# What are internal complete atomic boolean algebras, intuitively?

The category of complete atomic boolean algebras $$\mathbf{CABA}$$ is equivalent to $$\mathbf{Set}^{\mathrm{op}}$$ via $$\mathbf{Set}^{\mathrm{op}} \to \mathbf{CABA}, ~ X \mapsto (P(X),\bigcup,\bigcap).$$ The inverse takes the atoms. The functor $$P : \mathbf{Set}^{\mathrm{op}} \to \mathbf{Set}$$ is monadic, its left adjoint is $$P^{\mathrm{op}}$$. From this we deduce that $$\mathbf{CABA}$$ is monadic over $$\mathbf{Set}$$, and that the free CABA on a set $$X$$ is $$(P(P(X)),\bigcup,\bigcap)$$ with the inclusion $$\mathrm{prin} : X \to P(P(X)),~ x \mapsto \{A : x \in A\}.$$ One can also work out the following formula which expresses every element $$S \in P(P(X))$$ in terms of the generators: $$S = \bigvee_{A \in S} \left(\bigwedge_{x \in A} \mathrm{prin}(x) \wedge \bigwedge_{x \notin A} \neg \mathrm{prin}(x)\right)$$ Anyway, since $$\mathbf{CABA}$$ is monadic over $$\mathbf{Set}$$, we can define$$^{(1)}$$ for any complete category $$\mathcal{C}$$ (powers are sufficient, actually) the category of internal CABAs $$\mathbf{CABA}(\mathcal{C})$$.

Question. What is an intuitive way of thinking about internal CABA structures on an object $$B \in \mathcal{C}$$? More specifically, is there any description which (A) makes the forgetful functor $$\mathbf{CABA}(\mathcal{C}) \to \mathbf{CBA}(\mathcal{C})$$ to complete boolean algebras explicit (I hope that there is no problem with the non-existence of free CBAs), and (B) describes the image of this functor, i.e. saying what "being atomic" actually means for an internal complete boolean algebra?

Currently I only know the following descriptions, which do not answer the question so far.

1. We have $$\mathbf{CABA}(\mathcal{C}) \simeq \mathrm{Hom}_{\mathrm{cont.}}(\mathbf{Set},\mathcal{C})$$. An internal CABA structure on $$B$$ is a continuous functor $$F : \mathbf{Set} \to \mathcal{C}$$ with $$F(2)=B$$. This is very abstract, though.

2. An internal CABA structure on $$B$$ consists of natural maps $$P(P(X)) \to \hom(B^X,B)$$ which satisfy two axioms (more concisely, monad maps from $$P \circ P^{\mathrm{op}}$$ to the double dualization monad of $$B$$.) The unit axiom is just that $$\mathrm{prin}(x)$$ gets mapped to the projection $$p_x$$, but the other axiom looks awful. Is there any way we can simplify this? And how are the union, intersection and complement operators on $$B$$ derived from this?

Any references to the literature dealing with $$\mathbf{CABA}(\mathcal{C})$$ are appreciated as well.

$$^{(1)}$$If $$T$$ is a monad on $$\mathbf{Set}$$, a $$T$$-algebra structure on $$B \in \mathbf{Set}$$ can be described with a monad map $$T(X) \to \hom(B^X,B)$$. This description can therefore be generalized to objects $$B \in \mathcal{C}$$ of a category $$\mathcal{C}$$ with powers. So we actually can define $$\mathrm{Alg}_{\mathcal{C}}(T)$$, not just $$\mathrm{Alg}(T)$$. This can also be seen via the equivalence between monads and infinitary Lawvere theories.

