MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## projective generator in the category of left-exact functors

I expect this to be a small understanding problem and not a real interesting question.

In Gabriel's thesis you find a proof of the theorem that every small abelian category $C$ admits a faithful exact functor to the category of abelian groups. The proof goes like this: Consider the abelian category $Sex(C,Ab)$, which is the category of left-exact functors $C \to Ab$. By the way, does anybody know why Gabriel has chosen this notation? Why not "Lex" for left-exact or "Gex" for the french "exacte a gauche"? Anyway, it can be shown that $Sex(C,Ab)$ is a (nowadays called) Grothendieck category and has thus enough injective objects. Besides, every injective object is an exact functor $C \to Ab$. If we embed the generator $U=\sum_{X \in C} Hom(X,-)$ into an injective object, we are done.

Now Gabriels claims that $U$ is actually a projective generator. I doubt that this is used in the proof, nevertheless it is interesting. By Yoneda the functor $F \mapsto Hom(U,F)$ is isomorphic to $F \mapsto \prod_{X \in C} F(X)$, thus we have to prove that every epimorphism in $Sex(C,Ab)$ is actually pointwise an epimorphism of abelian groups. This is not clear to me since a cokernel in $Sex(C,Ab)$ is defined by the universal left-exact functor associated to the pointwise-defined cokernel (cf. Prop. 5 in II.2). The vanishing does not seem to imply the vanishing of the pointwise-defined cokernel (cf. Lemme 3 b).

Note that Grothendieck has supervised this thesis. Although there are many many typos, I don't think that such a statement will just be wrong. What am I missing?

-
I don't know about Gabriel, but Sex(C,AB) sounds a lot better. – Hailong Dao Oct 10 2010 at 16:57
The "S" comes from the Latin "sinister", which means "left" indeed. Bourbaki uses systematically the letter "s" to mean "left". For example $A_s$ mean "$A$ viewed as a left $A$-module". – Pierre-Yves Gaillard Jan 14 2012 at 19:51

The object $U$ is a projective generator in the category of additive functors $C\to Ab$. It is also a generator of the category of left exact functors $C\to Ab$. It is not projective in the latter category, though.
Generally, left exact functors are similar to sheaves (on the category opprosite to $C$; epimorphisms in the latter category, i.e., monomorphisms in $C$, play the role of the coverings). There are no projective sheaves, generally speaking. The object $U$ is like the direct sum of the constant (pre)sheaves on all open subsets, extended by zero to the outside. It is a projective presheaf. It is not supposed to be a projective sheaf.
Update: I have been asked to provide a specific counterexample. It suffices to present an example of a morphism of left exact functors whose object-wise cokernel is not left exact. Let us find such a morphism of functors among morphisms of representable functors. So we want a morphism $X\to Y$ in an abelian category such that the object-wise cokernel of the morphism $Hom(Y,{-})\to Hom(X,{-})$ is not left exact. This means that there is a commutative square of morphisms $X\to Y$, $A\to B$, $X\to A$, $Y\to B$ such $A\to B$ is a monomorphism and the morphism $X\to A$ does not factor through the morphism $X\to Y$. Start with any nonsemisimple abelian category and choose a nonsplit monomorphism $X\to Y$. Set $A=X$ and $B=Y$.
 Thanks! I understand the analogy. Nevertheless can you give a concrete counterexample to show that U is not projective? As I already indicated, this is basically the same as giving a non-zero $F$ with $R^0 F = 0$, where $R^0 F$ is the universal left-exact functor associated to $F$. – Martin Brandenburg Oct 10 2010 at 19:38 I've spelled out a counterexample. Now, just to save everybody a possible confusion, let me say that I do not understand your notation $R^0F$. If it is meant to imply that the functor associating to an additive functor its universal left exact approximation is not exact and has some higher derived functors, then the implication is incorrect. The sheafification functor is exact. – Leonid Positselski Oct 10 2010 at 20:38 I mean, the sheafification is exact as a functor from presheaves to sheaves. Though if one considers it as a functor from presheaves to presheaves, it is only left exact and has a right derived functor, indeed. Is that what you had in mind? – Leonid Positselski Oct 10 2010 at 20:49 For the notation, see Chapter II in Gabriel's thesis. I know that $R^0$ is an exact functor (this is also proved there), but I think we have to show that it is not faithful. Thank you for your example. If $X \to Y$ is a monomorphism in $C$, then $\eta : Hom(Y,-) \to Hom(X,-)$ is an epimorphism in $Sex(C,Ab)$ (I can show this with Lemme 3 b applied to the pointwise defined cokernel). But it is a pointwise epimorphism iff $X \to Y$ is split. – Martin Brandenburg Oct 10 2010 at 21:27