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A meromorphic map of complex spaces (in the sense of Remmert) $f: X \to Y$ is a multivalued map such that its graph $\Gamma$ is an analytic subset of $X \times Y$ and off some analytic subset $Z \subset \Gamma$, the projection on the first coordinate is a biholomorphic map. If additionally, off some analytic subset, the projection on the second coordinate is biholomorphic the map is called bimeromorphic.

Let $X,Y$ be two compact complex spaces, $A$ and $B$ their analytic subsets, and let $f: X\setminus A \to Y\setminus B$ be a biholomorphic map. Is it always possible to extend $f$ to a bimeromorphic map between $X$ and $Y$?

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up vote 1 down vote accepted

The answer is no. Please check this question I asked some time ago.

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Thanks! It's funny that an analogous question for algebraic varieties (with "compact" replaced by "proper") has a positive answer: take the Zariski closure of the graph of the isomorphism between the opens. I can't really point to one particular property of complex spaces that makes this statement false! My guess that the core reason is that Zariski topology is Noetherian, which makes it easy to extend ideal sheaves hence to take a Zariski closure. It doesn't work so smoothly in the complex topology, hence the counterexamples. – Dima Sustretov Mar 9 '12 at 16:17

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