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title: Boolean Algebras are regular and co-regular

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# Boolean Algebras are regular and co-regular

That the category BA of Boolean algebras is regular is a standard fact from universal algebra, see eg [here](https://hackmd.io/@alexhkurz/rJZq4jDfU). By duality, BA is co-regular since Stone is regular (see below). In particular, (Epi,RegMono) is a factorisation system in BA. In fact, in BA, Epi=RegEpi and Mono=RegMono (CHECK).

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I’ve been thinking in those terms too. But there is an important subtlety that I don’t fully understand. Priestley spaces are not simply 0d Nachbin spaces. Stralka gave two examples that are 0d Nachbin, but not Priestley. So the analogy is not as simple as it seems.

On the pro-completion angle: Your appendix mentions that monadicity of U may be sufficient, but that won’t help us. Our “forgetful” U is not monadic because it does not even have a left adjoint. But the situation may not so bad.

We are looking at a 2-category (Pos) that has all weighted limits, and full subcategory (FinPos) that has all finite weighted limits. So there is a unique functor U from Pro(FinPos) to Pos that commutes with the inclusions of FinPos is both categories, and preserves cofiltered limits. I have been thinking of U as a forgetful functor, but this now seems wrong. It is a lift.

Now Pro(FinPos) has all cofiltered limits by construction, and its generators have all finite limits. So Pro(FinPos) may have all weighted limits by associativity of limits (OK, I’m just improvising this — I haven’t checked details).  If those details check out, U preserves all weighted limits, so it has a right adjoint, and might be comonadic (it sort of looks promising to be). Concretely, the adjoint would send a poset X to P(X), the pro-finite poset (the Priestley space) obtained as the cofiltered diagram of all of X’s finite image quotients.

So it this what we want? "Pri is order-regular because it is Pro(A_0) for a finitely order-complete subcategory $A_0$ of an order-complete, -regular A."


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Pri is regular. I am sure there is some nicer way to see it, but essentially, the forgetful functor U to Pos creates everything one needs.

The forgetful functor creates all weighted limits.
More generally, given a jointly order monic family {f_i:X->U(Y_i)} of functions in Pos, the induced topology on X makes it a Priestley space. So one applies this to the projections of the weighted limit of a diagram $U\circ D$ to create the limit of D.
Coequalizers of kernel pairs are also created.
Basically, for f:X->Y in Pri, Im(U(f)) is compact as a subset of Y. Hence it is closed. So its induced subspace topology is also a Stone topology.  If x\nleq y in this image, then there is an upper clopen in Y separating them. The trace of this clopen in Im(U(f)) is upper clopen. So Im(f) (in Pri) is Im(U(f)) with induced topology.
Surely 1 can be formulated as a more general fact about pro-completions. In particular, the forgetful functor from Pri to Pos arises from the fact that Pos itself has all (cofiltered) limits and Pri is the pro-completion of FinPos. So U is just the lifting of the inclusion of FinPos into Pos guaranteed by the universal property of Pro(FinPos)=Pri. So (I don’t see the details yet), the creation of weighted limits by U should have to do with U being this lift.

It would be a lot to ask for 2 to be generalizable. But in the case of Priestley morphisms, images are nicely behaved. I suppose a close look at how images behave in pro-completions generally would be a help. My trouble is that morphisms in Pro(A) are really hard to think about concretely.


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On Aug 12, 2019, at 10:29, Moshier, M. Andrew <moshier@chapman.edu> wrote:

>  Stone is indeed regular, basically because the image of a continuous function is compact, hence closed, hence is Stone in its induced topology. 
Thus the coequalizer of a kernel pair is the image. 

another way of saying this is that if we factor a map between two stone spaces X-->Y in Top as X-->Z-->Y then Z is a Stone space, right? 

That can be said short and crisp: Stone is closed in Top under factorisations.

> I am now wondering whether this fact comes from pro-completion. That is, if A is regular, is Pro-A regular? Then Stone’s regularity is just due to FinSet’s regularity.

Yes, that would be nice to know.

> I’m not confident about Pri yet, but I think the same reasoning goes through.

Pri is also obtained by Pro-completion, this time of finite posets, is that right?

But we need to use the notion of order-regular category from appendix C as Pos is not regular. (BTW, every order-regular category with discrete homsets is also regular so Stone would remain a special case of Pri.)

The next question then is, whether both the spaces and the algebras being order-regular in a Stone duality is typical or exceptional. What about locales, do you know this?


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Stone is indeed regular, basically because the image of a continuous function is compact, hence closed, hence is Stone in its induced topology. 
Thus the coequalizer of a kernel pair is the image. 

I am now wondering whether this fact comes from pro-completion. That is, if A is regular, is Pro-A regular? Then Stone’s regularity is just due to FinSet’s regularity.

I’m not confident about Pri yet, but I think the same reasoning goes through.

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### Compact Hausdorff spaces are regular

A classic result from Manes states that compact Hausdorff spaces are the algebras for the ultrafilter monad on Set. Hence, [by the above](https://hackmd.io/@alexhkurz/rJZq4jDfU), they are regular. I add some comments to this.

1. It is easy to show that if $f:X\to Y$ is a morphism of algebras for a regular monad and $X\to Z\to Y$ is a regular factorisation in the base category then there is a unique algebra structure on $Z$ lifting $X\to Z\to Y$ to algebras.
2. In topological spaces, the above is not true since there are, in general, many topologies on $Z$ making the factor-maps continuous. 
3. But compact Hausdorff topologies cannot be made finer or coarser without loosing either compactness or Hausdorffness. It follows that on $Z$ the quotient topology equals the subspace topology.
