# Are compact objects in presheaf categories finite colimits of representables?

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?

• 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. – William Balderrama May 25 at 6:17
• @WilliamBalderrama: Retracts are colimits over a category with a single object and a single nonidentity morphism, which is idempotent. – Dmitri Pavlov May 25 at 6:24
• @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. – William Balderrama May 25 at 6:28
• Maybe a relevant motivational analogy is that compact objects in the poset $2^X$ are precisely finite subsets of $X$. – Ivan Di Liberti May 25 at 11:36
• 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.) – Dylan Wilson May 25 at 12:31

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.
• 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 $\kappa = \aleph_0$... – Tim Campion May 26 at 20:38
• The class of presheaves which are finite colimit of representables is stable under finite colimits: – Aurélien Djament May 27 at 5:55
• 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. – Aurélien Djament May 27 at 6:01
• I think I see how this works for finite coproducts -- but could you spell out the coequalizers for me? – Tim Campion May 27 at 17:10

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.

• Thanks for this answer that really works out what's going on here! – John Baez May 29 at 1:51

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.