Timeline for Are flat morphisms of analytic spaces open?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 4, 2014 at 15:59 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Oct 7, 2010 at 17:05 | comment | added | kaddar | 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... | |
| Oct 6, 2010 at 13:44 | comment | added | Francesco Polizzi | I did not know Douady's reference. Thank you for pointing it out! | |
| Oct 6, 2010 at 13:18 | comment | added | Laurent Moret-Bailly | About the non-archimedean analytic case (from Berkovich again, unpublished): Let $f:X\to Y$ be a morphism of non-Archimedean analytic spaces, and $F$ a coherent $\mathcal{O}_X$-module.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. | |
| Oct 6, 2010 at 13:11 | comment | added | Laurent Moret-Bailly | 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. | |
| Oct 5, 2010 at 15:43 | vote | accept | Laurent Moret-Bailly | ||
| Oct 5, 2010 at 15:43 | comment | added | Laurent Moret-Bailly | Thanks! In fact the proof gives something stronger, in the "equidimensional" style. | |
| Oct 5, 2010 at 14:57 | history | answered | Francesco Polizzi | CC BY-SA 2.5 |