# Indization as adjoint from finite colimits to all colimits

Where can I find a reference for the following fact:

If $C$ is a small category with finite colimits, then $\text{Ind}(C)$ is a category with all colimits, and it is the universal one in the following sense: If $D$ is a category with all colimits, then there is an equivalence of categories between finitely cocontinuous functors $C \to D$ and cocontinuous functors $\text{Ind}(C) \to D$.

This is a consequence of a combination of facts which can be found in SGA 4 I.8 and Section 6 in "Category of Sheaves" by Kashiwara, Shapira. I wonder if this is written down somewhere so that I can reference it in my work without spelling out the proof.

EDIT: Actually I don't know exactly how to prove it. Take a finitely cocont. functor $C \to D$. This extends to a cocont. functor $\widehat{C} \to D$ and may be restricted to $\text{Ind}(C)$. The resulting functor $\text{Ind}(C) \to D$ commutes with coproducts and with filtered colimits. The problem are the cokernels (or rather coequalizer, if we do not deal with linear categories). Namely, we should prove that every cokernel diagram in $\text{Ind}(C)$ is a filtered colimit of cokernel diagrams in $C$. Now in the book by Kashiwara, Schapira it is shown that (almost) every finite diagram in $\text{Ind}(C)$ is a filtered colimit of diagrams of the same shape in $C$. But we need a refinement of this!

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Every finite loopless diagram in Ind-$C$ can be represented as a levelwise diagram, i.e. as a diagram of functors $I \to C$ for some filtered $I$. (Artin-Mazur, App. of Etale Homotopy Theory, Prop. 3.3 for example.) In particular, this holds for coequalizer diagrams. Then the colimit can be computed levelwise (Prop. 4.1 in Artin-Mazur). See also D. Isaksen, "Calculating limits and colimits in pro-categorie", Fund. Math. 175 (2002).

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See the EDIT of my question, don't we need to do more than that? Take a coequalizer diagram X ==> Y -> Z in Ind(C). Then it is a filtered colimit of diagrams of the form X_i ==> Y_i -> Z_i in C. But why can we assume that these diagrams are coequalizer diagrams? They are only coequalizers in the colimit. –  Martin Brandenburg Jan 24 '11 at 13:46
Ok perhaps it is that easy: Just construct a filtered diagram of the form X_i ==> Y_i in C with colimit X ==> Y. Define Z_i to be the coequalizer of X_i ==> Y_i. Then the colimit of X_i ==> Y_i -> Z_i is the given diagram X ==> Y -> Z. And this is all what we need, right? –  Martin Brandenburg Jan 24 '11 at 14:29
I think so. (plus some more characters to make MO happy) –  Tilman Jan 24 '11 at 22:35

This is not what you asked (again!) but the derived ($\infty$-categorical) version of this statement is Proposition 5.3.5.10 in Lurie's Higher Topos Theory.

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:-) And Theo will answer the non-derived setting? –  Martin Brandenburg Jan 23 '11 at 21:56
I'd have loved to! Unfortunately, Appendix B of the same paper is missing this time... –  Theo Buehler Jan 23 '11 at 22:24

This book has a pretty thorough discussion of Ind-completion:

\bib{MR861951}{book}{
author={Johnstone, Peter T.},
title={Stone spaces},
volume={3},
note={Reprint of the 1982 edition},
publisher={Cambridge University Press},
place={Cambridge},
date={1986},
pages={xxii+370},
isbn={0-521-33779-8},
review={\MR{861951 (87m:54001)}},
}


However, I do not have my copy to hand, so I cannot guarantee that it contains precisely the result that you want.

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Thanks, this is a good reference. However, the fact in my question is not mentioned there. –  Martin Brandenburg Jan 23 '11 at 23:17

If your definition of $\operatorname{Ind}(C)$ is "those ($\operatorname{SET}$-valued) presheaves on $C$ that take finite colimits to limits in $\operatorname{SET}$" (so that it is a full subcategory of $\{\text{presheaves on }C\}$), then the fact follows from general facts about presentable categories, and a good reference is Adámek and Rosický, Locally presentable and accessible categories, 1994. But this probably is not your definition of $\operatorname{Ind}(C)$, and so you would have to prove it is equivalent, and by then you have probably proved the result you want.

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I know this description (for finitely cocomplete $C$). Which result about pres. cat. do you mean? –  Martin Brandenburg Jan 24 '11 at 0:58
Here's a general result: Recall that a colimit sketch is a small category $C$ along with some small diagrams in $C$ called "distinguished diagrams", and an assignment that to each distinguished diagram picks out a cocone for the diagram, called its "distinguished cocone". A functor $C \to D$ is sketch-compatible if it turns distinguished cocones into colimit cocones. A sheaf on $C$ is a sketch-compatible functor $C \to \text{SET}^{\rm op}$, and $\operatorname{Sheaves}(C)$ is the full subcategory of the presheaf category on the sheaves. (continued) –  Theo Johnson-Freyd Jan 24 '11 at 6:06
(continuation) Then the result is that $\operatorname{Sheaves}(C)$ is cocomplete, and for any cocomplete category $D$, there is an equivalence between sketch-compatible functors $C \to D$ and cocontinuous functors $\operatorname{Sheaves}(C) \to D$; in particublar, there is a canonical sketch-compatible functor $C \to \operatorname{Sheaves}(C)$. In general, the equivalence is given in one direction by pullback along the canonical functor, and in the other direction by some version of Kan extension. The forgetful map $\operatorname{Sheaves} \to \operatorname{Presheaves}$ has a left adjunct. –  Theo Johnson-Freyd Jan 24 '11 at 6:10
Anyway, proving this is probably useless --- you might as well just prove the result for your case. (It's not too hard --- anyway, Alex Chirvasitu and I include a proof in a paper we still haven't written yet. And some version is in any good book on presentable categories. I think calling these "sheaves" might be new to Alex and myself, but maybe not; it's an obvious name, but people usually talk about "models of a limit sketch" rather than "sheaves on a colimit sketch".) But if you ever need a generalization, it exists. –  Theo Johnson-Freyd Jan 24 '11 at 6:22
Ok, so in my case we just have the sketch of all finite diagrams. "You might as well just prove the result for your case" Yes and I have problems with that. ;-) In the EDIT of my question I mention the difficulty. –  Martin Brandenburg Jan 24 '11 at 9:21