# Connected component functors for distributive categories and Stone dualities

From this blog post by Qiaochu Yuan I learned that a distributive category $\mathsf C$ always admits a contravariant functor $\mathrm{hom}(-,\mathbf 2):\mathsf C^\text{op}\to \mathsf{BoolAlg}$ to Boolean algebras.

By taking opposites and composing with Stone duality one concludes any distributive category admits a "connected components functor" $\Pi_0:\mathsf C\to \mathsf{Pro}(\mathsf{FinSet})$ to profinite sets.

Particularly this applies to the category of topological spaces yielding a functor $\Pi_0:\mathsf{Top}\to \mathsf{Pro}(\mathsf{FinSet})$. This nicely resolves the problematic $0\cup \left\{ \frac 1n \right\} _{n\in \mathbb N}\subset \mathbb R$. However, profinite sets are equivalent to compact totally disconnected spaces and thus cannot include infinite discrete spaces. But then, what is e.g $\Pi_0(\mathrm{disc}\mathbb{N})$ where $\mathrm{disc}\mathbb N$ is the countable discrete space? It seems very strange for the space of connected components of $\mathrm{disc}\mathbb N$ to be compact - I'd expect it to be $\mathrm{disc}\mathbb N$ itself!

I've read that compact totally disconnected spaces are reflective in $\mathsf{Top}$ and so I take it $\Pi_0$ is the reflector, but I don't know what it is. I have also read that arbitrary totally disconnected spaces are reflective in $\mathsf{Top}$, so it seems there's an infinitary analogue of the latter $\Pi_0$, perhaps using the infinitary distributivity of $\mathsf{Top}$, but I don't know where it comes from.

Questions.

1. What is the functor $\Pi_0:\mathsf{Top}\to \mathsf{Pro}(\mathsf{FinSet})$, and does it coincide with the reflector of compact totally disconnected spaces?
2. "Why" does it make sense for $\mathrm{disc}\mathbb N$ to have a compact space of connected components via the latter $\Pi_0$?
3. What is the reflector of totally disconnected spaces?
4. Is there an "infinitary" analogue of the general construction, for infinitarily distributive categories?

I recall reading somewhere about an infinite stone duality describing an equivalent between the category of pro-sets and some horrifying creature involving filters, and hope this is not necessary to understand.

The functor $$\mathrm{hom}(-,\mathbf 2):\mathsf{Top}^\text{op}\to \mathsf{BoolAlg}$$ is very simple to describe explicitly: it takes a space $$X$$ to the Boolean algebra of all clopen subsets of $$X$$. So the functor $$\Pi_0$$ on $$\mathsf{Top}$$ just takes a space $$X$$ to the Stone space of its Boolean algebra of clopen sets. In the case $$\mathbb{N}$$, the Boolean algebra of clopen sets is the power set $$\mathcal{P}(\mathbb{N})$$, so $$\Pi_0(\mathrm{disc}\mathbb N)$$ is the space of all ultrafilters on the set $$\mathbb{N}$$. This space is also known as $$\beta\mathbb{N}$$, and is the Stone-Cech compactification of $$\mathrm{disc}\mathbb N$$.

The functor $$\Pi_0$$ is indeed the reflector of the subcategory of compact totally disconnected spaces, and you can use that to give an explicit formula for it. Since $$\mathbf 2$$ is a cogenerator for compact totally disconnected spaces, you can take $$\Pi_0(X)$$ to be the closure of the image of the canonical map $$X\to \prod_{\mathrm{hom}(X,\mathbf 2)}\mathbf{2}$$. (You may be more familiar with the similar formula for the reflector to compact Hausdorff spaces a.k.a. the Stone-Cech compactification, which is identical except with $$[0,1]$$ in place of $$\mathbf{2}$$.)

As for your question 2, I think you're going about this wrong. The functor $$\Pi_0$$ is meant to be a notion of "connected components" that makes sense in a broad context. It's not designed to always give the most sensible answer in every specific category that might have some more refined structure. In particular, this notion of "connected components" is by definition profinite, so you might think of it as a sort of profinite completion of the "true" connected components.

