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Let $C$ be a cocomplete category and $S$ a set of objects of $C$. We may assume, if need be, that the objects of $S$ are compact. Consider $S'$ the class of objects spanned by finite colimits of objects in $S$.

Is $S'$ closed under finite colimits? That is, if $F \colon I \to C$ is a diagram with $I$ a category with finitely many objects, and $Fi \in S'$ for all $i \in I$, then is necessarily $\mathsf{colim}_I \ F \in S'$?

This question was motivated by this answer to the characterization of compact objects in terms of finite colimits of a set of 'generators'.

Concretely: suppose that every object is a colimit of objects in $S$. Then every object in $C$ can be computed as a filtering colimit of objects in $S'$. If $K$ is compact and we describe it as such a colimit $K = \mathsf{colim}_{\alpha \in \Lambda} x_\alpha$ with $x_\alpha \in S'$, then it follows that the identity map $1_K$ factors through some component map $x_\alpha \to K$; hence $K$ is a retract of some $x_\alpha \in S'$. A retract can be seen as a particular case of a coequalizer, which is a finite colimit. It is claimed in loc. cit. that $S'$ is closed under retracts, together with a passing remark observing that these are finite colimits.

The question stems from trying to prove the aforementioned statement and, more generally than for retracts, that 'finite colimits are closed under finite colimits'.

I think this should be true for finite coproducts of finite limits, if $\{F_i \colon J_i\to C\}_{i=1}^n $ are diagrams with $J_1,\ldots, J_n$ finite, then $F = F_1 \sqcup \cdots \sqcup F_n \colon J_i \sqcup \cdots \sqcup J_n \to C$ satisfies $\mathsf{colim} F = \coprod_{i=1}^n \mathsf{colim} F_i$.

For coequalizers, the motivating example, I have been unable to do it. I suppose this would prove the claim in full generality, using that every (finite) colimit is a coequalizer between (finite) coproducts.

Any help would be much appreciated.

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  • $\begingroup$ I think it should be false in general, about coequalisers, since it really seems like you need to presume a compactness condition on $S$. Ok, if $f,g:A\rightrightarrows B$ for $A,B\in S'$ you can represent them by arrows $s\to B$ for some family of $s\in S$ but somehow you should need to get arrows not into $B$ but into the objects describing $B$. $\endgroup$
    – FShrike
    Dec 14, 2023 at 20:07
  • $\begingroup$ I would be happy to know the result (if true) assuming $S$ consists of compact objects. $\endgroup$
    – qualcuno
    Dec 14, 2023 at 20:27

3 Answers 3

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Hidden in this related MathOverflow thread there is a fairly simple counterexample. It is from the final paragraph of Section 5.9 of Basic Concepts of Enriched Category Theory by G.M. Kelly, but it is described there very briefly, so I have expanded a bit.

Let $C$ be the category of small categories, and let $S$ consist of the single category $\mathbf{2}$ (the category $0\xrightarrow{\alpha}1$ with two objects and just one nonidentity morphism between them).

The colimit of the diagram $$\left[0\xrightarrow{\alpha}1\right]\xleftarrow{s}\left[0\xrightarrow{\alpha}1\right] \xrightarrow{t}\left[0\xrightarrow{\alpha}1\right],$$ where $s(\alpha)=\operatorname{id}_{0}$ and $t(\alpha)=\operatorname{id}_{1}$, is the category $\mathbf{3}$ (the category $0\xrightarrow{\beta}1\xrightarrow{\gamma}2$ with three objects and just three nonidentity morphisms $\beta$, $\gamma$ and $\gamma\beta$).

The coequalizer of any pair of maps $$\coprod_{i}\mathbf{2}\rightrightarrows\coprod_{j}\mathbf{2}$$ is constructed from the copies of $\mathbf{2}$ in $\coprod_{j}\mathbf{2}$ by identifying some objects and some morphisms and contracting some morphisms to become identity morphisms. So the colimits of copies of $\mathbf{2}$ are the free categories on directed multigraphs. And the finite colimits are the free categories on finite directed multigraphs.

In particular, the category $D$ with one object and one nonidentity morphism $\varepsilon$ with $\varepsilon^{2}=\varepsilon$ is not a colimit of copies of $\mathbf{2}$.

However, $D$ is the coequalizer of the pair of maps $$\left[0\xrightarrow{\alpha_{1}}1\right]\sqcup \left[0\xrightarrow{\alpha_{2}}1\right] \xrightarrow{f,g} \left[0\xrightarrow{\beta}1\xrightarrow{\gamma}2\right]$$ where $f(\alpha_{1})=\beta$, $g(\alpha_{1})=\gamma\beta$, $f(\alpha_{2})=\gamma$ and $g(\alpha_{2})=\gamma\beta$.

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  • $\begingroup$ Thanks a lot for the detailed answer! $\endgroup$
    – qualcuno
    Dec 17, 2023 at 5:50
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Given the negative answer here, the "correct" definition to take is the smallest finitely-cocomplete full subcategory containing $S$. It is given explicitly by $S \cup S' \cup S'' \cup \cdots$. This construction is familiar from many other transfinite constructions (such as the algebraic closure).

