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

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  • $\begingroup$ 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, 2019 at 5:16
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    $\begingroup$ We are assuming that $\mathcal{F}$ is a coherent sheaf on $X$ that is flat over $C$. $\endgroup$
    – user61586
    Jan 13, 2019 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, 2019 at 20:05

1 Answer 1

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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, 2019 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. $\endgroup$
    – user61586
    Jan 12, 2019 at 22:02
  • $\begingroup$ (btw I also posted an answer, but didn't put the downvote on this one) $\endgroup$
    – Qfwfq
    Jun 12, 2019 at 18:44
  • $\begingroup$ (Ok, I deleted my answer because I made the exact same mistake sneak in. Sorry) $\endgroup$
    – Qfwfq
    Jun 12, 2019 at 18:55

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