Preimage of simply connected set Let $G\subset\mathbb{C}$ be a domain and $f:G\to\mathbb{C}$ a holomorphic, non-constant function.
Prove or disprove: If $f(G)$ is simply connected, then $G$ is simply connected.
My idea:
I know that for the complex exponential function $\exp(\mathbb{C})=\mathbb{C}^*$ holds (and $\mathbb{C}^*$ is not simply connected). I wanted to use the complex logarithm for my task, but I'm unsure, especially since it is defined across different branches.. So any hints are greatly appreciated!

Let $G=\mathbb C\setminus\{1\}$ and $f(z)=z^2$. Then, $f(G)=\mathbb C$ is simply connected, but $G$ is not.
