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Let $\varphi : U \rightarrow X$ be a holomorphic mapping of some open set $U\subseteq\mathbb{C}$ into a complex $n-$dimensional manifold $X$. If we know that this mapping is diffeomorphic onto its immage does it follow that it is also biholomorphic onto its image ?

edited Jun 12 2011 at 16:56

Qfwfq
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asked Jun 11 2011 at 6:25

martin
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I've also asked this here: mathoverflow.net/questions/44947/… – Martin Brandenburg Jun 11 2011 at 9:40

ulrich · Accepted Answer · 2011-06-11 06:36:05Z UTC

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Yes. Since the map is a diffeomorphism onto its image it means that the induced map on tangent spaces (thought of as $C^{\infty}$ manifolds) is an injection. But the tangent space as a complex manifold is the same space with an added complex structure, so it is still an injection. The (holomorphic) implicit function theorem then implies that the map is biholomorphic onlto its image.

answered Jun 11 2011 at 6:36

ulrich
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It is even sufficient that $\varphi$ be bijective and holomorphic for it to be biholomorphic. – Olivier Bégassat Jun 11 2011 at 9:12

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@Olivier: This is true only if we assume that the image is a submanifold. Consider for instance $t\mapsto (t^2,t^3)$ from $\mathbb C$ to $\mathbb C^2$. – Torsten Ekedahl Jun 11 2011 at 11:58

diffeomorphic, holomorphic, biholomorphic

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