# Is every complete Boolean algebra isomorphic to the quotient of a powerset algebra?

Is every complete Boolean algebra isomorphic to a quotient, as a Boolean algebra, of some powerset algebra $\wp(X)$?

It is not true for arbitrary Boolean algebras, see the comments, or see my MathSE question.

I am also aware of the Loomis-Sikorski Theorem: Every $\sigma$-complete Boolean algebra is isomorphic to the a quotient $\mathbf{F}/\mathcal{I}$, where $\mathbf{F}$ is a $σ$-field of sets and $\mathcal{I}\subseteq\mathbf{F}$ is a $\sigma$-ideal. So this can be viewed as asking whether there is an equivalent theorem for complete Boolean algebras.

• This is cross-posted to Math.SE and indeed more appropriate there, so I move my answer to a comment and vote to close. A free algebra on an infinite set is not complete (exercise). If it would be a quotient of a powerset it would be a retract of it (exercise), and a retract of a complete algebra is complete (exercise). – მამუკა ჯიბლაძე Jul 22 '18 at 9:24
• The Cantor set is not homeomorphic to a closed subset of a Stone-Cech compactification of a discrete set (since any converging sequence therein is eventually constant). By Stone duality, this reflects in the fact that the corresponding Boolean algebra is not quotient of a power set (@მამუკაჯიბლაძე 's argument yields something more general but I wanted to point out this topological interpretation). – YCor Jul 22 '18 at 9:34
• Link to MathSE post: math.stackexchange.com/questions/2859214/… – YCor Jul 22 '18 at 9:36
• @მამუკაჯიბლაძე That’s fine—then the question remains. Are all complete Boolean algebras isomorphic to quotients of powerset algebras? – Thomas Jul 22 '18 at 9:43
• @Thomas: ${\mathcal P}(\omega)/\textrm{fin}$ is an atomless quotient of a power set algebra. – Keith Kearnes Jul 22 '18 at 12:18

This variation of the question comes from the comments on the original question.

The question is whether all (complete) BAs are isomorphic in the category BA to a quotient of a powerset algebra.

The Sikorski extension theorem guarantees that every complete BA is a quotient of a power set algebra.

The Theorem. (Sikorski) Let $A$ be a subalgebra of a Boolean algebra $B$, and let $f:A\to C$ be a homomorphism from $A$ to a complete Boolean algebra $C$. Then $f$ can be extended to a homomorphism $\widehat{f}:B\to C$.

Application. Let $A=C$ be a complete Boolean algebra, and represent it as a subalgebra of a power set BA $B={\mathcal P}(X)$. (Use Stone duality or Birkhoff's subdirect representation theorem for this.) The Sikorski theorem guarantees that that the identity function ${\sf id}\colon A=C\to C$ can be extended to a homomorphism $\widehat{\sf{id}}: B={\mathcal P}(X)\to C$. Since $\widehat{\sf id}$ extends the identity function, it is surjective. \\\

This argument shows that any complete BA is in fact a retract of a power set BA. The converse is easily seen to be true, so the class of complete BA's is exactly the class of retracts of power set BA's.

Also, from the comments on the original question:

• There are incomplete BA's that are quotients of power set BA's. E.g. ${\mathcal P}(\omega)/\textrm{fin}$.
• Infinite free BA's are not quotients of power set algebras.

• Thank you! This is what I’m looking for. I have now found a side reference to this in my book on set theory. – Thomas Jul 22 '18 at 14:38