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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

Dontknowanything
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  • Let $U' = { z \in U : f'(z) \neq 0}$ and $V' = f(U')$. You know $f\lvert_{U'}\colon U' \to V'$ is biholomorphic. What kind of singularity could $f^{-1}$ have in a point $w \in V\setminus V'$? – Daniel Fischer May 15 '15 at 18:16
  • @DanielFischer I'm sorry.Can you please write it as a solution with some more words? – Dontknowanything May 15 '15 at 18:21

3 Answers3

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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 $U' = U$.)

Daniel Fischer
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  • +1..neat solution!If I recall correctly,sometime back I read a solution of this problem on MSE website which just used C-linearity condition of a R linear map.I think that solution was also given by you.But now I 'm unable to find that solution.Regards, – Dontknowanything May 15 '15 at 19:01
  • Dear If you know that solution can you please attach the link? – Dontknowanything May 15 '15 at 19:11
  • I'm sorry, @Learner, I don't remember such an argument, and searching on the site didn't turn up anything either. At the moment, I don't see how such an argument would look. – Daniel Fischer May 15 '15 at 19:51
  • Dear, Thank you! I accept your answer,If you recall that argument,don't forget to attach here.Regards, – Dontknowanything May 16 '15 at 03:21
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Hint: What happens in a small disk around $z_0\in U$ if $f'(z_0)=0$?

Hagen von Eitzen
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  • I see..So we don't need open mapping theorem ! I know that injective and holomorphic function is conformal.Can you please help me in recalling the proof? – Dontknowanything May 15 '15 at 16:35
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Following the hint by Hagen von Eitzen, we can show that, if there is a $z_0$ such that $f'(z_0)=0$, then $f$ is not injective. Consider a small circular path $\gamma$ around $z_0$. By Rouche's theorem, the winding number of $f(\gamma)$ around $f(z_0)$ is the order of the zero $z_0$ of $f(z)-f(z_0)$ which, by inspecting its Taylor series at $z=z_0$, is greater than 1. So $f(\gamma)$ must cross itself somewhere, making $f$ not injective. By contraposition, $f'$ is never 0. Now we have that $f^{-1}(z)$ is holomorphic because it's differentiable and \begin{eqnarray} \frac{df^{-1}(z)}{dz}=\frac{1}{f'(f^{-1}(z))}\end{eqnarray}

Alex Fok
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