TITLE: Is there an "Abelian envelope" 2-functor?
QUESTION [11 upvotes]: 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.

REPLY [13 votes]: 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.