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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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    @Gunnar, to get a feeling for the general case, you need to do at least $\mathrm{Ext}^2$.2010-12-09

2 Answers 2

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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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    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 $A$belian Group, i.e. a class with a group structure (another example: the Field o$f$ surreal numbers).2010-12-05
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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.