7
$\begingroup$

1) Suppose that I have a sifted diagram of categories $\mathcal{C}_i$, another of the same shape $\mathcal{D}_i$, and that I have a system $F_i:\mathcal{C}_i\to\mathcal{D}_i$ commuting with the morphisms in each diagram. Then I get a functor $$(\operatorname{colim} F_i):\operatorname{colim} \mathcal{C_i} \to \operatorname{colim}\mathcal{D}_i.$$ Suppose I know that each $F_i$ is fully faithful. Under what further assumptions may I conclude that $(\operatorname{colim} F_i)$ is fully faithful? For instance, if I would assume the $F_i$'s have right adjoints, so that fully faithfulness is equivalent to the unit for the adjunction being an isomorphism, does it help?

A question which seems like a de-categorified analog of this would be:

2) Fix some co-complete category $X$, and consider some sifted diagram of objects $V_i$ in $X$, another of the same shape $W_i$ in $X$. Suppose I have a system of split monomorphisms $f_i:V_i\to W_i$. Under what conditions do I know that the natural morphism, $$(\operatorname{colim} f_i): \operatorname{colim} V_i \to \operatorname{colim} W_i$$ is also a split monomorphism?

I really care about the answer to (1), but I offer (2) in my attempt to find an analogous proof one category number down, where there may be more examples/techniques/counter-examples.

Here, in motivating the attempted analogy between (1) and (2), I am trying to replace the 2-category $\operatorname{Cat}$ of categories by a 1-category $X$, and replace a fully faithful functor by a split monomorphism (I am imagining that the $F_i$'s have a right adjoint, which might behave like a splitting to make the analogy work, though it's a stretch).

I am aware of https://math.stackexchange.com/questions/1116101/colimit-preserves-monomorphisms-under-certain-conditions, where it is explained that filtered colimits preserve monomorphisms, in the 1-categorical setting. I am hoping, perhaps too optimistically, that if we move from filtered colimits to sifted colimits, we still have a chance to preserve split monomorphisms even though we (I suppose?) no longer preserve arbitrary monomorphisms. I am further hoping that if so, the argument would generalize to the 2-categorical setting.

I'd be most grateful for any hints, or references.

Disclaimer: I have simplified the above setup a bit to make a cleaner question. In case it could mislead an expert though, let me say that the version of (1) I need involves the $\mathcal{C}_i$ being Cauchy-complete, the $\mathcal{D}_i$ being presentable, and really the two colimits are taken in their respective 2-categories, in Cauchy-complete categories for the diagram of $\mathcal{C}_i$ and in presentable categories for the $\mathcal{D}_i$ (so in particular the functors appearing in each kind of colimit are suitable colimit preserving; the shape of the indexing diagram is the same, though). I assume these are technicalities and that if the question has a nice answer, I can do that kind of adaptation myself.

$\endgroup$
3
$\begingroup$

Reflexive coequalizers are examples of sifted colimits in a 1-category, and groupoids are examples of reflexive coequalizers. But quotients don't preserve monomorphisms in general.

This gives a counterexample to question (2) in the category Set:

Let $V = S \rightrightarrows S$ denote the trivial groupoid acting on a set $S$ with two elements, and $W = S\times \mathbb Z/2\mathbb Z \rightrightarrows S$ the action groupoid of $\mathbb Z/2\mathbb Z$ acting on the same set $S$. There is an inclusion map $V\to W$, which is injective on both objects and morphisms. But the induced map on the quotient sets is not injective.


In the $(2,1)$-category setting, according to this answer of Denis Nardin, one must truncate the simplicial diagram at the 2-simplices to get a sifted diagram (note that a reflexive coequalizer diagram is the simplicial diagram truncated at the 1-simplices).

To get a counterexample to question (1), first take the corresponding 2-truncated Cech simplicial sets of the diagrams above, i.e.

$$S \rightrightarrows^{\rightarrow} S \rightrightarrows S$$

and

$$S\times \mathbb Z/2 \times \mathbb Z/2 \rightrightarrows^\rightarrow S\times \mathbb Z/2 \rightarrow S$$

Then take the category of sheaves (of vector spaces, say) together with the pushforward functor on these diagrams of sets (thought of as discrete spaces), to get truncated simplicial diagrams of categories (all of which are direct sums of copies of Vect).

The diagrams are levelwise fully faithful, but the induced functor on the colimit is identified with the pushforward $Shv(S) \to Shv(pt)$ which is not fully faithful.

$\endgroup$
  • 2
    $\begingroup$ I don't think reflexive coequalizer are sifted in the (2,1) sense (for me sifted means I→I×I is cofinal, and the meaning of cofinal is different) $\endgroup$ – Denis Nardin Jun 12 '18 at 16:49
  • $\begingroup$ Ah, ok. To be honest, I haven't thought about what sifted would mean in a $(2,1)$-sense. But I had the impression that a simplicial diagram should be an example of a sifted diagram (I believe this is the case in the $(\infty,1)$-setting - perhaps you can truncate at the 2-simplices in the $(2,1)$-setting?). Can I just replace the groupoids in my example with the corresponding Cech simplicial diagrams to get a counterexample? $\endgroup$ – Sam Gunningham Jun 12 '18 at 17:05
  • $\begingroup$ Yes, I believe that will work. You can find the characterization of cofinality for (n,1)-categories worked out in the appendix to this answer of mine, and indeed the 2-truncated simplicial diagram is (2,1)-cofinal. $\endgroup$ – Denis Nardin Jun 12 '18 at 17:15
  • $\begingroup$ Cool, that's good to know! I edited my answer accordingly. Does it work now? $\endgroup$ – Sam Gunningham Jun 12 '18 at 17:25
  • $\begingroup$ Thanks Sam, I think I was being a bit naive with what I was assuming. Back to the drawing board... $\endgroup$ – David Jordan Jun 14 '18 at 10:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.