when is Lex[A,V] abelian?

when is Lex[A,V] abelian?

[Richard Garner]

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
>
>


[For admin and other information see: http://www.mta.ca/~cat-dist/ ]

when is Lex[A,V] abelian?

[Ignacio Lopez Franco]

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


[For admin and other information see: http://www.mta.ca/~cat-dist/ ]