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.

  • 2
    $\begingroup$ 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, 2021 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, 2021 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, 2021 at 20:43

1 Answer 1


(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 $$\bigvee_{i \in I} \bigwedge_{j \in J} v_{i,j} = \bigwedge_{s:I \to J} \bigvee_{i \in I} v_{i,s(i)}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.