Edit. Here is an idea: The universal property of $$P(P(X))$$ above implies that for every CABA $$B$$ and every map $$f : X \to B$$ the map $$\tilde{f} : P(P(X)) \to B$$ defined by

$$\tilde{f}(S) := \bigvee_{A \in S} \left(\bigwedge_{x \in A} f(x) \wedge \bigwedge_{x \notin A} \neg f(x)\right)$$

is a homomorphism of boolean algebras. It is clearly compatible with joins, so the statement reduces to the compatibility of complements. That is, for all $$S \in P(P(X))$$ $$\bigvee_{A \notin S} \left(\bigwedge_{x \in A} f(x) \wedge \bigwedge_{x \notin A} \neg f(x)\right) = \bigwedge_{A \in S} \left(\bigvee_{x \in A} \neg f(x) \vee \bigvee_{x \notin A} f(x)\right)$$ This equation can be written down for internal complete boolean algebras as well (replace $$f$$ by a generalized element of $$B^X$$), and it seems to be a necessary condition for being atomic.

• If $\mathcal{C}$ is a Grothendieck topos it seems that the opposite of these "internal CABA" is the category of Cosheaves of sets on $\mathcal{C}$. Cosheaves of sets are a bit weird sometimes, and cosheaves of abelian group have been studied much work, but they do some time appear in the litterature. May 16, 2021 at 19:16
• Is the following, somehow trivial, characterization an answer to your question: a Boolean algebra object B in C is a CBA if $Hom(X,B)$ is a CBA and $Hom(X,B) \to Hom(Y,B)$ is a CBA morphism for all $X$ and all $f:Y \to X$ and it is a CABA if in addition $Hom(X,B)$ is a CABA for all X ? if not, why ? May 16, 2021 at 20:33
• @SimonHenry Well this is the formal answer (valid for any infinitary Lawvere theory), but I would like to have a description with internal data. May 16, 2021 at 20:43

(A) Here is how to get an internal CBA-structure from an internal CABA-structure. A reference is Formula 1.5.22 in E. Manes, Algebraic theories.

Notice that for $$f : X \to Y$$ the induced map $$\tilde{f} : P(P(X)) \to P(P(Y))$$ is $$\tilde{f}(S) = \{A \in P(Y) : f^*(A) \in S\}$$.

Assume $$B \in \mathcal{C}$$ and we are given natural maps $$\alpha_X : P(P(X)) \to \hom(B^X,B)$$.

If $$X$$ is an empty set, then $$P(P(X)) = \{\emptyset,\{X\}\}$$, so we define $$0 := \alpha_{\emptyset}(\emptyset) \in \hom(B^0,B)$$ and $$1 := \alpha_{\emptyset}(\{X\}) \in \hom(B^0,B)$$.

If $$X$$ is a set with two elements, say $$X=\{u,v\}$$, then $$\eta(u) \vee \eta(v) \in P(P(X))$$ is actually $$\{\{u\},\{v\},\{u,v\}\}$$, and its image under $$\alpha$$ is an operation $$\vee : B^2 \to B$$.

More generally, If $$X$$ is any set, then $$\bigvee_{x \in X} \eta(x) \in P(P(X))$$ is $$\{A \in P(X) : A \neq \emptyset\}$$, and its image is an operation $$\bigvee : B^X \to B$$.

Likewise, if $$X$$ is any set, then $$\bigwedge_{x \in X} \eta(x) \in P(P(X))$$ is $$\{X\}$$, and its image is an operation $$\bigwedge : B^X \to B$$.

If $$X$$ is a set with just one element, say $$u$$, then $$\neg \eta(u) \in P(P(X))$$ is $$\{\emptyset\}$$, and $$\alpha$$ maps this to an operation $$\neg : B \to B$$.

If $$\alpha$$ is a monad morphism, then $$(B,\bigvee,\bigwedge,\neg)$$ is an internal complete boolean algebra.

(B) This is answered by Proposition VII.1.16 in Johnstone, Stone spaces: A complete boolean algebra is atomic iff it is completely distributive, i.e. we have $$\bigvee_{i \in I} \bigwedge_{j \in J} v_{i,j} = \bigwedge_{s:I \to J} \bigvee_{i \in I} v_{i,s(i)}.$$