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.
Best Answer
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 $g(\alpha_{2})=\gamma\beta$.