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An object $x$ in a category $\mathsf{C}$ is called compact or finitely presentable if $$\mathrm{hom}(x,-) : \mathsf{C} \to \mathsf{Set}$$ preserves filtered colimits. This concept behaves best when $\mathsf{C}$ has all filtered colimits, e.g. when it is the category of presheaves on some small category $\mathsf{X}$:

$$ \mathsf{C} = \mathsf{Set}^{\mathsf{X}^{\mathrm{op}}} $$

Every representable presheaf is compact. In general, any finite colimit of compact objects is compact. Thus, any finite colimit of representables is compact.

My question is about the converse: in the category of presheaves on a small category, is every compact object a finite colimit of representables?

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    $\begingroup$ If you don't want to assume that X is idempotent-complete, then I think you can get a counterexample by looking at retracts of representables that don't exist in your base category, since these will be compact and I don't think you can write them as finite colimits of representables in general. But a particular example with a proof that it isn't a finite colimit of representables isn't coming to me right now. $\endgroup$ – William Balderrama May 25 at 6:17
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    $\begingroup$ @WilliamBalderrama: Retracts are colimits over a category with a single object and a single nonidentity morphism, which is idempotent. $\endgroup$ – Dmitri Pavlov May 25 at 6:24
  • $\begingroup$ @DmitriPavlov Ah, so they are. For some reason I had in mind the sequential colimit along the given idempotent morphism as the way of splitting it. $\endgroup$ – William Balderrama May 25 at 6:28
  • $\begingroup$ Maybe a relevant motivational analogy is that compact objects in the poset $2^X$ are precisely finite subsets of $X$. $\endgroup$ – Ivan Di Liberti May 25 at 11:36
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    $\begingroup$ FWIW this is no longer true in the $\infty$-categorical setting, where "finite" has a slightly different meaning, and taking retracts is no longer a finite colimit. (It fails already when $C$ is a point, by the Wall finiteness obstruction.) $\endgroup$ – Dylan Wilson May 25 at 12:31
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Yes, it is. The reason is:

  • every object of your presheaf category is a colimit of representables;
  • so, every object is a filtered colimit of objects which are finite colimits of representables;
  • so, applying the definition of a compact object, you get a split monomorphism from your compact object $X$ to a finite colimit $T$ of representables. To conclude, write $X$ as the coequaliser of $Id_T$ and the idempotent of $T$ given by your split mono.
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    $\begingroup$ Hang on -- this shows that $X$ is a finite colimit of (finite colimits of representables) -- a "2-fold" finite colimit of representables. But how does one turn this into an actual finite colimit of representables? I believe that any retract of $\kappa$-small colimits of representables is a $\kappa$-small colimit of representables when $\kappa$ is an uncountable regular cardinal, but I'm not sure about the case $\kappa = \aleph_0$... $\endgroup$ – Tim Campion May 26 at 20:38
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    $\begingroup$ The class of presheaves which are finite colimit of representables is stable under finite colimits: $\endgroup$ – Aurélien Djament May 27 at 5:55
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    $\begingroup$ The class of presheaves which are finite colimit of representables is stable under finite colimits: the stability under finite coproducts is obvious and the stability under coequalisers is easily seen by hand using the universal property of colimits and the fact that Hom(x,-) commutes with colimits for x representable. $\endgroup$ – Aurélien Djament May 27 at 6:01
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    $\begingroup$ I think I see how this works for finite coproducts -- but could you spell out the coequalizers for me? $\endgroup$ – Tim Campion May 27 at 17:10
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I think that Aurelien Djament's answer is essentially correct, but I want to nitpick a bit.

  1. If $\mathcal A$ is any locally finitely presentable category and $\mathcal C \subseteq \mathcal A$ is any strong generator of finitely-presentable objects, then every finitely-presentable object $X \in \mathcal A$ lies in the closure of $\mathcal C$ under finite colimits. So $X$ is a finite colimit of finite colimits of ... of finite colimits of objects of $\mathcal C$ -- an "$n$-fold" finite colimit of objects of $\mathcal C$. But $X$ need not be a "1-step" finite colimit of objects of $\mathcal C$. For example, I don't think every finitely-presented group is a finite colimit of copies of $\mathbb Z$.

  2. One might strengthen the hypotheses and ask: if $\mathcal A$ is a locally finitely presentable category and $\mathcal C \subseteq \mathcal A$ is a dense generator, then is every finitely-presentable object $X \in \mathcal A$ a finite colimit of objects of $\mathcal C$? I don't know the answer to this.

