Boundary extension of conformal mapping

Question

Let $\phi:\Omega\to D$ be a conformal mapping, where $\Omega$ is a bounded domain in $\mathbb{C}$ and $D$ is the unit disk. Caratheodory's theorem says it can extend as a homeomorphism from $\overline{\Omega}$ to $\overline{D}$ if and only if $\Omega$ is a Jordan domain. What if I only know $\phi$ can extend continuously to $\overline{\Omega}$? Is there an example that $\phi$ can extend continuously but $\Omega$ is not a Jordan domain?

Supplements:

1. What I can see is that the conformal mapping from $D-[0,1)$ to $D$ cannot extend, as it goes forwards and backwards on the line [0,1] and gives multiple values at each point, but I don't know what happens if the boundary is more complicated.

2. I found this discussion about the inverse $\phi^{-1}:D\to\Omega$. It is said $\phi^{-1}$ can extend if and only if $\partial\Omega$ is locally connected. converse to the jordan curve theorem

Answer 1

There is a simple characterization of the points $w \in \partial \Omega$ where a conformal map onto the unit disc can be extended to be continuous, namely, they need to be simple boundary points, which means that any sequence $z_n \in \Omega, z_n \to w$ can be embedded in a path $\alpha : [0,1] \to \Omega \cup w$ all contained in $\Omega$ except for $\alpha(1)=w$, in such a way that $z_n=\alpha(t_n), t_n \to 1$.

One implication is obvious since any boundary point of $D$ is simple, so if $\phi(z_n) \to \phi(w)$ we get a path as required in $D \cup \phi(w)$ and pull it back to $\Omega$, the other requires some work but is not that hard by contradiction using that univalent functions on the disc have radial limits a.e. regardless if they are bounded or not since they live in some Hardy space $H^p, 0<p<\frac{1}{2}$

If $F$ is the set of simple points of $\partial \Omega$ and $\Omega$ bounded, a harder result to prove that uses boundness essentially is that $\phi$ is injective and continuos on $\Omega \cup F$. This answers the OP question since if $\phi$ extends continuosly to $\partial \Omega$ ($\Omega$ bounded), then indeed $\Omega$ must be a Jordan curve since the extension is injective automatically!