Round circles and straight lines are conformally removable

Let $F:\widehat{\mathbb{C}}\to \widehat{\mathbb{C}}$ be a homeomorphism conformal on $\widehat{\mathbb{C}}\setminus {\mathbb{S}}^1$. By [F4], around each point of this complement its local chart expression is a conformal homeomorphism between complex domains. By [F2] every such expression is analytically $1$-quasiconformal, so $F$ is analytically $1$-quasiconformal off ${\mathbb{S}}^1$. Countable Choice used in the analytic interface follows from AC by [F8].

The unit circle is a compact round circle by [F5]. Apply the sphere clause [F3] to $F$ and ${\mathbb{S}}^1$; it follows that $F$ is analytically $1$-quasiconformal on the whole sphere. The gluing interface carries the same AC/CC assumptions recorded in [F8].

Around any point of the sphere, choose source and target holomorphic charts and restrict them so the chart expression of $F$ is a homeomorphism between complex domains. By [F4] and step 2.1 it is analytically $1$-quasiconformal; [F2] makes it conformal. Thus $F$ is a biholomorphic self-map of the sphere, and [F6] makes it Möbius. Since $F$ was arbitrary, [F1] shows that ${\mathbb{S}}^1$ is globally conformally removable.

If $\mathrm{\Gamma }=\{c+rz:\mathrm{\mid }z\mathrm{\mid }=1\}$ with $r>0$, the affine map $A(z)=c+rz$ has determinant $r\mathrm{\ne }0$ and is Möbius by [F7], with $A({\mathbb{S}}^1)=\mathrm{\Gamma }$. Möbius invariance [F1] therefore makes every round circle $\mathrm{\Gamma }$ globally conformally removable.

Let $L=p+u\mathbb{R}$ be any straight line, with $p,u\in \mathbb{C}$ and $u\mathrm{\ne }0$. The map $M(z)=p+ui(1+z)\mathrm{/}(1-z)$ has coefficient determinant $2iu\mathrm{\ne }0$, hence is Möbius by [F7]. For $z\in {\mathbb{S}}^1\setminus \{1\}$, $i(1+z)\mathrm{/}(1-z)$ is real; conversely, for $t\in \mathbb{R}$, $z=(t-i)\mathrm{/}(t+i)$ lies on ${\mathbb{S}}^1$ and maps to $t$, while $z=1$ maps to $\mathrm{\infty }$. Hence $M({\mathbb{S}}^1)=L\cup \{\mathrm{\infty }\}$, which is globally conformally removable by [F1] and step 3.1.

A compact subset $K$ of a round circle or straight line is a compact subset of the corresponding globally removable sphere circle from steps 4.1–4.2. Monotonicity in [F1] makes $K$ globally conformally removable. This proves the Statement, including ${\mathbb{S}}^1$ and $\mathbb{R}\cup \{\mathrm{\infty }\}$.