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Oct 10, 2010 at 21:27 vote accept Martin Brandenburg
Oct 10, 2010 at 21:27 comment added Martin Brandenburg 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.
Oct 10, 2010 at 20:49 comment added Leonid Positselski 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?
Oct 10, 2010 at 20:38 comment added Leonid Positselski 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.
Oct 10, 2010 at 20:21 history edited Leonid Positselski CC BY-SA 2.5
a counterexample spelled out
Oct 10, 2010 at 19:38 comment added Martin Brandenburg 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 $F$.
Oct 10, 2010 at 17:59 history answered Leonid Positselski CC BY-SA 2.5