Let $AbCat$ denote the $2$-category of abelian categories with additive functors. Is the forgetful functor $AbCat \to Cat$ representable; i.e. is there an abelian category $T$ such that for every abelian category $A$, the category $Hom(T,A)$ is naturally isomorphic to (the category underlying) $A$?

This would be nice, because then it is possible to reconstruct an abelian category which has only a universal property.

I try to work out the structure of $T$: The isomorphism $Hom(T,T) \cong T$ maps $1_T$ to some object $x \in T$. If $A$ is arbitrary, then the isomorphism $Hom(T,A) \cong A$ is given by mapping a functor $F : T \to A$ to $F(x)$ and mapping a natural transformation $\eta : F \Rightarrow G$ to the morphism $\eta(x) : F(x) \to G(x)$.

Let $T'$ be the smallest full abelian subcategory of $T$, which contains $x$ (Existence: Construct inductively subcategories $T_{n+3k}$, which contain direct sums $(n=0)$, kernels $(n=1)$ and cokernels $(n=2)$ from $T_{n+3k-1}$.). Then $T'$ has the same universal property as $T$. Thus $T'=T$, and we see that $T$ is generated by $x$. Now we have to find such a $T$, which has no additional relations. In particular, we have to ensure that a natural transformation between additive functors on $T$ is determined by the morphism at $x$, although the functors don't have to be exact.

alladditive functors. I'm sure this example can be massaged into a rigorous no-go proof. On the other hand, if you put some conditions on which functors you consider then I think the answer may be yes. For example if you only consider right exact functors, then I think finitely presented abelian groups does the trick. Such a functor is entirely determined by where $\mathbb{Z}$ goes. – Chris Schommer-Pries Sep 15 '10 at 11:29