Which abelian categories possess an exact faithful functor into abelian groups that respects products? Let $A$ be an abelian category closed with respect to small coproducts (that is, and AB3 category). Which assumptions are sufficient to ensure the existence of an exact faithful functor from $A^{op}$ into abelian groups that respects products (i.e., it should send $A$-coproducts into the corresponding products)?
I suspect that in certain cases one can take a functor represented by an injective cogenerator in the category $\operatorname{Ind}-A$, but I don't understand when this works. Do exact $\alpha$-filtered colimits in $A$ (where $\alpha$ is a regular cardinal) help (say, if $A$ contains a generator)?
Upd. Sorry; I mistunderstood Qiaochu Yuan's answer below. His answer only treats the existence of functors that satisfy certain non-trivial additional conditions, whereas any functor from $A^{op}$ into abelian groups that respects products will be sufficient for my purposes. So, do there exist any re-formulations of this existence, or sufficient assumptions that imply it?
 A: I find it less confusing to work directly with $A^{op}$ so let me do that; I'll rename it $C$. We have a complete abelian category $C$ (completeness is equivalent to being closed under small products) and we want to know when it admits an exact faithful functor $G : C \to \text{Ab}$ which respects products (equivalently, in the presence of exactness, continuous). If $G$ satisfies the solution set condition then by the (general) adjoint functor theorem $G$ has a left adjoint $F : \text{Ab} \to C$. This left adjoint is determined (by cocontinuity) by $F(\mathbb{Z})$, which I'll name $P$. The adjunction gives that $P$ represents $G$, and exactness and faithfulness gives that $P$ is a projective generator. Conversely, given a projective generator $P$, $\text{Hom}(P, -)$ is an exact faithful functor which respects products. Going back to $A$, you want an injective cogenerator (in $A$, not in $\text{Ind}(A)$).
So if you want more general examples than this then $G$ can't satisfy the solution set condition.
