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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 $dim(M)=dim(N)$? |
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8
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Yes, $f^{-1}$ is holomorphic. In fact, the following result holds, see [Griffiths-Harris, Principles of Algebraic Geometry p. 19]. Proposition 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 $(0,0)$. |
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3
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Any non-constant surjective holomorphic map between connected compact complex manifolds of equal dimension is a ramified finite-sheet covering. If this map is in particular bijective, then there is only one sheet, and thus is a biholomorphism. |
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