# Does the category of Boolean algebras imbed in $\operatorname{Set}$

While trying to come up with some examples of functors, I realised that any function $$f:X\to Y$$ induces a function $$P_f: \mathbb{P}(X) \to \mathbb{P}(Y)$$ in a natural way, simply define $$f_p(A) = f(A)$$ for subsets $$A\subseteq X$$. Furthermore, every powerset of a set $$X$$ is a Boolean Algebra with join and meet given by intersection and union respectively. I believe (but am not certain) that every Boolean Algebra is also isomorphic to the powerset of some set. The next realisation was that the function $$f_p: \mathbb{P}(X)\to \mathbb{P}(Y)$$ is a homomorphism of Boolean Algebra's iff the function $$f$$ is injective. Clearly if $$f$$ is injective, then for any $$A,B\subseteq X$$ we have $$f_p(A\cap B) = f_p(A)\cap f_p(B)$$, and joins are preserved even without injectivity. Conversely, if $$f_p$$ is a homomorphism of Boolean Algebra's, then for any two distinct singletons we have $$f_p(\{x\}\cap\{y\}) = f_p(\{x\})\cap f_p(\{y\}) = \{f(x)\}\cap \{f(y)\} = \emptyset$$

Hence, $$f$$ is injective.

So we can make a sort of "partial functor" (I am not sure this is a thing) that sends objects and monomorphisms from $$\operatorname{Set}$$ to the category of Boolean Algebra's, and if we compose this with the forgetful functor from Boolean Algebra's back into sets, we have a natural way to categorically represent the powerset of a set.

Questions: I am trying to self study Category Theory, and do not feel confident with some of the concepts above, so my questions are

1. Is my thinking correct?

2. Is this a known construction, or is it a specific example of a class of constructions? If so, what are other examples?

Additionally, if I used any terms incorrectly, or unconventionally for the field, I would appreciate feedback regarding this!

• The title does not fit to the question. I suggest to change it. Commented Dec 26, 2023 at 17:40
• @MartinBrandenburg, I knew that it wasn't the best title, but I didn't know what else to call this, and I thought that the inverse of the functor that I described would be an imbedding, but seeing as I was wrong about every boolean algebra being a powerset, this is not the case. Is there a more descriptive name for the construction that you can think of? Commented Dec 26, 2023 at 17:47
• @Carlyle: The category of boolean algebras is dual to the category of compact totally disconnected Hausdorff spaces. See Stone's representation theorem for Boolean algebras. Commented Dec 26, 2023 at 17:49
• @Carlyle I mean the direction in the title is not the same as in the question. You go from sets to Boolean algebras. Commented Dec 26, 2023 at 20:59

Morphisms of Boolean algebras also need to preserve the top element. Here it means that the map needs to satisfy $$f_*(X)=Y$$ which means it is surjective. Combined with your observation this means that $$f$$ is bijective. So this is not very interesting, but yeah we have a functor from the category of sets with bijections to the category of Boolean algebras. The image consists of all complete atomic Boolean algebras (not every Boolean algebra is a power set).
If you want all maps, you need to dualize. If $$f : X \to Y$$ is any map, it induces a morphism of Boolean algebras $$f^* : P(Y) \to P(X)$$. This actually induces an equivalence of categories $$\mathbf{Set}^{\mathrm{op}} \cong \mathbf{CABA},$$ where $$\mathbf{CABA}$$ denotes the category of $$\mathbf{c}$$omplete $$\mathbf{a}$$tomic $$\mathbf{B}$$oolean $$\mathbf{a}$$lgebras. Interestingly, this provides a proof that $$\mathbf{Set}^{\mathrm{op}}$$ is monadic over $$\mathbf{Set}$$.
• Interesting, in this dualisation, I presume we define $f^{*}(A) = f^{-1}(A)$ ? Since then we have subjectivity, and preservation of joins and meets Commented Dec 26, 2023 at 17:45
To see that not all Boolean algebras are isomorphic to a power set, consider that because there is an infinite Boolean algebra, it follows by the Lowenheim-Skolem theorem that there is a countably infinite Boolean algebra (alternately, construct the free Boolean algebra on countably infinitely many elements, which will be countable). However, no power set is countably infinite. For if $$S$$ is a finite set, then $$P(S)$$ is also finite. And if $$S$$ is infinite, then $$|\mathbb{N}| \leq |S|$$, and therefore $$|\mathbb{N}| < |P(\mathbb{N})| \leq |P(S)|$$.