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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!

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  • $\begingroup$ The title does not fit to the question. I suggest to change it. $\endgroup$ Commented Dec 26, 2023 at 17:40
  • $\begingroup$ @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? $\endgroup$
    – Carlyle
    Commented Dec 26, 2023 at 17:47
  • $\begingroup$ @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. $\endgroup$
    – Chad K
    Commented Dec 26, 2023 at 17:49
  • $\begingroup$ @Carlyle I mean the direction in the title is not the same as in the question. You go from sets to Boolean algebras. $\endgroup$ Commented Dec 26, 2023 at 20:59

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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 $\mathbf{Set}$.

You have asked for generalizations, and sure there are, but I suggest that you make this question more specific.

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  • $\begingroup$ Interesting, in this dualisation, I presume we define $f^{*}(A) = f^{-1}(A)$ ? Since then we have subjectivity, and preservation of joins and meets $\endgroup$
    – Carlyle
    Commented Dec 26, 2023 at 17:45
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    $\begingroup$ Yes preimage. Not sure what you mean with subjective. $\endgroup$ Commented Dec 26, 2023 at 20:59
  • $\begingroup$ Unfortunately I cannot edit my comment, hence a bijection is injective and subjective $\endgroup$
    – Carlyle
    Commented Dec 26, 2023 at 21:03
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    $\begingroup$ Not clear what you are saying. The preimage operator is defined for every map, regardless of being injective or surjective. en.wikipedia.org/wiki/… $\endgroup$ Commented Dec 26, 2023 at 21:05
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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 $|\mathbb{N}| < |P(\mathbb{N})| \leq |P(S)|$.

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