Are flat morphisms of analytic spaces open? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:15:43Z http://mathoverflow.net/feeds/question/41158 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41158/are-flat-morphisms-of-analytic-spaces-open Are flat morphisms of analytic spaces open? Laurent Moret-Bailly 2010-10-05T14:25:38Z 2010-10-05T14:57:19Z <p>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?</p> <p>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.</p> <p>In the algebraic case it's also true (that's what led me to the question, see <a href="http://arxiv.org/abs/1010.0341" rel="nofollow">http://arxiv.org/abs/1010.0341</a>). 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).</p> <p>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.</p> http://mathoverflow.net/questions/41158/are-flat-morphisms-of-analytic-spaces-open/41163#41163 Answer by Francesco Polizzi for Are flat morphisms of analytic spaces open? Francesco Polizzi 2010-10-05T14:57:19Z 2010-10-05T14:57:19Z <p>The answer is <strong>yes</strong>.</p> <p>In fact, there is the following result, see Banica- Stanasila, "Algebraic methods in the global theory of Complex Spaces", Theorem 2.12 p. 180.</p> <p><strong>Theorem</strong> Let $f \colon X \to Y$ be a morphism of complex spaces and let $\mathcal{F}$ be a coherent analytic sheaf on $X$, which is flat with respect to $f$. Then the restriction of $f$ to supp($\mathcal{F}$) is an open map.</p> <p>In particular, every flat morphism is open.</p>