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Let $f:X\to Y$ be a morphism of complex analytic spaces. Assume $f$ is flat (or, more generally, that there is a coherent sheaf on $X$ with support $X$ which is $f$-flat). Is $f$ an open map?

The rigid-analytic analogue is true (via Raynaud's formal models): see Corollary 7.2 in S. Bosch, Pure Appl. Math. Q., 5(4) :1435–1467, 2009. I don't know about the Berkovich side.

In the algebraic case it's also true (that's what led me to the question, see http://arxiv.org/abs/1010.0341). Specifically, if $K$ is an algebraically closed field with an absolute value, and $f:X\to Y$ is a universally open morphism of $K$-schemes of finite type, the the induced map on $K$-points is open (for the strong topology).

Note that in the complex analytic case, I don't know any reasonable substitute for "universally open". If I believe in the analogy, the result ($f$ is open) should be true assuming for instance that $Y$ is locally irreducible and $f$ is "equidimensional" in some sense (e.g. surjective, $X$ irreducible and the fiber dimension is constant). In this setting the case of fiber dimension 0 is known.

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The answer is yes.

In fact, there is the following result, see Banica-Stanasila, Algebraic methods in the global theory of Complex Spaces, Theorem 2.12 p. 180.

Theorem. Let $f \colon X \to Y$ be a morphism of complex spaces and let $\mathscr{F}$ be a coherent analytic sheaf on $X$, which is flat with respect to $f$. Then the restriction of $f$ to supp($\mathscr{F}$) is an open map.

In particular, every flat morphism is open.

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  • $\begingroup$ Thanks! In fact the proof gives something stronger, in the "equidimensional" style. $\endgroup$
    – Laurent Moret-Bailly
    Oct 5, 2010 at 15:43
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    $\begingroup$ As V. Berkovich points out to me, the result seems due to Douady: see the final corollary in "Flatness and privilege", Ens. Math. 2, (14) fasc. 1 (1968), 47--74. But the proof in Banica-Stanasila appears more elementary. $\endgroup$
    – Laurent Moret-Bailly
    Oct 6, 2010 at 13:11
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    $\begingroup$ About the non-archimedean analytic case (from Berkovich again, unpublished): Let $f:X\to Y$ be a morphism of non-Archimedean analytic spaces, and $F$ a coherent $\mathcal{O}_X$-module. Suppose that $F$ is $f$-flat and $f$ has no boundary. Then the restriction of $f$ to the support of $F$ is an open map. $\endgroup$
    – Laurent Moret-Bailly
    Oct 6, 2010 at 13:18
  • $\begingroup$ I did not know Douady's reference. Thank you for pointing it out! $\endgroup$
    – Francesco Polizzi
    Oct 6, 2010 at 13:44
  • $\begingroup$ It seems to me that the notion of " universally open" agree with the notion of "open" in the complex analytic case but it is no true for universally equidimensional and equidimensional unless the base space is locally irreducible... $\endgroup$
    – kaddar
    Oct 7, 2010 at 17:05

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