# Ext & Complexes

I have heard that given two sheaves $A$ and $B$ on a variety, one can identify elements of $Ext^d(A,B)$ with complexes of sheaves $$0\to B \to C_1 \to \cdots \to C_d \to A \to 0.$$

My questions are,

How do I see that this is true?

and

If I have obtained an element of $Ext^n$ by some other method, can I explicitly construct the $C_j$ sheaves and the differentials?

I am sure this is well-known, so I'm marking it also as "reference-request".

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Are you familiar with how to do this in the setting without sheaves? For example $Ext$ of $R$-modules and how to get extensions from cocycles? – Sean Tilson Dec 5 '10 at 18:34
I've never gone through the general case, but I found that working out the special case of exact sequences $0 \to S \to E \to Q \to 0$ of vector bundles (i.e. of $Ext^1(S,Q)$) gives a pretty good idea of what's going on (it also makes you never want to check the details in the general case). – Gunnar Þór Magnússon Dec 5 '10 at 18:43
@Gunnar, to get a feeling for the general case, you need to do at least $\mathrm{Ext}^2$. – Mariano Suárez-Alvarez Dec 9 '10 at 1:51

## 2 Answers

For modules, Weibel discusses this in "Introduction to homological algebra." Section 3.4 deals with d=1 case and in Vista 3.4.6 is about the general case. He gives no proof for d>1 and refers to Bourbaki "Algebre homologique" 7.5 and Maclane "Homology" pp82-87.

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In Vista 3.4.6, Weibel says "... the set of equivalence classes ... (if this is indeed a set)", and then claims that $Ext^1(A,B)$ is an abelian group. How can $Ext^1$ not be a set, and still be an abelian group? – James Davidoff Dec 5 '10 at 19:13
I think (I could be wrong) that the point of that vista is that the Baer sum on representatives of equivalence classes of extensions is always well-defined in an abelian category, and always such that modulo the equivalence it is associative, has an identity, every element has an inverse and is commutative. If the collection of equivalence classes formed as set, we would have an abelian group. If the collection of equivalence classes did not form a set, but rather a class, then we would have an Abelian Group, i.e. a class with a group structure (another example: the Field of surreal numbers). – Vladimir Sotirov Dec 5 '10 at 22:13

Extremely late answer, but Eisenbud discusses this in his appendix on homological algebra in "Commutative algebra with a view towards algebraic geometry".

This is the Yoneda interpretation of Ext, and gives rise to a multiplicative structure on Ext-modules.

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