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$\newcommand{\I}{\mathscr{I}}\newcommand{\C}{\mathsf{C}}\newcommand{\colim}{\operatorname{colim}}$It's true if $S$ consists only of compact finite? objects. By remarks in this answer we may replace finite colimits with finite filtered colimits (appeal to the category of elements).

As you observe, it's easy to cook up finite coproducts, so the only thing to do is show $S'$ is closed under binary coequalisers. Well,... if you change your notion of "finite category" to not just be a category with finitely many objects but a category with finitely many arrows. Else, inspecting the coproduct-and-coequaliser recipe for colimits, we see that we might need to have $S'$ closed under infinite coproducts and this cannot be guaranteed.

Let $A,B\in S'$ and $f,g:A\rightrightarrows B$. We want to show there is a coequaliser of $f,g$ which is in $S'$. Choose finite diagrams $D_1:\I_1\to\C,D_2:\I_2\to\C$ and colimiting cocones $\lambda_{\bullet,1}:D_1\implies\Delta A$, $\lambda_{\bullet,2}:D_2\implies\Delta B$. Much of what I'm about to say is ""obvious"" and makes sense if you first draw a picture of what we're trying to do. It's mostly notational bookkeeping to make sure everything actually works (and hopefully it does and I didn't mess up).

Each $f\lambda_{i,1}$ factors as $\lambda_{j(i),2}\alpha_i$ for some $\alpha_i:D_1(i)\to D_2(j(i))$ by compactness; each $g\lambda_{i,1}$ factors as some $\lambda_{k(i),2}\beta_i$. Construct $\I_3$ to be a category with objects equal to $\I_1\sqcup\I_2$ (and yes I want that to be a disjoint union, use some trick like $\I_1\times\{0\},\I_2\times\{1\}$ if you have to) and with all the same arrows and composition law for objects of $\I_2$ and with two new formal arrows $i\to j(i),i\to k(i)$ for each $i\in\I_1$ whose composition with other arrows in $\I_2$ is freely defined. Formally, $\I_3(i,i')$, for $i\in\I_1$ and $i'\in I_2$, has no elements, one element tagged "$j$" or one element tagged "$k$" or both according to whether or not there exists at least one arrow $j(i)\to i'$ or $k(i)\to i'$ in $\I_2$ (or neither or both).

There is a diagram $D_3:\I_3\to\C$ which acts like $D_1,D_2$ on objects, like $D_2$ on arrows of $\I_2$ and maps the arrow $i\to j(i)$ to $\alpha_i$, the arrow $i\to k(i)$ to $\beta_i$. Let $\lambda_{\bullet,3}$ denote the colimiting cocone for $\colim D_3$. I claim that $\colim D_3\in S'$ is a coequaliser of $f,g$.

First of all, there is a map $B\to\colim D_3$ induced by the legs $\lambda_{i',3}:D_2(i')\to\colim D_3$. It makes $f$ agree with $g$ since maps out of $A$ are determined by their components on each $D_1(i)$ and all roads lead to Rome, namely to $\lambda_{3,i}:D_1(i)\to\colim D_3$ in this case.

Now say $h:B\to C$ makes $hf=hg$, $C$ any object of $\C$. Then in particular $h\lambda_{2,j(i)}\alpha_i=hf\lambda_{1,i}=hg\lambda_{1,i}=h\lambda_{2,k(i)}\beta_i$. It follows that $h$ induces a cocone $D_3\implies\Delta C$ since, as an arrow $B\to C$, there is already an associated cocone from $D_2$ and we just checked $h$ is compatible with the other arrows of the diagram $D_3$. It is now fairly clear (but check it! We need to use the universal property of $B=\colim D_2$ again) $h$ factors uniquely through $B\to\colim D_3$ and it follows $S'$ is contained under coequalisers.

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  • $\begingroup$ I did not follow the existence of the $\alpha_i$. Maps from compact objects factor when they land in filtering colimits, but these are arbitrary finite ones, no? $\endgroup$
    – qualcuno
    Dec 14, 2023 at 23:08
  • $\begingroup$ @qualcuno Perhaps I misremembered the definition. Rip. That said, I think we can replace a finite colimit with a finite filtered one. Adjoin a terminal object to the indexing category and map the terminal object to the colimit of the diagram; then map each arrow to the terminal object to the corresponding leg of the colimiting cocone. This new diagram should have the same colimit and the new diagram is still finite but now it is filtered! $\endgroup$
    – FShrike
    Dec 14, 2023 at 23:15
  • $\begingroup$ The problem with that 'fix' is that the colimit of the original diagram need not lie in $C'$ --that is precisely what we are trying to prove. $\endgroup$
    – qualcuno
    Dec 14, 2023 at 23:17
  • $\begingroup$ @qualcuno The colimit lies in $S'$ by definition of $S'$, but you're right there's an issue in that it doesn't necessarily lie in $S$, which is a problem $\endgroup$
    – FShrike
    Dec 14, 2023 at 23:28
  • $\begingroup$ That is what I meant, indeed; sorry for the misunderstanding. $\endgroup$
    – qualcuno
    Dec 14, 2023 at 23:30

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