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](https://math.stackexchange.com/questions/2859214/is-every-boolean-algebra-isomorphic-to-the-quotient-of-a-powerset-algebra).




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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.



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




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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. 




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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.



