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!



Equality on $\partial \mathbb{H}$ of lifts for isotopy to a conformal map
I found a proof and post it here for completeness/further reference.


Suppose $f \circ g^{-1}$ is isotopic to a conformal map $h \colon \mathbb{H} / \Gamma\_g \rightarrow \mathbb{H} / \Gamma\_f$. Abbreviate $g\_0 = f \circ g^{-1}$. Let $[g\_0]\_\*$ and $[h]\_\*$ denote the induced maps $\Gamma\_g \rightarrow \Gamma\_f$ on the fundamental groups. Suppose the isotopy is realized by homeomorphisms $G\_t$. Then the induced maps $[G\_t]\_\* \colon \mathbb{H} / \Gamma\_g \rightarrow \mathbb{H} / \Gamma\_f$ depend continuously on $t$ and, hence, must be the same for all $t$. In particular, $$[g\_0]\_\* = [G\_0]\_\* = [G\_1]\_\* = [h]\_\*.$$ Fix a compact fundamental domain $F$ for $\Gamma\_g$ in $\mathbb{H}$. Let $g\_0^\*$ and $h^\*$ be lifts of $g\_0$ and $h$, respectively. Given any $w \in \mathbb{H}$, take $\gamma \in \Gamma\_g$ and $z \in F$ with $\gamma(z) = w$. Then $$ d(g\_0^\*(w),h^\*(w)) = d(g\_0^\* \circ \gamma(z),h^\* \circ \gamma(z)) = d([h]\_\*(\gamma) \circ g\_0^\*(z),[h]\_\*(\gamma) \circ h^\*(z)) = d(g\_0^\*(z),h^\*(z)). $$
Since $F$ is compact, we see that $ d(g\_0^\*(w),h^\*(w)) $ is uniformly bounded in $\mathbb{H}$. As discussed in the question, we have $h' \circ \overline{g\_0^\*} = g\_0^\*$ on $\mathbb{H}$ for some conformal map $h'$. The uniform bound tells us that $h' \circ \overline{g\_0^\*}$ and $h^\*$ are the same on $\partial \mathbb{H}$. Thus, $\overline{g\_0^\*} = h'^{-1} \circ h^\*$ is a M{\"o}bius transformation on $\partial \mathbb{H}$ that fixes 0, 1 and $\infty$, i.e. it is the identity on $\partial \mathbb{H}$. Lastly, we only need to see that $\overline{g\_0^\*} = \overline{f^\*} \circ (\overline{g^\*})^{-1}$.


