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I'm looking for a notion of an Abelian category $\mathcal{A}$ "generated" by a given category $\mathcal{C}$

More precisely I need something along the following lines. Denote $\mathcal{Ab}_2$ the 2-category of Abelian categories and $\mathcal{Cat}$ the 2-category of categories. We have the forgetful 2-functor $\mathcal{F}: \mathcal{Ab}_2 \rightarrow \mathcal{Cat}$. Is there an adjoint 2-functor $\mathcal{G}: \mathcal{Cat} \rightarrow \mathcal{Ab}_2$ ?

I suspect the answer is "yes" because it can be constructed along the following lines. Denote $\mathcal{Ab}$ the category of Abelian groups. For any category $\mathcal{C}$, the category $\mathcal{Hom(C, Ab)}$ is Abelian. Moreover, we have the natural functor $\mathcal{i:C \rightarrow Hom(Hom(C,Ab),Ab)}$. Thus $\mathcal{C}$ is embedded in the Abelian category $\mathcal{D:=Hom(Hom(C,Ab),Ab)}$ and we can take the Abelian category generated by $\mathcal{C}$ within $\mathcal{D}$. The result is supposed to be $\mathcal{G(C)}$

However, the only construction I managed to search up is the Karoubi envelope which generates a pseudo-Abelian category. So either my purported construction is wrong or simply not popular. Which is it?

EDIT: I realized my construction amounts to $\mathcal{G(C):=Hom(C,Ab^{op})}$. At least for small $\mathcal{C}$ this is indeed adjoint to $\mathcal{F}$, provided we interpret $\mathcal{Ab_2}$ as having right exact functors for 1-morphisms. Here $\mathcal{C}$ embeds by applying opposite Yoneda and taking the freely generated Abelian group.

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That depends on what you mean by the 2-category of abelian categories. –  Fernando Muro Feb 11 '12 at 21:51
    
I very recently had a similar idea. The main subtlety that I found was with the concept of "generated". Suppose you generate an object by two different methods, but do not generate an isomorphism between them. In the subcategory, will the two objects be isomorphic? If no, are you sure that the category is well-defined? You might have to be careful. If yes, then which isomorphism? Secondarily, it seems that all functors in $Hom(C,Ab)$, allowed to act on the generated category, are exact. But shouldn't there be non-exact functors on the generated category sometimes? –  Will Sawin Feb 11 '12 at 22:21
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$Hom(Ab,Ab)$ is already quite a complex category. Which subcategory is generated by the identity? –  Will Sawin Feb 11 '12 at 22:24
    
I may be barking completely up the wrong tree, but I vaguely recall the construction of some kind of Abelian envelope for an additive category (Freyd?) - is something like that along the lines you seek? –  Yemon Choi Feb 12 '12 at 1:06
    
@Fernando, you are right I need to specify what counts as morphisms. I think 1-morphisms are exact functors and 2-morphisms are arbitrary natural transformations. When you say "it depends" you mean in some case it exists while in another it doesn't? Can you detail? –  Squark Feb 12 '12 at 8:34

1 Answer 1

up vote 11 down vote accepted

I'd be inclined to go in a slightly different direction in constructing the free abelian category generated by a category. I would do it in stages:

  • First construct the free $Ab$-enriched category $F_1(C)$ generated by a category $C$. This would have the same objects as $C$, but the hom $F_1(C)(a, b)$ between two objects would be the free abelian group generated by the hom $C(a, b)$ in $C$.

  • Next, to any $Ab$-enriched category $C$, we may freely adjoin finite limits. We can do this by embedding $C$ into $(Ab^C)^{op}$ by the dual of the Yoneda embedding, and then cutting down to the full subcategory $F_2(C)$ of $Ab$-enriched functors which arise as limits of finite diagrams of opposites of representables $C(c, -)^{op}$ in $(Ab^C)^{op}$.

  • Next, to any finitely complete $Ab$-enriched category $C$, pass to the so-called ex/lex completion $F_3(C)$. The "lex" here refers to left exactness (i.e., closed under finite limits, and functors preserving such), whereas the "ex" refers to Barr exactness and functors preserving such.

The point is that an abelian category is precisely a Barr-exact (finitely complete) $Ab$-enriched category (see Freyd-Scedrov's Categories, Allegories, page 90, where a Barr-exact category is what they call an effective regular category). So the free abelian category generated by a category $C$ is $(F_3 \circ F_2 \circ F_1)(C)$. I did not check carefully that the ex/lex completion, as described in the nLab article, lifts from the $Set$-enriched world to the $Ab$-enriched world, but I think it's okay.

Not yet sure how this relates to your proposed construction (there are some size considerations to, er, consider), but I think the answer to your first question is 'yes'.

Addendum: In response to Martin's comment, and as confirmation that the description above is correct, here is a paper by Carboni and Magno which in the course of things constructs the free abelian category generated by an additive category: see page 300. The free abelian category on the terminal category $\mathbf{1}$ is $C_{ex}$, where $C$ is the category opposite to that of finitely generated abelian groups. There may be simpler descriptions, but I'm not prepared to say more on that now.

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Thx Todd, looks very nice! I have to chew on it a bit more. In the meanwhile, can you explain what you mean by size considerations? –  Squark Feb 12 '12 at 8:30
    
I meant that your $\hom(\hom(C, Ab), Ab))$ is raising a large category $Ab$ to the power of another large category $Ab^C$. This operation is illegitimate within the confines of ZFC or GBN set theory. At this point many people consider strengthening their set theory to deal with this annoyance, allowing some hierarchies of inaccessible cardinals (cf. universes), but subtleties can still arise, and I hadn't thought those through yet. –  Todd Trimble Feb 12 '12 at 12:53
    
@Todd: What about a specific example, say the abelian category which is free on one object? My first guess was the category of finitely generated abelian groups. But this seems to be the free finitely cocomplete category on one object and not every right exact functor on it is exact. So this free abelian category will be more complicated! –  Martin Brandenburg Feb 12 '12 at 14:56
    
@Martin: yes, you're right, it is more complicated. Please see the addendum above. –  Todd Trimble Feb 12 '12 at 16:00
    
@Will: There is no embedding $C \to \mathrm{Ab}^C$. @Todd: The link is broken. What are the objects of the free abelian category on one object? –  Martin Brandenburg Feb 12 '12 at 16:04

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