I have a question on biholomorphic functions. I saw the following theorem in lecture:

Let $U,V\subseteq \mathbb{C}$ open subsetes and $f:U\to V$ holomorphic, bijective and $f'(z)\neq 0$ for all $z\in U$. Then $f$ is biholomorphic and it is $$(f^{-1})'=\frac{1}{f'(f^{-1}(w))}$$ for all $w\in V$.

My question is: I thought that holomorphic+bijective is enough to say that $f$ is biholomorphic. Do I need the condition $f'(z)\neq 0$ for all $z\in U$ that the inverse map is holomorphic too? I first thought if a function is holomorphic and bijective, then it follows automatically, that the inverse map is holomorphic.

  • $\begingroup$ I changed the title $\endgroup$ – dgl Jan 8 '16 at 7:57
  • $\begingroup$ If $f'(z_0) = 0$ for some $z_0 \in U$, then there exists a $w_0 \in V$ with $f^{-1}(w_0) = z_0$ and therefore the denominator is not defined for $w = w_0$. $\endgroup$ – user302234 Jan 8 '16 at 8:20
  • $\begingroup$ Related: math.stackexchange.com/questions/1048204/… $\endgroup$ – Watson Nov 24 '18 at 16:36

The assumption $f'(z) \neq 0$ is indeed superfluous. If $f'(a) = 0$, then locally near $a$, $f(z) = z^k\,g(z)$ for some holomorphic $g$ with $g(a) \neq 0$ and $k > 1$ which implies that $f$ is locally $k$-to-$1$ (in particular, not bijective).


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