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If $f$ ist holomorphic on the domain $D \subset \mathbb{C}$ and not constant then $f(D)$ is also a domain. This is the open mapping theorem.

Now I would like to know if the following statement is true:

$D \subset \mathbb{C}$ simply connected, $f:D \rightarrow \mathbb{C}$ holomorphic in $D \quad \Rightarrow \quad f(D)$ is also simply connected.

If this might be true can you actually proove it?

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It's not true. The function $f(z)=e^z$ maps $\mathbb{C}$ to $\mathbb{C}\setminus\{0\}$ which is not simply connected.

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Consider $f(z) = z^3$ on the open unit half-disk. Its image is the punctured disk.

More generally, but as simple, is $f(z)=z^n$ on any open sector spanning an angle $\theta$ with $2\pi/n < \theta < 2\pi$ (here, $n$ is a positive integer). The image is again the punctured unit disk.

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