Let $\mathsf{Stone}$ denote the category of Stone spaces (compact, totally disconnected Hausdorff spaces) and continuous maps. The forgetful functor $U : \mathsf{Stone} \to \mathsf{Set}$ has a left adjoint $F : \mathsf{Set} \to \mathsf{Stone}$ which maps a set $X$ to the space $F(X)$ of ultrafilters on $X$ (i.e. the Stone-Čech compactification of the discrete space $X$). 


**Question.** Is $F$ comonadic?


I have tried to use Beck's monadicity criterion in its dualized form: $F$ is comonadic iff $F$ is a left adjoint ($\checkmark$), $F$ is conservative ($\checkmark$) and $\mathsf{Set}$ has ($\checkmark$) and $F$ preserves equalizers of $F$-split pairs. Only the last point is unclear to me. I hope that it is true, and that actually $F$ preserves much more equalizers.


Here is some background: In his (fantastic!) paper on [codensity](http://www.tac.mta.ca/tac/volumes/28/13/28-13abs.html), Tom Leinster mentions an interesting monad on the category of commutative rings $\mathsf{CRing}$ defined by $T(A) = \prod\_{\mathfrak{p} \in \mathrm{Spec}(A)} Q(A/\mathfrak{p})$ and [asks](https://golem.ph.utexas.edu/category/2020/01/codensity_monads.html) for a description for its category of algebras. Each $T(A)$ is von Neumann regular, so I thought it would be good to first look at the restriction to the category of Boolean rings and determine the algebras for the induced monad. By Stone duality, this is monad is dual to the comonad $FU : \mathsf{Stone} \to \mathsf{Stone}$ from above, and comonadicity of $F$ would imply that the boolean $T$-algebras are just products of copies of $\mathbb{F}\_2$.



Ultrafilter comonad on the category of Stone spaces
I believe the answer is *yes*: via the ultrafilter functor, $Set$ is comonadic over $Stone$ (and also over $CpctHaus$). Instead of the plain comonadicity theorem, let's use the crude comonadicity theorem, which says the following:



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> Let $F: C \to D$ be a conservative left adjoint such that $C$ has and $F$ preserves equalizers of coreflexive pairs. Then $F$ is comonadic.
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> 


So consider a coreflexive pair in $Set$, i.e. we have $f,g: A \rightrightarrows B$ which are both sections of a map $p: B \to A$. Let $E = \{a \in A \mid f(a) = g(a)\}$ be the equalizer. Let $\mathcal F$ be an ultrafilter on $A$ such that $f\_\ast(\mathcal F) = g\_\ast(\mathcal F)$, i.e.



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> $(\ast)$ For all $V \subseteq B$ we have $f^{-1}(V) \in \mathcal F \Leftrightarrow g^{-1}(V) \in \mathcal F$.
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It will suffice to show that $E \in \mathcal F$. Suppose not. Then $(A \setminus E) \in \mathcal F$. Let $V \subseteq B$ be the set $V = \{f(a) \mid a \in (A \setminus E)\}$. Then we have $f^{-1}(V) = (A \setminus E) \in \mathcal F$, but $g^{-1}(V) = \emptyset \not \in \mathcal F$, contradicting $(\ast)$. Thus we must have $E \in \mathcal F$ after all, and so the ultrafilter functor preserves this coreflexive equalizer.


Note that the same proof shows that the ultrafilter functor $\beta: Set \to \mathcal C$ preserves coreflexive equalizers for any of $\mathcal C \in \{ Stone, CpctHaus, Set\}$. In particular, the ultrafilter functor exhibits $Set$ as comonadic over the category of compact Hausdorff spaces as well as over the category of Stone spaces.


