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Sep 29, 2011 at 1:17 comment added J.C. Ottem I corrected my answer in accordance with your comments. Thanks!
Sep 29, 2011 at 1:14 history edited J.C. Ottem CC BY-SA 3.0
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Sep 28, 2011 at 18:47 comment added Donu Arapura Sure, with smooth maps there is no problem. It is also OK for semistable maps, which is why I remembered it that way.
Sep 28, 2011 at 18:39 comment added J.C. Ottem Yes you are right, I guess I misquoted the theorem. But the OP wanted the morphism to be smooth, so the Du Bois condition with the fibers should be satisfied, right?
Sep 28, 2011 at 18:30 comment added Donu Arapura JC: I'm sorry if this sounds critical. In fact, I would be happier if it was true unconditionally since I use these sorts of results, but note the sentence: "Si pour tout point geometrique s de la fleche $O_{X_s}\to \Omega_{X_s}^0$ est isomorphisme" Nowadays, we would say the fibres are Du Bois. It is quite a strong condition.
Sep 28, 2011 at 18:24 comment added user2035 In Theorem 4.6, the condition is on the geometric fibers, not on $X$ itself.
Sep 28, 2011 at 18:03 comment added J.C. Ottem @Donu: I don't think so? This is Theorem 4.6 in the paper I linked to.
Sep 28, 2011 at 18:01 comment added Will Sawin Since every vector bundle is locally trivial, one should be able to apply this result locally.
Sep 28, 2011 at 17:54 comment added user2035 See also mathoverflow.net/questions/23891/…
Sep 28, 2011 at 17:45 comment added user2035 For $i=0$ only, a similar result is given in EGA III, 7.8.7.
Sep 28, 2011 at 17:41 comment added Donu Arapura I haven't looked at that paper in a while, but don't need to assume that $f$ is semistable? Should be OK in general when $\dim Y=1$, by semistable reduction.
Sep 28, 2011 at 16:22 history edited J.C. Ottem CC BY-SA 3.0
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Sep 28, 2011 at 15:59 history edited J.C. Ottem CC BY-SA 3.0
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Sep 28, 2011 at 15:51 history answered J.C. Ottem CC BY-SA 3.0