Existence of surjective homomorphism between Boolean algebras $\Lambda\subset\mathscr P(\mathscr B)\to\mathscr B$ (in ZF)

Question

I am trying to prove the following theorem, due to Tarski according to W. A. J. Luxemburg on Reduced powers of the real number system and equivalents of the Hahn-Banach extension theorem:

Given a Boolean algebra $\mathscr B$, there is a subalgebra $\Lambda$ of the Boolean algebra $\mathscr P(\mathscr B)$ and a surjective homomorphism $h: \Lambda \to \mathscr B$.

I need it in order to prove that $\Lambda/\ker h \cong \mathscr B $, but have no proper experience on Boolean algebras, have not found any reference, and trying to prove it have some doubts.

It is clear that Tarski proved it using only Zermelo-Fraenkel's axioms, and I need not to depend on any extra axiom such as AC or BPI (that excludes the possibility of using Stone's representation theorem).

Answer 1

For $x\in\mathscr{B}$, let $I(x)=\{y\in\mathscr{B}:y\leq x\}$. Let $\Lambda$ be the subalgebra of $\mathscr{P}(\mathscr{B})$ generated by the sets $I(x)$ for all $x\in\mathscr{B}$. Then I claim there exists a homomorphism $h:\Lambda\to\mathscr{B}$ such that $h(I(x))=x$ for all $x$.

The key to proving this is noticing that if such an $h$ did not exist, this failure would be detected by a finite subalgebra of $\mathscr{B}$. Indeed, since $\Lambda$ is generated by the sets $I(x)$, there is at most one way to define $h$, and the only thing that can go wrong is if it is not well defined. That means that there are two Boolean combinations of sets of the form $I(x)$ which are equal in $\mathscr{P}(\mathscr{B})$ but such that the corresponding Boolean combinations of the elements $x$ in $\mathscr{B}$ are not equal. Such an occurrence is witnessed by finitely many elements of $\mathscr{B}$, namely the finitely many $x$'s such that $I(x)$ appears in either of the two Boolean combinations. Letting $\mathscr{B}_0\subseteq\mathscr{B}$ be the subalgebra generated by these finitely many elements, then the corresponding $h$ for the algebra $\mathscr{B}_0$ will also fail to be well-defined.

This means that we may assume $\mathscr{B}$ is finite. But now we can just use the classification of finite Boolean algebras to check that our $h$ is well-defined. Indeed, we have $\mathscr{B}\cong\mathscr{P}(A)$ for some finite set $A$; let us identify $\mathscr{B}$ with $\mathscr{P}(A)$. It is easy to check that the subalgebra $\Lambda$ is all of $\mathscr{P}(\mathscr{P}(A))$. Let $g:A\to \mathscr{P}(A)$ be the map $g(a)=\{a\}$. It is then straightforward to verify that if we define $h:\mathscr{P}(\mathscr{P}(A))\to\mathscr{P}(A)$ by $h(S)=g^{-1}(S)$, then $h$ is a homomorphism satisfying $h(I(x))=x$ for each $x\in\mathscr{P}(A)$.

By the way, if I'm not mistaken, $\Lambda$ can be considered as the quotient on the free Boolean algebra of the set $\mathscr{B}$ by relations saying that meets of the generators are computed as in $\mathscr{B}$ (clearly these relations hold in $\Lambda$ among the generators $I(x)$; conversely, you can use the explicit description when $\mathscr{B}$ is finite to show that all relations among the generators are implied by the meet relations). The surjective homomorphism $\Lambda\to\mathscr{B}$ is then the quotient map that further imposes the relations that joins are computed as in $\mathscr{B}$.