# Inverse of power-set functor?

In his answer to this Q: How to interpret $1 \to 0$ in $\mathbf {Set^{op}}$, and $\mathbf {Set^{op}}$ itself? Zhen Lin proposed that $\mathbf {Set^{op}}$ is naturally equivalent to the category of complete atomic boolean algebras via the contravariant power set functor.

As a step in understanding this equivalence, what's the inverse of the contravariant set functor?

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The set of atoms? –  Hurkyl Feb 26 '13 at 5:51
Yes, some kind of projection. But what exactly? What are atoms in sets, elements? –  alancalvitti Feb 26 '13 at 5:52
I don't understand the question in your comment. –  Qiaochu Yuan Feb 26 '13 at 6:20
Atoms in a Boolean algebra are elements that minimal non-zero, that is, $a$ is an atom if $a\neq 0$ and if $0\leq b\leq a$ then either $b=0$ or $b=a$. The atoms in the power set thought of as a Boolean algebra are exactly the singletons, hence Hurkyl's description of the inverse. –  Shawn Henry Feb 26 '13 at 7:11
For the purposes of this question atom definitely means an element that is minimal in the sub-poset excluding the bottom element. The general definition is harder, yes. –  Zhen Lin Feb 26 '13 at 13:15

The equivalence between $\mathbf{Set}^{op}$ and $\mathbf{CABA}$ is induced by an adjointness between $\mathbf{Set}^{op}$ and $\mathbf{CBA}$, where the latter is the category of complete Boolean algebras (i.e. without any assumption about atoms).

First a recall the general principle:

Suppose $F: \mathcal{X} \to \mathcal{A}$ is left adjoint to $G: \mathcal{A} \to \mathcal{X}$ with unit $\eta: Id_{\mathcal{X}} \to GF$ and counit $\varepsilon: FG \to Id_{\mathcal{A}}$.

Then the two full subcategories $\{ X \in \mathcal{X} \mid \eta_X \text{ is an isomorphism} \}$ and $\{ A \in \mathcal{A} \mid \varepsilon_A \text{ is an isomorphism} \}$ are equivalent and the equivalences are given by the restrictions of F and G. (this follows from the triangular equalities).

So now let's proceed to construct such adjoint functors. I will do this with $\mathcal{X}=\mathbf{Set}$ and $\mathcal{A}=\mathbf{CBA}^{op}$, but it is of course only a matter of taste where to put the $op$. Anyway I will draw the maps in $\mathbf{Set}$ and $\mathbf{CBA}$. The crucial ingredient is the two element set $2 = \{ 0,1 \}$, which is of course a set, but also a complete Boolean algebra under the ordering with $0 < 1$.

Given a set $X$, set $P(X) = \mathbf{Set}(X,2)$. Given a complete boolean algebra $B$, set $Q(B) = \mathbf{CBA}(B,2)$.

On morphism $P$ and $Q$ act via precompositon, i.e. for $f: X \to Y$ the map $P(f): P(Y) \to P(X)$ takes $h \in P(Y)$ to $h\circ f \in P(X)$. Similar for $Q$.

Then that gives two functors $P: \mathbf{Set} \to \mathbf{CBA}^{op}$ and $Q: \mathbf{CBA}^{op} \to \mathbf{Set}$. This is clear for $Q$. For $P$ observe that the power set $P(X)$ is indeed a complete Boolean algebra, regardless of whether you view its elements as subsets of $X$ or as characteristic maps from $X$ to $2$.

For a set $X$ define $\eta_X : X \to Q(P(X)$ via $\eta_X(x)\sigma := \sigma(x)$ for all $x\in X$, $\sigma\in P(X)$. This gives the unit of the adjunction: given $f: X \to Q(B)$, the corresponding map $\hat{f}: B \to P(X)$ must satisfy

$$\matrix{ f(x)b = 1 &\iff& (\eta_X(x)\circ \hat{f})b = 1 \cr &\iff& \eta_X(x)(\hat{f}b) = 1 \cr &\iff& \hat{f}(b)x = 1 \cr }$$ and this can be taken to define $\hat{f}$ via $\hat{f}(b)x := f(x)b$.

In particular, for $X=Q(B)$ and $f=id_{Q(B)}$ the counit $\varepsilon_B: B \to P(Q(B))$ is given by $\varepsilon_B(b)h := h(b)$ for all $b \in B$, $h\in Q(X)$.

For a complete Boolean algebra $B$, write $at(B)$ for the set of its atoms and given $b\in B$ write $at(B/b)$ for the atoms below $b$. No matter what precise definition of 'atom' is used, the only facts to recall are that the meet of two different atoms is the bottom element and that if $B$ is atomic then every $b\in B$ satisfies $b = \bigvee(at(B/b))$.

Now the following steps pin down the full subcategories:

(1) If $B$ is atomic, then every $h\in Q(B)$ is completely determined by its restriction to $at(B)$. Moreover the preimage $h^{-1}(1)$ contains exactly one atom $a_h$ and for each $b \in B$ we have the equivalence $h(b)=1 \iff a_h\leq b$. Conversely, given an $a\in at(B)$, this equivalence defines an element of $Q(B)$. Therefore the assignment $h \mapsto a_h$ gives a bijection between $Q(B)$ and $at(B)$.

(2) $P(X)$ is always atomic and the atoms are the singleton subsets of $X$. From (1) we have that for every $h \in Q(P(X))$ there is a unique $x\in X$ such that $h = \eta_X(x)$. In particular $\eta_X$ is always an isomorphism.

(3) If $\varepsilon_B: B\to P(Q(B))$ is an isomorphism then $B$ must be atomic because of (2). Conversely suppose $B$ is atomic. Given $\sigma \in P(Q(B))$, let $S = \sigma^{-1}(1) \subseteq Q(B)$ be the corresponding subset of morphisms from $B$ to $2$. Set $b = \bigvee \{a_s\mid s \in S \}$. Then $$\matrix{ \varepsilon_B(b)h = 1 &\iff& h(b) = 1 \cr &\iff& \exists s\in S: h(a_s) = 1 \cr &\iff& \exists s\in S: a_h \leq a_s \cr &\iff& \exists s\in S: a_h = a_s \cr &\iff& h \in S \cr &\iff& \sigma(h) = 1 \cr }$$ and this $b$ is the unique $b\in B$ with $\varepsilon_B(b)=\sigma$. Therefore $\varepsilon_B$ is an isomorphism iff $B$ is atomic.

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Thanks you for the thorough answer. I'm accepting it even if I don't understand many if not most of the details. But I'm interested in this subject so surely I'll ask additional related Q's based on this. –  alancalvitti Mar 16 '13 at 1:38