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Suppose that $\mathcal{E}$ is a (Grothendieck) topos. It carries a (large) Grothendieck topology (in fact the canonical topology), a basis of which is determined by declaring a collection of morphisms $\left(f_\alpha:E_\alpha \to E\right)$ to be a covering family if the induced map $\coprod E_\alpha \to E$ is an epimorphism. On one hand, it is well known result (and I know some "by hand" proofs) that a functor $$F:\mathcal{E}^{op} \to Set$$ is a sheaf for this topology if and only if $F$ is representable. (I.e. $\mathcal{E}$ is sheaves over itself.) On the other hand, a functor $$F:\mathcal{E}^{op} \to Set$$ is representable if and only if $$F^{op}:\mathcal{E} \to Set^{op}$$ has a right adjoint, and since $\mathcal{E}$ is locally presentable, by the adjoint functor theorem this is if and only if $$F:\mathcal{E}^{op} \to Set$$ preserves limits. It is easy to show that every representable is a sheaf for this topology (since every epi is an effective epi), hence any limit preserving functor $$F:\mathcal{E}^{op} \to Set$$ is a sheaf. Is there a slick way of seeing the converse?

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Well the first two facts of single line arguments, so not so "unslick". What about coequalizers? – David Carchedi Mar 5 '13 at 23:57
*of=are ( ) – David Carchedi Mar 5 '13 at 23:58
Yes, I see the problem now. Hrm. Nevermind! – Zhen Lin Mar 6 '13 at 0:06
@Zhen: I wish you didn't delete your comment, since now the comments make no sense. Anyway, the two facts above were that $F^{op}$ preserves the initial object, and arbitrary coproducts. If we knew if preserve coequalizers, we'd be done. Notice however, that it does preserve coequalizers by equivalence relations, since these are covers. – David Carchedi Mar 6 '13 at 20:34

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