1
$\begingroup$

Disclaimer: I first tried to ask this question on stackexchange https://math.stackexchange.com/questions/4577320/sufficient-conditions-for-a-right-exact-functor-to-be-a-left-adjoint but I did not get an answer there.

I know that by the adjoint functor theorem, for a right exact functor between Abelian categories to have a right adjoint, it is sufficient to check a certain "smallness condition" and the cocontinuity of the functor.

Question 1: What would be a good approach to proving that the functor indeed preserves small colimits?

In particular, an answer to Left/right exact functor "in nature" which is not a right/left adjoint states that for a left exact functor, to obtain continuity, it is enough (in most cases) to check that it preserves all products.

Question 2: Do we have a similar condition for right exact functors and preserving coproducts (again, in sufficiently well-behaved cases)?

Question 3: If the answer to question 2 is positive, and if the domain of the functor only contains finite coproducts, do we automatically get cocontinuity from right exactness?

$\endgroup$
0

1 Answer 1

2
$\begingroup$

Any additive functor between additive categories preserves finite coproducts and hence finite products, since they can be characterized as biproducts.

A right exact functor preserves all colimits iff it preserves coproducts.

Coproducts can be described as filtered colimits of finite coproducts. Hence, a right exact additive functor preserves all colimits iff it preserves filtered colimits. This is often useful to check in practice.

Often you can turn the adjoint functor Theorem around and only use the trivial direction: a left adjoint preserves all colimits. For example, it immediately follows that the tensor product of abelian groups preserves colimits in each variable, using the hom-tensor adjunction. This is much easier than working with the colimits directly.

The assumption in your second question is a bit unclear to me. What do you mean by saying that only finite coproducts exist? If an initial object exists, which is the empty and thus a finite coproduct, arbitrary direct sums of it exist. So perhaps you want to assume that if a coproduct exists, then almost all summands are initial? Yes in this case, finite coproduct preservation is enough.

$\endgroup$
1
  • $\begingroup$ Thank you so much for the answer Martin! In the third question I totally forgot the initial object (as a "trivial case"). $\endgroup$
    – Adelhart
    Commented Nov 20, 2022 at 7:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .