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It is well-known that the category CH of compact Hausdorff spaces has a strong categorical flavor (e.g. Properties of the category of compact Hausdorff spaces, which includes Manes' theorem asserting that CH is the category of algebras over the ultrafilter monad).

On the other hand CH contains the full subcategory ProFin of profinite spaces, which in turn admits a purely categorical interpretation as the Pro-completion of the category of finite sets.

Is there a way in which CH arises by ProFin by freely adding some sort of colimits?

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    $\begingroup$ The ultrafilter monad can also be defined “purely categorically theoretically”. Look up codensity monads. $\endgroup$
    – Zhen Lin
    Commented Feb 28, 2023 at 12:56
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    $\begingroup$ See this earlier question. I intend to expand my comments to an answer but I've been quite busy lately. In short, the category of compact Hausdorff spaces is the free $\aleph_1$-cofiltered completion of the category of compact metrizable spaces (or closed subsets of $[0,1]^{\mathbb{N}}$ if you need a small rather than just essentially small category when you take a completion). This is best viewed under Gelfand duality - a commutative unital C$^*$-algebra is an $\aleph_1$-filtered colimit of separable commutative unital C$^*$-algebras. $\endgroup$
    – Robert Furber
    Commented Feb 28, 2023 at 15:50
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    $\begingroup$ You're never going to get all compact Hausdorff spaces as a limit of finite spaces because Stone spaces are closed under limits (because the product of Stone spaces is Stone and a closed subspace of a Stone space is Stone). $\endgroup$
    – Robert Furber
    Commented Feb 28, 2023 at 16:03
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    $\begingroup$ @Jakob Your specific question is a lot narrower than the title question… but in some sense the answer must be yes, if the category of compact Hausdorff spaces can be embedded in some category of sheaves over profinite sets. The colimits would not be free (corresponding to the non-trivial Grothendieck topology) and it is not obvious to me what diagram shapes would be needed. $\endgroup$
    – Zhen Lin
    Commented Mar 1, 2023 at 2:16
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    $\begingroup$ @მამუკაჯიბლაძე : compact Hausdorff spaces indeed embed fully faithfully in the free sifted-cocompletion of extremally disconnected sets, which is the category of condensed sets. $\endgroup$
    – Maxime Ramzi
    Commented Mar 1, 2023 at 12:06

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The category of compact Hausdorff spaces is the pretopos completion of the category of profinite sets. It means that it is obtained by adding freely all quotients by equivalence relations, which are colimits. It is mentioned and proved here: A characterisation of the category of compact Hausdorff spaces (Marra, Reggio - 2020)

The proof uses the characterization of the article but it can be seen more directly. More generally, if $\mathbf{C}$ is a pretopos and if $\mathbf{D} ⊆ \mathbf{C}$ is a full subcategory stable under subobjects and finite limits, and if every object of $\mathbf{C}$ is a quotient of an object of $\mathbf{D}$, then $\mathbf{C}$ is the exact completion of $\mathbf{D}$.

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    $\begingroup$ To be precise. $KH$ is the ex/reg completion of the category of profinite sets. So it is the pretopos completion only whem the category of profinite set is considered as a regular and extensive category. (i.e. a functor fron profinite set to a pretopos extent to a pretopos morphism on $KH$ if it is a regular and coproduct preserving functor...) $\endgroup$
    – Simon Henry
    Commented Nov 12 at 18:49

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