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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.

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    $\begingroup$ 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. $\endgroup$ May 16 at 19:16
  • $\begingroup$ 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 ? $\endgroup$ May 16 at 20:33
  • $\begingroup$ @SimonHenry Well this is the formal answer (valid for any infinitary Lawvere theory), but I would like to have a description with internal data. $\endgroup$ May 16 at 20:43
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(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 $$\bigvee_{i \in I} \bigwedge_{j \in J} v_{i,j} = \bigwedge_{s:I \to J} \bigvee_{i \in I} v_{i,s(i)}.$$

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