Suppose $U$ and $V$ be domains(i.e., open and connected) in $ \mathbb C$.Let $f\colon U \to V$ be a bijective holomorphic function. Show that the inverse of $f$ is also holomorphic.
By Open Mapping Theorem it is clear that $f^{-1}$ is also continuous. Please give some ideas to complete the proof.
Edit:I'm interested in a proof which comes as a corollary of open mapping theorem
Best Answer
By the open mapping theorem, $f$ is a homeomorphism. When $f'(z_0) \neq 0$, the complex differentiability of $f^{-1}$ at $f(z_0)$ follows in the usual way.
Now consider $U' = \{ z\in U : f'(z) \neq 0\}$ and $V' = f(U')$. By the above, $f^{-1}$ is holomorphic on $V'$. But since $U\setminus U'$ consists only of isolated points, and $f$ is a homeomorphism, $V\setminus V'$ consists only of isolated points. Thus a $w\in V\setminus V'$ would be an isolated singularity of $f^{-1}$. Since $f^{-1}$ is continuous, it would be a removable singularity. Hence $f^{-1}$ is holomorphic on all of $V$. (And consequently we have $U' = U$.)