Let $R$ be a Riemann surface, and define two quasiconformal maps $f_i : R \rightarrow R_i, i=1,2$ to be equivalent if there exists a conformal map $c : R_0 \rightarrow R_1$ and a homotopy $g_t : R \rightarrow R$ through quasiconformal maps, satisfying $g_0 = \text{id}_{R}$, $g_1 = (f_1)^{-1} \circ c \circ f_0$.
Now consider $\inf_{f \in [f_1]} \log K(f)$, where $K(f) \geq 1$ is $f$'s dilatation, and $[f_1]$ is the equivalence class of the quasiconformal map $f_1 : R \rightarrow R_1$ with respect to the equivalence relation defined above.
It is a fact that $\inf_{f \in [f_1]} \log K(f) = \log K(f^{\ast})$ for some $f^{\ast} \in [f_1]$. The proof according to most sources I have been able to find is a standard argument involving normal families.
This is what I have been able to do so far:
Let us assume that $R = \tilde{R} / \Gamma$, so that $R_1 = \tilde{R} / \Gamma_0$, for some deck transformation groups $\Gamma, \Gamma_0 < \text{Aut}(\tilde{R})$ (here $\tilde{R}$ is $R$'s universal cover) and $\Gamma_0 = \tilde{f_1} \Gamma (\tilde{f_1})^{-1}$ where $\tilde{f_1} : \tilde{R} \rightarrow \tilde{R}$ is $f_1$'s quasiconformal lift.
Then wlog we may take a sequence $f_n^{\ast} \in [f_1]$, $f_n^{\ast} : \tilde{R} / \Gamma \rightarrow \tilde{R} / \Gamma_n$, where $\Gamma_n = \tilde{f_n^{\ast}} \Gamma (\tilde{f_n^{\ast}})^{-1}$, and $\tilde{f_n^{\ast}} : \tilde{R} \rightarrow \tilde{R}$ is $f_n^{\ast}$'s quasiconformal lift fixing $0,1,\infty$, such that $\log K(f_n^{\ast}) \rightarrow \inf_{f \in [f_1]} \log K(f)$.
It follows that (some subsequence of) $\tilde{f_n^{\ast}}^\rightarrow_\rightarrow \tilde{g} $ (i.e. converges uniformly to $\tilde{g}$ on compact sets), and $\tilde{g}$ descends to a quasiconformal map $g : \tilde{R} / \Gamma \rightarrow \tilde{R} / \tilde{\Gamma}$, where $\tilde{\Gamma} = \tilde{g} \Gamma (\tilde{g})^{-1}$.
Then we necessarily have $\log K(g) = \inf_{f \in [f_1]} \log K(f)$.
My issue now is I have no idea how to show that $g \in [f_1]$.