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Is anyone aware of a good resource which deals with how the cup/cap products of sheaf cohomology classes are a generalization of those in singular cohomology? I would say that I already understand the basics of both constructions.

Specifically, I am asking how the cup product from singular cohomology is actually a special case of cup-product-as-Yoneda-product from sheaf cohomology.

Eric Auld
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    Are you asking why they agree for constant sheaves? –  Jul 07 '16 at 15:58
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    Sheaf cohomology for constant sheaves is not that much different from the simplicial cohomology (the only essential difference is taking a limit) and the cup-product works the same way. I suggest you first work out this case by yourself. Then read Chapter 2.7 of Bredon's "Sheaf Theory" for the case of general sheaves. – Moishe Kohan Jul 07 '16 at 16:26
  • @MikeMiller Yes. – Eric Auld Jul 07 '16 at 23:30
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    @studiosus Do you mean "singular cohomology" when you say "simplicial cohomology"? The cup product I'm familiar with on sheaves is in terms of $\operatorname{Ext}^j(\underline{k}, \underline{k})$, and I was trying to bring that down to earth a bit. I know how to show that singular equals sheaf cohomology in nice cases. – Eric Auld Jul 07 '16 at 23:34
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    I mean simplicial. – Moishe Kohan Jul 07 '16 at 23:54

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