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.