Does a complex function is 1-1 and continuous on a open and connected set imply its inverse function is continuous?

Question

It is well known that if a real-valued function $f: I \to \mathbb{R}$ is continuous and 1-1, then $f^{-1}$ is also continuous where $I$ is an interval in $\mathbb{R}.$ I wonder if it is also true for complex function, i.e., if $f : D \to \mathbb{C}$ is continuous and 1-1, then $f^{-1}$ is also continuous. Here $D$ is open and connected in $\mathbb{C}.$

Thanks in advance!

Answer 1

Yes. This is true more generally for $\mathbb{R}^n$: if $U\subseteq\mathbb{R}^n$ is open and $f:U\to\mathbb{R}^n$ is injective and continuous, then the image $V$ of $f$ is open and the inverse map $f^{-1}:V\to U$ is continuous. This is known as "invariance of domain". The proof for $n>1$ is considerably more difficult than the proof for $n=1$, though, and requires tools from algebraic topology.