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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 ?

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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.

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    $\begingroup$ It is even sufficient that $\varphi$ be bijective and holomorphic for it to be biholomorphic. $\endgroup$ – Olivier Bégassat Jun 11 '11 at 9:12
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    $\begingroup$ @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 $\mathbb C^2$. $\endgroup$ – Torsten Ekedahl Jun 11 '11 at 11:58

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