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Let $i:C\subset S$ where $S$ a smooth projective complex surface and $C$ a smooth projective complex curve, then the functor $i_*:\textbf{Coh}(C)\rightarrow\textbf{Coh}(S)$ is fully faithful, i.e. $$\text{Hom}_C(\mathcal{E},\mathcal{F})\cong\text{Hom}_S(i_*\mathcal{E},i_*\mathcal{F})$$ Can we show that $\text{Ext}^1_C(\mathcal{E},\mathcal{F})\cong\text{Ext}^1_S(i_*\mathcal{E},i_*\mathcal{F})$ in general or with some additional (but not vary special) conditions?

This post Significance of adjoint relationship with Ext instead of Hom may be related.

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EDIT: The argument below is wrong, $\text{Ext}^i(\mathcal{F},-)$ is not computed using flasque resolutions. As a counterexample, a skyscraper sheaf is always flasque, but $\text{Ext}^1(k(x),k(x))$ typically does not vanish.

This is true with no further assumptions.

In this case $i_*$ is an exact functor. You compute $\text{Ext}^1_C(\mathcal{E},\mathcal{F})$ by taking a flasque resolution of $\mathcal{F}$, applying $\text{Hom}(\mathcal{E},-)$ and taking the first cohomology. Now if you apply $i_*$ to this resolution you get a flasque resolution (check that $i_*$ preserves flasque sheaves!). So $\text{Ext}^1_S(i_*\mathcal{E},i_*\mathcal{F})$ may be computed by applying the functor $\text{Hom}(i_*\mathcal{E},i_*(-))$ to your original flasque resolution.

If you know about effaceable functors, you can just argue that $\text{Ext}^k_S(i_*\mathcal{E},i_*(-))$ vanishes on flasque sheaves.

SVG
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  • Thank you, so basically we use the isomorphism $\text{Hom}(\mathcal{E},\mathcal{F}^{\bullet})\cong \text{Hom}(i_\mathcal{E},i_\mathcal{F}^{\bullet})$ and apply the cohomology. –  Dec 22 '21 at 21:50
  • @Liu, right, but it's critical that $i_*$ takes flasque resolutions to flasque resolutions. – SVG Dec 22 '21 at 21:59
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    The argument is definitely incorrect, because there is a simple counterexample --- take $E = F$ to be the structure sheaf of a point, then $Ext^1$'s are the tangent spaces at that point to $C$ and $S$ respectively, hence they are not isomorphic. – Sasha Dec 23 '21 at 04:30
  • @Sasha. Thanks! I did indeed goof up. Of course, flasque sheaves are simply very much not acyclic for Ext. – SVG Dec 23 '21 at 07:38
  • @Liu. I am sorry, but I made a very dumb mistake. Please unaccept the answer. I am going to leave it up as a warning for future users :) – SVG Dec 23 '21 at 07:44
  • @SergeyGuminov I see, so the problem is that we still need use injective resolution to compute Ext but not flasque resolutions. –  Dec 23 '21 at 09:44
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To compare $\mathrm{Ext}^1(E,F)$ to $\mathrm{Ext}^1(i_*E,i_*F)$ one can proceed as follows. First, there is adjunction isomorphism $$ \mathrm{Ext}^\bullet(i_*E,i_*F) \cong \mathrm{Ext}^\bullet(Li^*i_*E,F). $$ Next, since $i$ is an embedding of a Cartier divisor, there is a distinguished triangle $$ Li^*i_*E \to E \to E(-C)[2], $$ where $E(-C) = E \otimes \mathcal{O}_C(-C)$. Applying the functor $\mathrm{Ext}^\bullet(-,F)$ to this triangle, we obtain a long exact sequence $$ \dots \to \mathrm{Ext}^{i-2}(E(-C),F) \to \mathrm{Ext}^{i}(E,F) \to \mathrm{Ext}^{i}(Li^*i_*E,F) \to \mathrm{Ext}^{i-1}(E(-C),F) \to \dots $$ Since $\mathrm{Ext}^i(-,-)$ between two sheaves vanishes for $i \not\in \{0,1,2\}$, using the above adjunction isomorphisms, we obtain an isomorphism $$ \mathrm{Hom}(E,F) \to \mathrm{Hom}(i_*E,i_*F), $$ a short exact sequence $$ 0 \to \mathrm{Ext}^{1}(E,F) \to \mathrm{Ext}^{1}(i_*E,i_*F) \to \mathrm{Hom}(E(-C),F) \to 0 $$ and yet another isomorphism $$ \mathrm{Ext}^{2}(i_*E,i_*F) \to \mathrm{Ext}^1(E(-C),F). $$ In particular, the morphism $\mathrm{Ext}^{1}(E,F) \to \mathrm{Ext}^{1}(i_*E,i_*F)$ is an isomorphism if and only if $\mathrm{Hom}(E(-C),F) = 0$.

Sasha
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  • Thank you for your answer Sasha, I wonder if '$E(-C)[2]$' in the exact triangle is a typo (should it be $E(C)[2]$?). Also, I do not know how to get that exact triangle. –  Dec 23 '21 at 10:20
  • I found the exact triangle in Lemma 3.3 of paper Semiorthogonal decompositions for algebraic varieties. –  Dec 23 '21 at 15:47
  • @Liu: You are right, sorry for the typo ($E(-C)$ was correct, but $E(C)$ in all other cases was incorrect). I think you can also find this triangle in Huybrechts "Fourier-Mukai transforms" and in many other places. – Sasha Dec 23 '21 at 18:06