Continuous extensions of Riemann mappings

Question

Let $K$ be a compact set in $\mathbb C$ without interior. Suppose, additionally, that $K$ is a retract (or equivalently $K$ connected, $K$ locally connected and $\mathbb C\setminus K$ connected). Then $G:=S_2\setminus K$ is a simply connected domain in the Riemann sphere $S_2\sim \widehat{\mathbb C}$. It is known that under these conditions any Riemann map $f$ from the exterior (within $S_2$) of the closed unit disk $D$ onto $G$ with $f(\infty)=\infty$ has a continuous extension $F$ to the unit circle $\mathbb T$ with $K=F(\mathbb T)$. Can this be deduced from the "usual" version of bounded simply connected domains whose boundary is a curve? This is easy if $K$ has interior points. Note that the usual "trick" in the proof of Riemann's mapping theorem by considering on $G$ functions of the form $\sqrt{1/ (w-a)}$, $a\in K$, gives a priori no information on the boundaries (which get split). Why local connectedness of the boundaries is an invariant?

Answer 1

This is true, and can be proved as you suggest, and a convenient reference is Milnor, Dynamics in one complex variable. The conditions you stated are necessary and sufficient for the conformal map to be continuous in the closed disk.

Remark on terminology. This function is the inverse to the Riemann map.