# holomorphy of inverse map

Let $M,N$ be complex manifolds and $f : M \to N$ be a bijective holomorphic map. Is then $f^{-1}$ also holomorphic?

The open mapping theorem implies that $f^{-1}$ is continuous. In order to apply the inverse function theorem, we need that the differential of $f$ is invertible. This is the case if $M,N$ are open subsets of $\mathbb{C}$. Can we generalize this do higher dimensions? If not, what happens if we assume $dim(M)=dim(N)$?

• Sorry, I do not understand your last question. It seems to me obvious that one must require $\dim(M)=\dim(N)$. – Francesco Polizzi Nov 5 '10 at 14:49
• Can you give a example of a bijective holomorphic $f : M \to N$, such that $f^{-1}$ is not holomorphic (in particular, $dim(N) \neq dim(N)$)? – Martin Brandenburg Nov 5 '10 at 14:53
• Under your assumptions ($M$ and $N$ smooth complex manifolds, $f$ one-to-one and holomorphic), in any case $f^{-1}$ is continuous, so $f$ is a homeomorphism. By Brouwer's invariance of domain, it follows $\dim(M)=\dim(N)$ as real manifolds, hence also as complex manifolds. Or am I missing something? – Francesco Polizzi Nov 5 '10 at 15:14

Yes, $f^{-1}$ is holomorphic. In fact, the following result holds, see [Griffiths-Harris, Principles of Algebraic Geometry p. 19].
If $f \colon U \to V$ is a one-to-one holomorphic map of open sets in $\mathbb{C}^n$, then $|J_f| \neq 0$, that is $f^{-1}$ is holomorphic.
The fact that $N$ is smooth is crucial. For instance, if $N \subset \mathbb{C}^2$ is the cuspidal cubic curve of equation $y^2=x^3$ and $f \colon M \to N$ is the normalization map, then $f$ is bijective and holomorphic but it is not a biholomorphism, since $f^{-1}$ is not holomorphic at the point $(0,0)$.