Let $f:X\rightarrow C$ be a morphism, where $C$ is a smooth curve. For $t\in C$ let $i_t:X_t = f^{-1}(t)\rightarrow X$ be the inclusion of the fiber of $f$ over $t$, and let $\mathcal{F}$ a coherent sheaf on $X$ that is flat over $C$.

Does there exist an isomorphism $i_t^{*}\mathcal{E}xt^1(\mathcal{F},\mathcal{O}_X)\cong \mathcal{E}xt^1(i_t^{*}\mathcal{F},\mathcal{O}_{X_t})$ ?

  • $\begingroup$ Take $f =Id_C$ and $\mathcal{F} = \mathcal{O}_t$. The RHS vanishes whereas $\mathcal{E}xt^1(\mathcal{O}_t, \mathcal{O}_X) \simeq \mathcal{O}_t$. So there's no isomorphsm in general. $\endgroup$ – HYL Jan 13 '19 at 5:16
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    $\begingroup$ We are assuming that $\mathcal{F}$ is a coherent sheaf on $X$ that is flat over $C$. $\endgroup$ – gxg Jan 13 '19 at 11:27
  • $\begingroup$ The identity map is flat and the structure sheaf is flat, but if you want a more 'complicated' example how about $\mathcal{F} = \mathcal{O}_X$ and X = Proj(E) where E is a rank 2 vb over a smooth non-rational curve. $\endgroup$ – meh Jun 12 '19 at 20:05

Yes. Here's some extra words necessary for MO to allow this as an answer.

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  • $\begingroup$ I may have been a bit too hasty in my response, but I believe the statement in your setup should be true, see e.g. mathoverflow.net/questions/37889/… . Usually some flatness assumption of i_t would be necessary (which does not hold here), or as in loc. cit. flatness of the sheaf (which is one of your assumptions), or some regularity of the embedding which'd ensure finiteness of tor dimension (you haven't specified if this is true). $\endgroup$ – Frank Jan 12 '19 at 16:45
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    $\begingroup$ It is seems that in order to use Lemma 15.87.2 the flatness of the morphism is fundamental. In my case the inclusion $i_t:X_t\rightarrow X$ is not flat. $\endgroup$ – gxg Jan 12 '19 at 22:02
  • $\begingroup$ (btw I also posted an answer, but didn't put the downvote on this one) $\endgroup$ – Qfwfq Jun 12 '19 at 18:44
  • $\begingroup$ (Ok, I deleted my answer because I made the exact same mistake sneak in. Sorry) $\endgroup$ – Qfwfq Jun 12 '19 at 18:55

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