TITLE: Is the category of abelian presheaves on a topos closed?
QUESTION [6 upvotes]: Take the category of presehaves of abelian groups on a topos $\mathcal{C}$. That is, an object of our category is a functor $F: \mathcal{C}^{\operatorname{op}} \to \operatorname{Ab}$.
We have a clear definition of a symmetric tensor product by $(F \otimes G)(S) = F(S) \otimes G(S)$.
However, defining an appropriate internal hom seems trickier. Taking inspiration from the internal sheaf hom, it seems like it would make sense to define
$${\mathscr{H}\kern -.5pt om} (F,G)(S) = \operatorname{Hom}(F_{| \mathcal{C}/S}, G_{|\mathcal{C}/S})$$
where $\mathcal{C}/S$ is the slice category. However, I can't seem to prove the adjunction (if it is even true). The main problem is that the slice category is more difficult to work with than in a general topos. Could there be some stronger requirements on the topos to make sure the adjunction is satisfied?

REPLY [3 votes]: The category of abelian presheaves is locally finitely presentable, and for each abelian presheaf $F$, the functor $F \otimes {-}$ preserves colimits, so we may apply the accessible adjoint functor theorem to obtain a right adjoint $\mathscr{H}om (F, -)$.
Representability of the functor $H \mapsto H (S)$ can be used to determine the values of $\mathscr{H}om (F, G)$:
$$\mathscr{H}om (F, G) (S) \cong \textrm{Hom} (F \otimes \mathbb{Z} h_S, G)$$
Here $h_S = \mathcal{C} (-, S)$ and $\mathbb{Z} h_S$ is the free abelian presheaf generated by $h_S$.
Actually, your formula is also correct.
The point is that you need to verify that
$$\textrm{Hom} (F \otimes \mathbb{Z} h_S, G) \cong \textrm{Hom} (F |_{\mathcal{C}_{/ S}},  G |_{\mathcal{C}_{/ S}})$$
which can be done essentially the same way as the Yoneda lemma.