Suppose $\Omega \subset \mathbb{C}$ is open, bounded, and connected, and its boundary $\partial \Omega$ is also connected.
If it helps, we may assume $\partial \Omega$ is a Jordan curve or even that $\Omega$ is a disk.
Let $f \colon \overline{\Omega} \rightarrow \mathbb{C}$ be continuous with $f|_{\Omega} \colon \Omega \rightarrow \mathbb{C}$ holomorphic.
My question is whether $f$ necessarily sends the boundary $\partial \Omega$ into the boundary of $f(\overline{\Omega})$.
If $f$ happens to be injective, then this is part of Carathéodory's theorem. However, if $f$ is not injective, I was not able to verify whether $f(\partial \Omega) \subset \partial f(\overline{\Omega})$ must hold. (To be perfectly clear: that $f$ continuously extends to the boundary is assumed.)
Certainly, the pre-image of $\partial f(\overline{\Omega})$ lies in $\partial \Omega$ by the open mapping theorem, but this does not suffice.
If it is true, I expect it to be a topological property, i.e. it suffices to assume $f$ is an open map and does not need analyticity, but I included it in the problem statement to be on the safe side.
Any ideas or counter-examples would be greatly appreciated! Cheers.
Best Answer
It is not true. Let $f(z) = z^2$ and $\Omega = \{z \in \mathbb C \mid \lvert z \rvert < 1, \text{Im} z > 0 \}$.
Then $\partial \Omega = S^1_+ \cup L$, where $S^1_+ = \{ z \in \mathbb C \mid \lvert z \rvert = 1, \text{Im} z \ge 0 \}, L = \{ z \in \mathbb C \mid \lvert z \rvert \le 1, \text{Im} z = 0 \}$, and $\overline \Omega = \{ z \in \mathbb C \mid \lvert z \rvert \le 1, \text{Im} z \ge 0 \}$. It is easy to see that $f(\overline \Omega ) = D = \{ z \in \mathbb C \mid \lvert z \rvert \le 1 \}$ whose boundary is $\partial D = S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1 \}$.
But $f(L) = \{ z \in \mathbb C \mid 0 \le \text{Re} z \le 1, \text{Im} z = 0 \}$ which is certainly no subset of $S^1$.
In this example we have $\partial f(\Omega) = S^1 \cup L$. Thus $f(\partial \Omega) \subset \partial f(\Omega)$. However, consider the function $f(z) = z^3$ on the same $\Omega$ to see that $f(\partial \Omega) \subset \partial f(\Omega)$ is not true in general.