Let $\mathbb{H} \subset \mathbb{C}$ be the upper half plane. First recall the following statement: if $f^* \colon \mathbb{H} \rightarrow \mathbb{H}$ is quasi-conformal (qc), then there exists an extension $\overline{f^*} \colon \overline{\mathbb{H}} \rightarrow \overline{\mathbb{H}}$ of $f^*$. Note that this is an extension in the following sense: we do not (neccessarily) have $\overline{f^*} = f^*$ on $\mathbb{H}$ but they are conformally eqiuvalent, i.e. they are equal after composing with a conformal map.

Now fix a Fuchsian group $\Gamma$ so that $\mathbb{H} / \Gamma$ is a compact Riemann surface. A qc map $f \colon \mathbb{H} / \Gamma \rightarrow \mathbb{H} / \Gamma_f$ to some other Riemann surface has a qc lift $f^* \colon \mathbb{H} \rightarrow \mathbb{H}$. This map has an extension $\overline{f^*}$ in the sense above. After composing it with a conformal map, we can assume that $\overline{f^*}$ fixes 0, 1 and $\infty$. By a version of the measurable Riemann mapping theorem for the upper half plane, this map $\overline{f^*}$ is then unique.

I am intersted in the following statement: for two qc maps $f \colon \mathbb{H} / \Gamma \rightarrow \mathbb{H} / \Gamma_f$ and $g \colon \mathbb{H} / \Gamma \rightarrow \mathbb{H} / \Gamma_g$ with the same domain but possibly different images we have the following: the composition $f \circ g^{-1}$ is isotopic to a conformal map if and only if the lift extensions $\overline{f^*}$ and $\overline{g^*}$ agree on the real line.

I proved the backwards direction but assuming that $f \circ g^{-1}$ is isotopic to a conformal map, I do not know how to deduce equality of the extended lifts on $\mathbb{R}$.

Remark: I do not know for sure whether the statement is true but I stronly believe so.

Cheers!

newversion of $\overline{f^*}$ whose restriction to $\mathbb{H}$doesequal $f^*$. $\endgroup$ – Lee Mosher May 22 '18 at 14:56