# Simply connected space and open mapping theorem

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?

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