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Le $X_1$, $X_2$ and $Z$ be smooth quasi-projective connected varieties defined over $\mathbf{C}$. Let $p_1:X_1\rightarrow Z$ and $p_2:X_2\rightarrow Z$ be finite etale maps. Assume that $f:X_1\rightarrow X_2$ is an analytic isomorphism such that $p_2\circ f=p_1$.

Q1: Does it follow that $f$ is regular?

Note that if the answer to Q1 is positive then because of the symmetry of the problem $f$ is automatically biregular. A positive answer to Q1 would give "in some sense" a strengthening of Proposition 9 on p. 13 of GAGA. This proposition says that if one has a regular map $f:X_1\rightarrow X_2$ such that $f$ is an analytic isomorphism then $f$ is biregular.

So basically, I'm asking if it is possible to replace the regularity assumption on $f$ by the weaker data of two finite etale maps over a base $Z$ which are compatible with $f$ in order to be able to deduce that $f$ is biregular.

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up vote 1 down vote accepted

This (from SGA 1) is the proof that the functor is fully faithful: We may suppose that $X$ is connected. To give an $X$-morphism $Y\rightarrow Y^{\prime}$ is to give a section to $Y\times_{X}Y^{\prime}\rightarrow Y$, which is the same as to give a connected component $\Gamma$ of $Y\times_{X}Y^{\prime}$ such that the morphism $\Gamma\rightarrow X$ induced by the projection $Y\times_{X}Y^{\prime}\rightarrow Y$ is an isomorphism. But the connected components of $Y\times_{X}Y^{\prime}$ coincide with the connected components of $Y^{an}\times_{X^{an}}Y^{\prime an}$, and if $\Gamma$ is a connected component of $Y\times_{X}Y^{\prime}$, then the projection $\Gamma\rightarrow X$ is an isomorphism if and only if $\Gamma^{an}\rightarrow Y^{an}$ is an isomorphism.

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Typo: $\Gamma \to Y$ instead of $\Gamma \to X$? – name Apr 23 '12 at 7:26
Ok I got it. So by taking the fiber product you can bring it back to the case where you only have to deal with one arrow (modulo the simple but important observation that the connected components in the analytic and algebraic category agree). Many thanks! – Hugo Chapdelaine Apr 23 '12 at 12:35

RIEMANN EXISTENCE THEOREM: For any nonsingular algebraic variety $X$ over $\mathbb{C}$, the functor $Y\mapsto Y^{an}$ defines an equivalence between the categories of finite etale coverings of $X$ and $X^{an}$.

So the answer is yes.

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So taking for granted Grauer-Remmert, I would you prove the Riemann Existence theorem? – Hugo Chapdelaine Apr 21 '12 at 23:46
For proofs, see: Grauert and Remmert, Math. Ann. (1958), 245--318; SGA 4, XI.4.3; and SGA 1, XII. If you are willing to use resolution of singularities, the proof is not too hard (see SGA 1). – anon Apr 22 '12 at 0:00
So I never read Grauert und Remmert, but are they proving the existence of a finite analytic ramified covering over suitable compactifications of a single arrow $f:X\rightarrow Y$ or they also do it for a commutative triangle like in my question? – Hugo Chapdelaine Apr 22 '12 at 1:04
So using Grauert-Remmert and GAGA we have: for every finite analytic unramified covering $f:X_1^{an}\rightarrow X_2$ with $X_2$ algebraic, there exists a unique algebraic structure un $X_1$ such that $f$ is finite etale. But In my question I'm also requiring that this regular map $f$ is compatible with the algebraic structure of $Z$. – Hugo Chapdelaine Apr 22 '12 at 11:57
Look at the graph of $f$ (or the references I gave you). – anon Apr 22 '12 at 15:40

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