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.



Classifying functors of abelian categories
If we require an isomorphism (as opposed to an equivalence) the answer is no, for reasons having little to do with the (interesting aspect of) the question.


Assume there exists an abelian category $T$ and an object $x$ of $T$ such that for every abelian category $A$ and object $a$ there is a unique functor $F\_a:T\to A$ sending $x$ to $a$. Clearly, $x$ is nonzero, so the object $x^2$ has a nontrivial automorphism $\sigma$. Consider the functor $\Phi:T\to T$ defined as follows:


$\Phi$=identity on objects,


for a map $f:u\to v$, put:


$\Phi(f)=f$ if $u\neq x^2\neq v$ or $u=x^2=v$,


$\Phi(f)=\sigma f$ if $u\neq x^2$ and $v=x^2$,


$\Phi(f)=f\sigma^{-1}$ if $u=x^2$ and $v\neq x^2$.


Now $\Phi$ is indeed a functor (case-by-case inspection) sending $x$ to $x$. So it should be equal to the identity functor but isn't.


