Let $X$ be a topological space. Let $Op(X)$ be the category of open sets. A sheaf of abelian groups is a (contravariant) functor $F:\mathcal{C} \to Ab$ satisfying the sheaf condition: the following sequence is exact $$ 0 \to F(U)\to \prod_i F(U_i) \to \prod_{i,j} F(U_{ij})$$ in the category of abelian groups.

Now, one can replace "category of abelian groups" by any abelian category $\mathcal{A}$ (or category where exact sequences make sense) to get the definition of a sheaf with values in $\mathcal{A}$.

But the category of rings is not abelian and exact sequences don't really make sense (since ideals aren't rings in my opinion).

So does one define a sheaf of rings to be a presheaf of rings which is a sheaf of abelian groups? (So you only ask the sequence to be exact in the category of abelian groups.)

  • $\begingroup$ For rings, and certain other "algebraic structures," you can replace the sequence for abelian groups with a diagram where there are two arrows coming out of $\prod_iF(U_i)$, and you ask that the arrow $F(U)\rightarrow\prod_iF(U_i)$ be an equalizer of the two arrows. It turns out that, for the category of rings and similar categories, this is equivalent to asking that the underlying presheaf of sets be a sheaf of sets (using the same definition). $\endgroup$ – Keenan Kidwell Apr 9 '12 at 14:04

There is no condition that the category is abelian for a sheaf to be defined. You may as well define a sheaf of sets. The exact sequence in the definition of a sheaf is merely a bookkeeping device to encode the fact that the presheaf must satisfy the gluing condition in order to be a sheaf, and it is not an exact sequence in the ordinary sence.


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