{-1}a_{2j}^{-1}$. A map between marked Riemann surfaces $f:(S, \alpha) \rightarrow (S',\alpha')$ is a homeomorphism that respects the markings. Finally, call a pair of marked Riemann surfaces $(S, \alpha), (S', \alpha')$ to be similarly oriented if there is an orientation-preserving homeomorphism $f: S \rightarrow S'$.
Now, Bers goes on to make the following claim:
If $(S, \alpha)$ and $(S', \alpha')$ are similarly oriented marked Riemann surfaces, then there is a quasiconformal mapping $f: (S, \alpha) \rightarrow (S', \alpha')$.
Ok, this sounds plausible enough to me. On the other hand, it is by no means obvious. Here, I should mention Bers has earlier stated the equivalence of a few (standard) definitions of quasiconformality for a mapping $f: S \rightarrow S'$, including the following:
(i) $\hat{f}_\bar{z} = \mu \hat{f}_{z}$, in terms of weak derivatives, for any coordinate representation $\hat{f}$, where $\mu \in L^{\infty}$
(ii) any coordinate representation $\hat{f}$ has bounded dilatation across all quadrilaterals.
Specifically, Bers goes on to say
Had we demanded that the homeomorphism $f$ be continuously differentiable everywhere the proof would be somewhat laborious. Since we use a very general definition of quasiconformality, the proof presents no difficulties and may be omitted.
My question is thus, what is the idea that Bers' has in mind for the proof? By hypothesis, we've got our hands on a homeomorphism $f$ that respects the desired marking, but a priori $f$ might have unbounded dilatation. So, we need some clever way to homotope $f$ so that its dilatation becomes bounded across the whole surface, at which point we are done. However, I see no intuitive way to make this work. Technical details aside, I am very interested to see what ideas anyone has, or better yet, what idea they imagine Bers thought would easily yield a proof.
Bers, Lipman, Quasiconformal mappings and Teichmüller’s theorem, Princeton Math. Ser. 24, 89-119 (1960). ZBL0100.28904.
These things are easier said than done. The relevant theorem is:
Theorem 1. Let $S$ be a closed connected oriented smooth surface. Then every orientation-preserving homeomorphism $f: S\to S$ is homotopic (equivalently, isotopic) to an (orientation-preserving) diffeomorphism $g: S\to S$. Alternatively, $g$ can be taken to be a piecewise-linear homeomorphism provided that $S$ is equipped with a PL structure.
If, in addition, $S$ is equipped with the structure of a Riemann surface, then $g$ is quasiconformal. (This is the easy part.)
The part about piecewise-linear homeomorphism is proven in Appendix to
D. B. A. Epstein, Curves on 2-manifolds and isotopies. Acta Math. 115 (1966), 83–107.
From that result, one obtains (by one more isotopy) a diffeomorphism, according to
J. Munkres, Obstructions to the smoothing of piecewise-differentiable homeomorphisms. Ann. of Math. (2) 72 (1960), 521–554.
One can prove more:
Theorem 2. Let $S$ be as above. Then every automorphism $\phi: \pi_1(S)\to \pi_1(S)$ is induced by a diffeomorphism $S\to S$.
In fact, one can also detect (group-theoretically) if $\phi$ is "orientation-preserving," but I will not get into this.
There are several other arguments (neither one is easy) for proving this theorem, you can find these in
B. Farb, D. Margalit, A primer on mapping class groups. Princeton Mathematical Series, 49. Princeton University Press, Princeton, NJ, 2012. Chapter 8
and (an analytical proof)
J. Hubbard, Teichmüller theory and applications to geometry, topology, and dynamics. Vol. 1. Matrix Editions, Ithaca, NY, 2006. Section 5.1.
With more work, these proofs apply to compact Riemann surfaces with finitely many punctures as long as certain conditions on $\phi$ are satisfied (these conditions are automatically satisfied in Theorem 1). The proof given in Hubbard's book requires very little extra work.