# Why is $Lex(\mathcal{A},\mathcal{Ab})$ abelian? Does $Lex(\mathcal{A},\mathcal{Ab})\rightarrow Func(\mathcal{A},\mathcal{Ab})$ admit a left-adjoint?

What is the best way to show, that $Lex(\mathcal{A},\mathcal{Ab})$ is abelian, where $\mathcal{A}$ is an abelian category and $\mathcal{Ab}$ is the category of abelian groups from scratch?

There is the canonical inclusion functor $S\colon Lex(\mathcal{A},\mathcal{Ab})\rightarrow Func(\mathcal{A},\mathcal{Ab})$, which should have a nice left-adjoint $T$ (somehow like sheafication). How does one construct this? What is the kernel of the left-adjoint?

I think this left adjoint should be exact and $T\circ S\cong id_{Lex(\mathcal{A},\mathcal{Ab})}$, such that $Lex(\mathcal{A},\mathcal{Ab})$ is equivalent to the gabriel quotient $Func(\mathcal{A},\mathcal{Ab})/ker(T)$.

I would be glad if I get some answers to this questions, but I would love to hear some references from you, with wich I can prove this rigorously. Thanks.

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It would be helpful that you explained the notation you're using. – Fernando Muro May 11 '13 at 17:45
Isn't this notation quite common? $Lex(A,B)$ denotes the category of all left−exact functors from $\mathcal{A}$ to $\mathcal{B}$ and $Func(\mathcal{A},\mathcal{B})$ the category of all functors. – archipelago May 11 '13 at 17:55
Laumon (MR0726427) interpretes Lex as a category of sheaves on a site - whence it is abelian. The proof in Gabriel's dissertation I couldn't entirely follow. – Matthias Künzer Oct 24 '13 at 9:34
Yes, the class of epimorphisms in an abelian categroy forms a site and the sheaves of abelian groups on site site are exactly the left exact functors. Gabriel's elementary constructed left adjoint I mentioned in my answer then corresponds to the abstractly constructed sheafification-functor. – archipelago Dec 28 '13 at 22:33

Almost all of your questions are answered in Pierre Gabriel's dissertation "Des catégories abéliennes".

He shows in a more general case, that the left exact functors between nice abelian categories are abelian und constructs indeed an exact "sheafification"-functor $T$, which is the left-adjoint of the inclusion functor. Have a look at Proposition 4 on page 348.

After that, Proposition 5 on page 374 implies, that $Func(\mathcal{A},\mathcal{Ab})/ker(T)\cong Lex(\mathcal{A},\mathcal{Ab})$, what you already mentioned.

I think $ker(T)$ are the so called "weakly effaceable" functors, but I am neither completely sure nor do I know any reference for that. Maybe someone can help you with this point.

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I just wanted to take a second to post this paper: http://arxiv.org/abs/1211.0054

In it, the author defines zeroth derived functors in abelian categories that do not have enough projectives or injectives.

A brief piece of it: Let $C$ and $D$ be abelian categories. Take the functor category $(C,D)$ and consider any subcategory $S\subset (C, D)$ and then the full subcategory of $S$ consisting of left exact functors in $S$, $Lex(S)$. There is the canonical inclusion $s: Lex(S) \rightarrow S$. We say $S$ admits a zeroth right derived functor if there is a left adjoint (say $r^0: S \rightarrow Lex(S)$) satisfying two properties (we'll call the map $s\circ r^0$ the zeroth right derived functor and relabel it $R^0$):

1) The unit of adjunction $u: 1_s \rightarrow R^0$ is an isomorphism on injectives, and

2)The 'other'composition, $r^0\circ s$ is isomorphic to the the identity functor of the full subcategory of left exact functors of $S$. (The paper says something different, namely that $r^0\circ s \cong 1_S$. Based on the direction of the arrows, that composition should land in the $Lex(S)$.

In general, the unit of adjunction has a kernel and by evaluating the exact sequence

$0 \rightarrow \ker u \rightarrow 1_S \rightarrow R^0$

at $F$, a functor from $C$ to $D$, we recover a definition for the injective stabilization of a functor $F$, as defined by Auslander and Bridger in Stable Module Theory.

The author goes on to say that $r^0$ is a functor which takes functors $F$ and produces left exact functors $r^0F$. Then the inclusion $s(r^0F)=R^0F$ is the zeroth right derived functor of $F$, the result of making F left exact while changing as little as possible about the functor.

Now, all this being said, I did not write the paper, and I do not know much more about the details. I just think the paper is an interesting one.

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