# Colimits in colimits of categories

I was considering the following situation. I have a directed/filtering system of categories $C_i$. I understand how take its direct limit (aka colimit) $C$ in the category of categories. My question is:

If $C_i$ is cocomplete is $C$ also cocomplete?

I think in general the answer is no. For example, taking $C_i = C_0 \times C_0 \cdots \times C_0$ for some fixed category $C_0$ and taking $C_i \to C_{i+1}$ to be the obvious inclusion. I don't think the limiting category $C$ has coproducts.

Can I add (non trivial) assumptions so that $C$ suddenly has colimits? (if so, how do I compute them?)

In the situation I am considering I have a bunch more of assumptions, which might be handy. All the categories $C_i$ are Grothendieck abelian. The functors $C_i \to C_j$ are exact, essentially surjective and have right adjoints. The indexing system has an initial element $0$, so any object in $C_i$ comes from $C_0 \to C_i$. Also, all I really want in $C$ are direct limits (ie filtered colimits), not necessarily arbitrary ones.

• If the diagram is $\kappa$-filtered, any structure of size less than $\kappa$ (like e.g. $\kappa$-colimits) that is shared by all categories and preserved by the transition functors is inherited by the $\kappa$-filtered colimit. – godelian Jan 21 '17 at 16:00
• @godelian that's great! how do you prove it? I can't see what the recipe is to form colimits in the limiting category. – Yosemite Sam Jan 21 '17 at 20:50
• It follows from the construction of $\kappa$-filtered colimits of categories. If you have a diagram in the colimit category, you can find representatives of each arrow in the categories involved in the $\kappa$-filtered colimit. Because of $\kappa$-filteredness, you can find a unique category where all representatives lie. Then you just take the colimit in this category and it will be a representative of the colimit in the colimiting category. – godelian Jan 21 '17 at 21:01
• @godelian I'm probably not seeing something obvious. Take the example of a diagram $x_i$ which can be entirely represented in the first category $C_0$. I would like to conclude that $f(colim_i x_i) = colim_i f(x_i)$ where $f\colon C_0 \to C$ is the functor from $C_0$ to the colimit category $C$. But if I try and verify the universal property, ie $C(f(colim_i x_i), y) = lim_i C(x_i, y)$ I get stuck, essentially because limits and colimits do not commute in general. Am I missing something? If so, would you care to write an answer? – Yosemite Sam Jan 21 '17 at 21:17
• More explicitly, say $x,y \in C$ can be represented in the initial category $C_0$, we have $C(x,y) = colim_j C_j(f_j(x), f_j(y))$ where $f_j \colon C_0 \to C_j$ is the transition functor. For the universal property to be satisfied I would have to have $colim_j C_j( f_j colim_i x_i, y) = lim_i colim_j C_j(f_j(x_i), f_j(y))$. Assuming $f_j$ commutes with colimits (which I'd be happy to), this amounts to $colim_j lim_i C_j(f_j(x_i), f_j(y)) = lim_i colim_j C_j(f_j(x_i), f_j(y))$ which doesn't hold in general. – Yosemite Sam Jan 21 '17 at 21:20

Let $C_i$ a diagram of categories on a filtred small category $I$, $C:=\varinjlim_{i} C_i$, let write $[A, B]$ the category of funtors (and natural transformation) between categories $A, B$, observe that a colimit of categories is also a 2-colimit (i.e. it is natural respect natural transformations).

If $J$ is a finite category then $\varinjlim_{i} [J, C_i]\cong [J, C]$ (a functor $J \to C$ as a lifting on some $C_i$, and two such lifting have another one with usual coherence). Consider the colimit adjunction $[J, C_i]\leftrightarrows [\ast, C_i]$ then getting the colimit we have the adjoint diagram $[J, C]\leftrightarrows [\ast, C]$.

THere is a easy generalization to more high cardinal..