My problem relates to the construction of the quasiconformal map $T:\overline{\mathbb{D}} \longrightarrow \overline{\mathbb{D}}$ from beginning of the proof of Lemma 2.2. in Marshall-Rhode.
Question
The text in question writes as follows:
Let $T:\overline{\mathbb{D}} \longrightarrow \overline{\mathbb{D}}$ be a quasiconformal map with $T(i)=\alpha ^+, T(-i)=\alpha ^-, T(1)=1$ and $T^{-1}(0)\in (-1,1)$. For instance, $T$ can be constructed as a composition of a quasiconformal map that sends $\alpha^+$ and $\alpha^-$ to symmetric points, followed by a Möbius transformation.
Here $\alpha ^+$ and $\alpha ^-$ are fixed points on $\partial \mathbb{D}$ such that the oriented arc from $\alpha ^-$ to $\alpha ^+$ (denoted $\langle \alpha ^-,\alpha^+\rangle$ in the text) on $\partial \mathbb{D}$ is seperated by 1 (they are defined above Lemma 2.2, but I don't think that it is relevant how).
My questions are these: Why does the quasiconformal map sending $\alpha^-$ and $\alpha^+$ to symmetric points exist? Is it symmetric in the real axis? And how does it look?
Attempt to solve
To me it would seem as though we need to either
1) Use some sort of "quasiconformal stretching" of the the arcs $\langle \alpha^-, 1\rangle$ and $\langle 1, \alpha^+\rangle$, where we somehow try to preserve the disk and map $\alpha^{-}$ and $\alpha^+$ into $c$ and $\bar{c}$ respectively for some $c\in \partial \mathbb{D}$. Then we could use the Möbius transformation which maps $(\bar{c},c,1)$ into $(-i,i,1)$ and so kinda ``push'' $0$ around on the real line (satisfying $T^{-1}(0)\in (-1,1)$), or
2) Quasiconformally stretch $\overline{\mathbb{D}}$ to an elliptical disk such that the two points become symmetric and then map the elliptical disk conformally to $\overline{\mathbb{D}}$ with a Möbius transformation.
Any help would be greatly appreciated.
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