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Let $S$ be a scheme and the category of $S$-schemes be equipped with one of the standard Grothendieck topologies, say étale or fppf.

Let $G \rightarrow H$ be a morphism of abelian sheaves on this site and assume that $H$ is actually an $S$-scheme. But $G$ is only a sheaf.

Now define on the category of $H$-schemes the presheaf

$T \mapsto Hom_H(T,G)$.

This should be a sheaf on the category of $H$-schemes, but why is this?

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up vote 2 down vote accepted

Your claim is basically the internal version of the well-known fact that the presheaf $$U \mapsto \textbf{Top}_B (U, E)$$ where $U$ varies over the open sets of $B$ and $E \to B$ is any continuous map, is a sheaf on $B$ in the traditional sense of algebraic topology.

Now, let $\mathcal{S}$ be an (infinitary) pretopos (i.e. a category satisfying all the exactness requirements in Giraud's theorem but without a small generating family). Let $B$ be an object in $\mathcal{S}$ and let $\mathcal{B} = (\mathcal{S} \downarrow B)$. Let $Y$ be an object in $\mathcal{B}$, and consider the presheaf $$X \mapsto \mathcal{B}(X, Y)$$ on $\mathcal{B}$. I claim this is a sheaf in the canonical topology for $\mathcal{B}$. Indeed, suppose we have an epimorphism $U \to X$ in $\mathcal{B}$, and $f : U \to Y$ is a morphism in $\mathcal{B}$ with the property $f \circ p_1 = f \circ p_2$, where $p_1, p_2 : U \times_X U \to U$ are the projections of the fibre product over $X$. Now, fibre products in $\mathcal{B}$ agree with fibre products in $\mathcal{S}$, so it doesn't matter too much which category we work in. However, because $\mathcal{S}$ is a pretopos, every epimorphism in $\mathcal{S}$ is the coequaliser of its kernel pair, i.e. the canonical diagram $$U \times_X U \rightrightarrows U \to X$$ is a coequaliser in $\mathcal{S}$ whenever $U \to X$ is an epimorphism in $\mathcal{S}$. (In other words, every epimorphism in $\mathcal{S}$ is effective.) It is not hard to show that epimorphisms and coequalisers in $\mathcal{B}$ agree with epimorphisms and coequalisers in $\mathcal{S}$; hence, $f : U \to Y$ factors through $U \to X$ in $\mathcal{B}$ in a unique way. A more complicated version of this argument shows that $\mathcal{B}(-, Y)$ satisfies the sheaf condition with respect to general jointly epimorphic families. Thus $\mathcal{B}(-, Y)$ is indeed a sheaf on $\mathcal{B}$.

Now, to get the claim for sheaves of abelian groups, we must show that when all of the objects in question are internal abelian groups, then a homomorphism $f : U \to Y$ descends to a homomorphism $X \to Y$, but this is straightforward enough.

Addendum. I forgot to explain what this has to do with sheaves on a site. Let $\mathcal{C}$ be a subcanonical site, and let $\mathcal{S}$ be the category of sheaves on $\mathcal{C}$. By definition of subcanonical, $\mathcal{C}$ embeds as a full subcategory of $\mathcal{S}$, and it can be shown that the category of sheaves on the slice site $(\mathcal{C} \downarrow B)$ is equivalent to the slice category $(\mathcal{S} \downarrow B)$. (See [Johnstone, Sketches of an elephant, Part C, Lemma 2.2.17].)

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