Let $ \Omega _1, \Omega _2$ be domains and let $f:\Omega _1\rightarrow \Omega _2$ be biholomorphism. Is this $f$ an open function?

Let $ \Omega _1, \Omega _2$ be domains and let $f:\Omega _1\rightarrow \Omega _2$ be biholomorphism. Is this $f$ an open function?

I want to show that $f$ is continuous, bijective, open to conclude that $ \Omega _1, \Omega _2$ are homeomorphic.

Continuity and bijectivity are obvious, but I guess $f$ has to be an open function as well.

But, to show that, I could not use the open mapping theorem which only shows that $f(U)\subseteq _{op}\mathbb{C}$ for every $U\subseteq _{op}\Omega _1$, which doesn't tell $f(U)\subseteq _{op}\Omega _2$.

I also tried picking one element $ z_0$ from $f(U)$ s.t $B(z_0,\varepsilon )\nsubseteq f(U)$ for all $\varepsilon >0$, but couldn't go any further.

Is $f$ really an open function? If so, how can I show this? Any help would be appreciated.

Solution 1:

Yes, by the classic open mapping theorem this holds. More trivially, a biholomorphic function has a continuous inverse (so is a homeomorphism) and is a fortiori open.