projective generator in the category of left-exact functors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:03:30Z http://mathoverflow.net/feeds/question/41693 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41693/projective-generator-in-the-category-of-left-exact-functors projective generator in the category of left-exact functors Martin Brandenburg 2010-10-10T16:40:59Z 2010-10-10T20:21:24Z <p>I expect this to be a small understanding problem and not a real interesting question.</p> <p>In <a href="http://ncatlab.org/nlab/show/Des+Cat%25C3%25A9gories+Ab%25C3%25A9liennes" rel="nofollow">Gabriel's thesis</a> 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.</p> <p>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).</p> <p>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?</p> http://mathoverflow.net/questions/41693/projective-generator-in-the-category-of-left-exact-functors/41699#41699 Answer by Leonid Positselski for projective generator in the category of left-exact functors Leonid Positselski 2010-10-10T17:59:30Z 2010-10-10T20:21:24Z <p>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.</p> <p>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.</p> <p>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 $B=Y$. </p>