# First Ext group of a sheaf

Let $F$ be a quasicoherent sheaf on a scheme $X$, which is supposed to be sufficiently nice.

Does one then have a canonical isomorphism

$Ext^1(F,F) \simeq H^1(X, \underline{End}(F))$,

where with $\underline{End}(F)$ I denote the sheaf of endomorphisms of $F$.

I know that this holds for $F$ locally free, but I read an article where this iso was also used for quasicoherent $F$, but I don't see how to prove it.

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## 1 Answer

There is a spectral sequence $$E_2^{i,j} := H^i(X,\underline{Ext}^j(\mathcal E,\mathcal F)) \implies Ext^{i+j}(\mathcal E,\mathcal F),$$ for two sheaves $\mathcal E$ and $\mathcal F.$ (Here $\underline{Ext}$ denotes sheaf Ext.) Looking at the the low degree implications of this, one finds an exact sequence $$0 \to H^1(X, \underline{Hom}(\mathcal E,\mathcal F)) \to Ext^1(\mathcal E,\mathcal F) \to H^0(X,\underline{Ext}^1(\mathcal E,\mathcal F)) \to H^2(X,\underline{Hom}(\mathcal E,\mathcal F)) \to \cdots.$$ Taking $\mathcal E = \mathcal F$, the first map is the one you are asking about. If $\mathcal F$ is locally free than $\underline{Ext}^1$ vanishes, and one gets the isomorphism that you recalled. In general, this $\underline{Ext}^1$ need not vanish, the corresponding map in the exact sequence also need not vanish, and so there will be an injection $$H^1(X,\underline{End}(\mathcal F)) \hookrightarrow Ext^1(\mathcal F)$$ which is not surjective.

[Added: A typical example would come from taking $\mathcal F$ to be a skyscraper at a point, with $X$ positive dimensional.]

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Thank you, Matt, that's an answer to dream of. –  Veen Oct 16 '11 at 13:56
@Veen: Dear Veen, You're welcome. Regards, –  Matt E Oct 16 '11 at 19:22
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