How to compute (co)continuity of functors

Question

In these days I am thinking to a problem on accessible categories. I have a faithful functor $F : \mathcal{K} \to \text{Set}$ and I want to know if it preserves directed colimits. More generally I want to know if such a functor exists.

This was just context and motivations. Now comes the question.

For a diagram $D : I \to \mathcal{K}$ and a functor $F: \mathcal{K} \to \mathcal{A}$, one would like to have a machinery to understand if the $F$ preserves $D$.

Such a question is very vague but in fact is much studied. For the special case of D = diagram of (co)kernel, the theory of derived functors $F^i$, gives a gadget - an abelian group - which is $0$ precisely when $F$ preserves $D$.

This theory uses many hypotesis of $\mathcal{A}$, such as the existence of any variant of snake lemma. And uses hypotesis on $\mathcal{K}$, too, such as existence of enough projectives. I am not allergic to any kind of hypotesis.

All in all, my question is:

• Did anyone try to develop a tool which is sensitive to (co)continuity of a functor on some (co)limits, such as it was done for the (co)kernel - finite limits, in fact - in the case of abelian categories?
• Is there any evident obstruction that makes the (co)kernel a very lucky case?

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As a first try, a friend of mine, told me that Freyd and Kelly gave some very natural conditions under which category of continous functors with respect to a class of diagrams is a reflexive subcategory of the category of all functors. This is interesting but I does not seem to me to answer the question.

Surely this proves that for an accessible category $\mathcal{K}$ there is a functor $F: \mathcal{K} \to$ Set which preserves many colimits, but I need this functor to be faithful too, thus this characterization is too blind for me. Something more computational would be much better. One could hope that the reflection takes faithful functors into faithful functors, but this is just superstition.

Any reference as the one I mentioned would be welcome.

Answer 1

The "reason" that continuity of abelian (and also stable/triangulated) functors can be detected by zeroness of objects is that more generally, invertibility of morphisms in an abelian/stable/triangulated category is detected by zeroness of objects. That is, $f:A\to B$ is an isomorphism exactly when its kernel and cokenel are zero, or in the stable/triangulated case exactly when its cone/cocone is zero. This is not true in the unstable case, so we have to consider morphisms directly; and cocontinuity of a functor is literally the statement that a certain morphism $\mathrm{colim}(F D) \to F(\mathrm{colim} D)$ is an isomorphism.

That said, since you phrased your question so generally, I would argue that technically the answer is yes, since this morphism $\mathrm{colim}(F D) \to F(\mathrm{colim} D)$ is itself a "gadget" and "being an isomorphism" is a condition on it, just like "being zero" is a condition on the derived functor objects. The analogy is not very far either; a morphism $f:A\to B$ is an isomorphism just when it is a terminal object of the slice category $\mathcal{A}/B$, and just when it is an initial object of the co-slice category $A/\mathcal{A}$. Internally in $\mathcal{A}$, the former condition says that "every fiber of $B$ is trivial", so we can see that the only difference in the unstable case is that we have to consider all the fibers rather than just the fiber over the basepoint (the kernel/cone).