Finally, for your last question, suppose $$\mathsf{C}$$ is an infinitely distributive category in the sense that it has all coproducts and products and all products distribute over all coproducts: that is, $$\prod_{j\in J}\coprod_{i\in I_j} X_{ij}\cong\coprod_f\prod_{j\in J}X_{f(j)j}$$ via the natural map, where $$f$$ ranges over all functions that choose an element of $$I_j$$ for each $$j$$. Then I believe the following should be true (but have not checked the details): the functor $$\mathrm{hom}(-,\mathbf 2)$$ can be taken to land not just in the category of Boolean algebras but the category of completely distributive complete Boolean algebras (and complete homomorphisms). But this latter category is equivalent to $$\mathsf{Set}^\text{op}$$, since every such Boolean algebra is a power set and all complete morphisms between them are induced by functions of sets. So for such a category, you can get a set of connected components in an infinitary sense.

However, this does not apply to $$\mathsf{Top}$$! Indeed, the strong infinite distributive law given above fails: when the $$X_{ij}$$ are all 1-point spaces, it would say an infinite product of discrete spaces is discrete, which is false.

I don't think there's any nice construction of the reflector onto totally disconnected spaces along these lines. It does have a very straightforward construction though: given a space $$X$$, say $$x\sim y$$ if $$x$$ and $$y$$ are in the same connected component of $$X$$, and take the quotient $$X/{\sim}$$. This quotient is totally disconnected (see Does collapsing the connected components of a topological space make it totally disconnected?) and is the totally disconnected reflection of $$X$$.

• As always, thank you for the instructive answer. Will the profinite $\Pi_0$ generally be a fibration? For family fibrations the fibers are functor categories from a discrete category, so perhaps in the profinite case the fibers might also be functor categories? – Arrow Dec 12 '17 at 11:41
• @Arrow: I don't see any reason to expect that would be true in general, but I haven't thought about it deeply. – Eric Wofsey Dec 12 '17 at 18:16

Eric already basically said this, but it's worth emphasizing. This $\pi_0$ functor should be thought of as a lower categorical analogue of the etale fundamental group: it is built from finite sets (of size a power of $2$) and so can at best see the profinite completion of the "true" $\pi_0$, in the same way that the etale fundamental group can at best see the profinite completion of the "true" $\pi_1$ obtained from looking at complex points. For example, $\pi_0$ of a discrete space $X$ will always be the profinite completion $\beta X$.

Nonetheless, in some categories the "true" $\pi_0$ is already profinite; the simplest example is the category of profinite sets itself, and the example of affine schemes given in the post you linked also has this property. The point is that the category of affine schemes is naturally a pro category: since commutative rings are ind-(finitely generated commutative rings), affine schemes are pro-(affine schemes of finite type).

• I guess it is also relevant that a finite type affine scheme has only finitely many connected components. This is a nontrivial algebraic input that makes the "true" connected components of affine schemes profinite. If, for instance, you instead considered the category of affine morphisms over some base scheme which was infinite and discrete, then this would no longer be true. – Eric Wofsey Dec 12 '17 at 9:09
• @EricWofsey I don't understand this point well. What about an infinite coproduct of $\operatorname{Spec}\mathbb Z$'s? It is an affine scheme and it seems intuitive that its "true" $\pi_0$ should be the countable discrete space, instead of a profinite space. Perhaps I am missing some geometric intuition? – Arrow Dec 12 '17 at 11:28
• @Arrow: An infinite coproduct of $\operatorname{Spec}\mathbb Z$'s is not affine! It is just an infinite disjoint union of copies of $\operatorname{Spec}\mathbb Z$, which is not quasicompact and so cannot be affine. If you're instead taking the coproduct in the category of affine schemes, you get Spec of an infinite product of copies of $\mathbb{Z}$, which is a rather monstrous object and quite different from just a disjoint union of copies of $\operatorname{Spec}\mathbb Z$. – Eric Wofsey Dec 12 '17 at 18:11
• @Arrow: Right, so it depends whether you are computing the coproduct in all schemes or just in affine schemes. In all schemes, coproducts are nicely geometric and are just disjoint unions. In affine schemes, that's only true for finite coproducts and infinite coproducts are much more complicated. – Eric Wofsey Dec 12 '17 at 18:14
• A topological analogue here is that the inclusion of profinite sets into topological spaces also does not preserve infinite coproducts. – Qiaochu Yuan Dec 12 '17 at 19:25