most recent 30 from http://mathoverflow.net 2013-06-18T07:00:18Z http://mathoverflow.net/feeds/question/44947 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44947/holomorphy-of-inverse-map Martin Brandenburg 2010-11-05T14:05:44Z 2010-11-06T03:17:25Z

<p>Let $M,N$ be complex manifolds and $f : M \to N$ be a bijective holomorphic map. Is then $f^{-1}$ also holomorphic?</p> <p>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 $dim(M)=dim(N)$?</p>

http://mathoverflow.net/questions/44947/holomorphy-of-inverse-map/44950#44950 Colin Tan 2010-11-05T14:34:54Z 2010-11-06T03:17:25Z

<p>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.</p>

http://mathoverflow.net/questions/44947/holomorphy-of-inverse-map/44952#44952 Francesco Polizzi 2010-11-05T14:45:08Z 2010-11-05T16:01:05Z

<p>Yes, $f^{-1}$ is holomorphic. In fact, the following result holds, see [Griffiths-Harris, Principles of Algebraic Geometry p. 19].</p> <p><strong>Proposition</strong></p> <p>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. </p> <p>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 <strong>not</strong> a biholomorphism, since $f^{-1}$ is not holomorphic at the point $(0,0)$. </p>