  3. But let's focus on the question at hand, i.e. the case where $\mathcal A = \hat {\mathcal C}$ is a presheaf category and $\mathcal C$ is the representables. Let $\tilde {\mathcal C}$ comprise the finite colimits of representables. Then indeed, $\tilde {\mathcal C}$ is closed under finite colimits. This is clear for finite coproducts -- just take the coproduct of the indexing diagrams for the colimits. Now let $A\rightrightarrows B \to C$ be a coequalizer where $A,B \in \tilde {\mathcal C}$. Then there is an epimorphism $\amalg_i X_i \to A$ and a coequalizer diagram $\amalg_j Y_j \rightrightarrows \amalg_k Z_k \to B$ where $X_i,Y_j,Z_k \in \mathcal C$ and the coproducts are finite. The composite maps $\amalg_i X_i \to A \rightrightarrows B$ lift to maps $\amalg_i X_i \rightrightarrows \amalg_k Z_k$. Then we have that $C$ is the coequalizer of the two induced maps $(\amalg_i X_i) \amalg (\amalg_j Y_j) \rightrightarrows \amalg_k Z_k$.

    Now I claim that if $f,g \amalg_{i \in I} X_i \rightrightarrows \amalg_{k \in K} Z_k$ are two maps with coequalizer $C$, and if the $X_i$ are representable, then $C$ is the colimit of the following diagram. Indeed, for each $i \in I$, there is a unique $k = k_0(i) \in K$ such that $X_i \to \amalg_{i \in I} X_i \xrightarrow f \amalg_{k \in K} Z_k$ factors through $Z_k$, and similarly a $k_1(i)$ for $g$. The indexing set for our diagram has object set $I \amalg K$, and the nonidentity morphisms are a map $i \to k_0(i)$ and a map $i \to k_1(i)$ for each $i \in I$. Then $C$ is the colimit of the obvious diagram sending $i \mapsto X_i$ and $k \mapsto Z_k$. This diagram is finite if $I$ and $K$ are.

    Thus in our case, $C \in \tilde{\mathcal C}$ as desired.

I want to emphasize that here we heavily used the fact that we're in a presheaf category.

  1. I agree that any category which has finite colimits and filtered colimits has all colimits. But Aurelien's second bullet seems to suggest something stronger -- that if $X$ is a colimit of objects of $\mathcal C$, then $X$ is a filtered colimit of finite colimits of objects of $\mathcal C$. I don't have a counterexample, but I'm not sure this is true. The closest I can convince myself of is that $X$ is a coequalizer of coproducts of objects of $\mathcal C$, and therefore a coequalizer of filtered colimits of finite coproducts of objects of $\mathcal C$ -- but this only ensures that $X$ is a finite colimit of filtered colimits of finite colimits of objects of $\mathcal C$.

  2. But using (3), Aurelien's third bullet goes through with some modification. As in any locally finitely presentable category $\mathcal A$ with strong generator $\mathcal C$, any finitely-presentable object is in the closure of the $\mathcal C$ under finite colimits. By (3), in the case $\mathcal A = \hat{\mathcal C}$, the closure of $\mathcal C$ under finite colimits consists exactly of $\tilde{\mathcal C}$, the objects which are "1-step" finite colimits of representables. Here, (3) is actually used in 2 places: first to ensure that the category $\tilde C \downarrow X$ is filtered (this being the diagram which indexes the canonical colimit for $X$), and second to ensure that $\tilde{\mathcal C}$ is closed under retracts.

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    $\begingroup$ Thanks for this answer that really works out what's going on here! $\endgroup$ – John Baez May 29 at 1:51
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Here is another perspective on the problem, using some big guns (Gabriel-Ulmer duality).

Given a small category $C$, let $K$ be its free finite cocompletion. This means $K^{op}$ is the free finite completion of $C^{op}$, which means in turn that for any functor $F: C^{op} \to \mathbf{Set}$, there is a finitely continuous (or left exact) functor $\tilde{F}: K^{op} \to \mathbf{Set}$ that extends $F$ along the canonical inclusion $i: C^{op} \to K^{op}$, and this extension is unique up to unique isomorphism. Put differently, restriction along $i$ induces an equivalence

$$\mathrm{Lex}(K^{op}, \mathbf{Set}) \to \mathrm{Cat}(C^{op}, \mathbf{Set}).$$

In particular, the presheaf category $\mathrm{Cat}(C^{op}, \mathbf{Set})$ is locally finitely presentable. By the way, it's well known that the free finite cocompletion $K$ of a small category $C$ is simply the category of finite colimits of representables: see section 5.9 of Kelly's Basic Concepts of Enriched Category Theory.

On the other hand, Gabriel-Ulmer duality assures us that given a locally finitely presentable category $A$, there is up to equivalence only one finitely complete category $L$ for which $A \simeq \mathrm{Lex}(L, \mathbf{Set})$. Even better, Gabriel-Ulmer duality gives a recipe for obtaining $L$: it is the dual of the category of compact objects in $A$, meaning objects $a$ such that $A(a, -): A \to \mathbf{Set}$ preserves filtered colimits.

Putting all this together, this shows that the category of compact objects in the category of presheaves over $C$ is equivalent to the free finite cocompletion of $C$, or to the category of finite colimits of representable presheaves.

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