Hi Ignacio, If C has finite limits (in fact kernels would do) as well as finite colimits, then Lex[C^op, V] abelian actually implies C abelian. Indeed, the restricted Yoneda embedding C -> Lex[C^op, V] then preserves limits and finite colimits, and is fully faithful; whence C may be identified with a full subcategory of L closed under finite limits and finite colimits. But any such subcategory of an abelian category is itself abelian. This then means that Lex[C^op, V] abelian => C abelian => Lex[C^op, V] Grothendieck abelian. When V = Ab, this is sufficient to ensure that Lex[C^op, V] is lex-reflective in [C^op, V]. Off the top of my head, I don't know if the same is true when V = k-Mod; I feel like it might be necessary to assume that, for every f.p. flat k-module M, the functor M * (-) : C -> C preserves monomorphisms. Maybe that's automatic; I don't know enough algebra to say for sure. If C doesn't have kernels, then the situation is more interesting. I believe it should still be possible to give elementary conditions on C which are equivalent to L's being abelian. The point is that kernels do exist in L and so one can work out what it means for L to satisfy the kernel-cokernel exactness conditions for maps between representables in terms of structure in C. This will give some necessary conditions on C for L to be abelian. With any luck they will also be sufficient, though it might be necessary to consider a wider class of maps in L than merely those between representables. A relevant article, I think, is C. Centazzo, R.J. Wood An extension of the regular completion J. Pure Appl. Algebra, 175 (2002), pp. 93–108 That deals with the non-additive context but the same ideas should apply. I am also wondering if Mike Shulman's work on unary sites is relevant; see his CT2011 slides. Richard On 22 February 2012 11:31, Ignacio Lopez Franco <ill20@cam.ac.uk> wrote:
Dear all, may be some of the readers of this list will know the answer to the following question.
Let V be the category k-Mod for commutative ring k. For a finitely cocomplete V-category C, when is L = Lex[C^{op},V] abelian?
I know some cases: 1. When C is abelian so is L. 2. When C is a free completion under finite colimits of a small category, L is abelian (because it's equivalent to a presheaf V-category). 3. L is reflective in the abelian [C^{op},V]. When the reflection is left exact L is abelian. However I don't any conditions that guaranty that the reflection is left exact.
I would like to know some other conditions that ensure that L is abelian, and perhaps an example where L is not abelian.
Thanks Ignacio
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