Equality on $\partial \mathbb{H}$ of lifts for isotopy to a conformal map

Question

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!

Answer 1

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 $\overline{g_0^*} = \overline{f^*} \circ (\overline{g^*})^{-1}$.