Welding uniqueness for conformally removable curves

Let ${\mathrm{\Omega }}_0,{\mathrm{\Omega }}_1$ be the components of $\widehat{\mathbb{C}}\setminus \mathrm{\Gamma }$ parameterized by $f,g$, and let ${\mathrm{\Omega }}_0^{\mathrm{\prime }},{\mathrm{\Omega }}_1^{\mathrm{\prime }}$ be the corresponding components for $f^{\mathrm{\prime }},g^{\mathrm{\prime }}$. By [F1]–[F2], all four maps extend to homeomorphisms of the closures, with their boundary maps taking values in $\mathrm{\Gamma }$ and ${\mathrm{\Gamma }}^{\mathrm{\prime }}$. The Countable Choice interface used by the boundary supplier follows from AC by [F7].

Equality of the two welding maps gives $(\bar{f}{\mathrm{\mid }}_{{\mathbb{S}}^1}{)}^{-1}\circ (\bar{g}{\mathrm{\mid }}_{{\mathbb{S}}^1})=({\bar{f}}^{\mathrm{\prime }}{\mathrm{\mid }}_{{\mathbb{S}}^1}{)}^{-1}\circ ({\bar{g}}^{\mathrm{\prime }}{\mathrm{\mid }}_{{\mathbb{S}}^1})$. Composing with ${\bar{f}}^{\mathrm{\prime }}$ on the left and $(\bar{g}{)}^{-1}$ on the right yields ${\bar{f}}^{\mathrm{\prime }}\circ (\bar{f}{)}^{-1}={\bar{g}}^{\mathrm{\prime }}\circ (\bar{g}{)}^{-1}$ on the common boundary $\mathrm{\Gamma }$.

Define $F$ on $\overline{{\mathrm{\Omega }}_0}$ by $F={\bar{f}}^{\mathrm{\prime }}\circ (\bar{f}{)}^{-1}$ and on $\overline{{\mathrm{\Omega }}_1}$ by $F={\bar{g}}^{\mathrm{\prime }}\circ (\bar{g}{)}^{-1}$. These closed sets cover the sphere, and their intersection is $\mathrm{\Gamma }$; step 2.1 makes the definitions agree there. Each branch is a homeomorphism onto the corresponding primed closure. The inverse branches likewise agree on ${\mathrm{\Gamma }}^{\mathrm{\prime }}$, so the closed-set pasting argument applied to both maps shows that $F$ is a sphere homeomorphism.

On ${\mathrm{\Omega }}_0$ and ${\mathrm{\Omega }}_1$, respectively, $F$ is the conformal composition $f^{\mathrm{\prime }}\circ f^{-1}$ and $g^{\mathrm{\prime }}\circ g^{-1}$; hence it is conformal on $\widehat{\mathbb{C}}\setminus \mathrm{\Gamma }$. If $\mathrm{\Gamma }$ is globally conformally removable, [F3] makes $F$ a Möbius transformation $A$. Restricting to each component gives $f^{\mathrm{\prime }}=A\circ f$ and $g^{\mathrm{\prime }}=A\circ g$, so ${\mathrm{\Gamma }}^{\mathrm{\prime }}=A(\mathrm{\Gamma })$.

Let $h$ be quasisymmetric. By [F5], choose a welding $({\mathrm{\Gamma }}_0,f_0,g_0)$ whose curve is a quasicircle; [F6] makes ${\mathrm{\Gamma }}_0$ globally conformally removable. For any other welding $({\mathrm{\Gamma }}_1,f_1,g_1)$ of $h$, apply the conclusion of step 4.1 with $({\mathrm{\Gamma }}_0,f_0,g_0)$ first and $({\mathrm{\Gamma }}_1,f_1,g_1)$ second. Thus a Möbius map carries ${\mathrm{\Gamma }}_0$ to ${\mathrm{\Gamma }}_1$ and postcomposes both parameter maps. This proves the uniqueness claim in part (b); Countable Choice conditions on the existence route follow from AC by [F7].