Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
The answer is yes if you replace "abelian" by "additive": take for $T$ the category $F$ of free finitely generated $\mathbb{Z}$-modules, which can be viewed as a "free additive category on one generator". So the question is: does $F$ have an "abelian envelope"? The obvious candidate would be the category $G$ of finitely generated $\mathbb{Z}$-modules (forgetting "free") but that doesn't work: the "bidual" functor $G\to G$ sends $\mathbb{Z}$ to $\mathbb{Z}$ but is not the identity. In view of this, my guess is no to the original question. –  Laurent Moret-Bailly Sep 15 '10 at 9:11
I agree with Laurent Moret-Baily's reasoning that it seems very unlikely that there is such a category when you consider all additive 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
@Chris: In your modification, $Hom(T,A)$ is only equivalent to $A$, not isomorphic. In my application, I need isomorphism. –  Martin Brandenburg Sep 15 '10 at 12:20
Martin, do you mean it has to be bijective on objects? That's a wild requirement. –  Laurent Moret-Bailly Sep 15 '10 at 12:57
I guess if "free abelian category on an additive category" was a well-known construction, someone would have applied it to the category of Chow-motives (but maybe someone did and I don't know it)... –  Peter Arndt Sep 15 '10 at 13:25
show 1 more comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
I don't understand. As far as I can see by now, this functor (if well defined) is not additive (i.e. does not preserve finite sums), so that there is no contradiction: in fact, your construction uses only the existence of finite sums in abelian categories, so that your argument should also hold for additive categories; but, as you said yourself, the free additive category on one object exists... –  Denis-Charles Cisinski Sep 16 '10 at 15:04
This $\Phi$ is isomorphic to identity: for an object $y$ define $s(y):y\to\Phi(y)$ by: $s(y)=id_y$ if $y\neq x\times x$, $s(x\times x)=\sigma$. By inspection, $s$ is an isomorphism from $id_T$ to $\Phi$. The universal property of the cat. $F$ of free f. g. ab. groups holds "up to equivalence": $\mathrm{Hom}(F,A)$ is only equivalent to $A$. The "isomorphism" requirement is wild because it violates the implicit rule for using categories: only consider properties and constructions which are invariant under equivalence. My construction is not (hence is hard to understand!). –  Laurent Moret-Bailly Sep 16 '10 at 16:49
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.