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Let $\mathbb{B}$ be a boolean algebra. Then we know that $\mathbb{B}$ is isomorphic to an algebra of sets, viz. ${h:\mathbb{B}\longrightarrow \mathrm{Clopen}(\mathrm{Ult}(\mathbb{B}))\subseteq\wp(\mathrm{Ult}(\mathbb{B}))}$ via $h(b)=U_{b}=\{\mathfrak{F}\in\mathrm{Ult}(\mathbb{B})\mid b\in\mathfrak{F}\}$.

I want to know, whether there is a powerset algebra $\wp(X)$ and an epimorphism, ${g:\wp(X)\longrightarrow\mathbb{B}}$. This is equivalent to the question: is there a powerset algebra $\wp(X)$ and an ideal $\mathcal{I}\subseteq \wp(X)$, such that $\mathbb{B}\cong\wp(X)/\mathcal{I}$.

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. Is there an equivalent theorem for complete boolean algebras?

UPDATE: Answered by @Keith Kearnes here [https://mathoverflow.net/questions/306578/is-every-complete-boolean-algebra-isomorphic-to-the-quotient-of-a-powerset-algeb].

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    $\begingroup$ Simultaneous crosspost to MathOF: mathoverflow.net/questions/306578/… $\endgroup$
    – YCor
    Commented Jul 22, 2018 at 9:36
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    $\begingroup$ 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). $\endgroup$
    – YCor
    Commented Jul 22, 2018 at 9:36
  • $\begingroup$ @YCor on which site is it better to post? $\endgroup$
    – Thomas
    Commented Jul 22, 2018 at 9:44
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    $\begingroup$ Not on both. I'd say, on MathSE, and then if no answers after a few days, on MathOF, with links in both directions (and taking into account relevant comments to the MathSE post, if applicable). $\endgroup$
    – YCor
    Commented Jul 22, 2018 at 10:08

1 Answer 1

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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).

Practically the same argument shows that the same algebra is also not a quotient of any countably complete one.

Another argument uses the fact that images of atoms under surjective homomorphisms are either 0 or atoms themselves.

Since there is some controversy in comments, let me try to supply an argument for that. Take $\pi:P\twoheadrightarrow B$ and suppose $a\in P$ is an atom. Take any $x\leqslant\pi(a)$. Choose $y\in P$ with $\pi(y)=x$. Then also $\pi(a\land y)=x$ (since $\pi(a\land y)=\pi(a)\land\pi(y)=\pi(a)\land x=x$).

Now since $a$ is an atom, either $a\land y=a$ or $a\land y=0$, so that either $x=\pi(a)$ or $x=0$.

However to the contrary of what I stated, this does not imply that an atomless algebra cannot be quotient of a powerset; thanks to JDH for noticing that.

In fact for complete algebras the answer seems to be positive. In fact there is a theorem by Sikorski that injective objects in Boolean algebras are precisely the complete ones. Thus any embedding $i:B\hookrightarrow P$ of a complete Boolean algebra $B$ as a subalgebra of any Boolean algebra $P$ admits a retraction $\pi:P\twoheadrightarrow B$ (i. e. $\pi i$ is identity). Thus in fact any complete Boolean algebra is a retract of a powerset.

A proof can be found on planetmath. Note that it uses AC; that planetmath page claims that it is not known whether AC is unavoidable.

I believe also that not every quotient of a powerset is complete, i. e. retract - e. g. I think quotient by the ideal of finite subsets is not complete.

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    $\begingroup$ In the final remarks of your answer, you seem to suggest that $P(X)/I$ can not be an atomless complete Boolean algebra. But this is not true, since for example the instance of $P(\omega_1)$ modulo the non-stationary ideal NS is studied in set theory, and under certain large cardinal assumptions this can be complete (and it is atomless). Have I misunderstood something? $\endgroup$
    – JDH
    Commented Jul 22, 2018 at 11:07
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    $\begingroup$ @მამუკაჯიბლაძე your argument holds in terms of complete Boolean algebras (cBA). That is, there is no cBA-epimorphism $h:\wp(X)\longrightarrow\mathbb{B}$, if $\mathbb{B}$ is atomless, since then necessarily $h(\{x\})=0$ for all $x\in X$ and since $h$ is a cBA-morphism $1=h(X)=\bigvee\{h(x):x\in X\}=0$, which is a contradiction. However this does not exclude morphisms in the category BA, since the last argument cannot be executed for general BA-morphisms. Thus the question remains: is every (complete) BA isomorphic to a quotient of a powerset algebra within the category BA. $\endgroup$
    – Thomas
    Commented Jul 22, 2018 at 11:59
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    $\begingroup$ I wasn't objecting to your claims about images of atoms, which are fine. I am objecting to your claims that P(X)/I cannot be atomless and complete, since this isn't true. $\endgroup$
    – JDH
    Commented Jul 22, 2018 at 12:06
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    $\begingroup$ @JDH Ok hopefully this is now correct. Many thanks for pointing it out! $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 22, 2018 at 12:48
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    $\begingroup$ @Thomas The paper "Extremally disconnected spaces, subspaces and retracts" by Dow and Vermeer (Top. Appl. 50 (1993), 263-282) contains a characterization of compact subspaces of extremally disconnected spaces, so dually could give a characterization of Boolean algebra quotients of powersets $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 22, 2018 at 14:54

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