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The question may be trivial, but has eluded me, may be it is more appropriate for mathstack-exchange.

Let $B$, $C$ be boolean algebras and $i:B\to C$ be an homomorphism. By Stone duality to each such $i$ corresponds a continuos map $\pi:St(C)\to St(B)$ defined by $G\mapsto i^{-1}[G]$ for any $G$ ultrafilter on $B$ (where $St(B)$ is the compact $0$-dimensional space of ultrafilters on $B$ with topology given by the base of clopen sets $N_c$ given by ultrafilters to which $c$ belongs, as $c$ ranges in $B$).

It is easy to check that if $\pi$ is open then $i$ is a complete homomorphism (i.e. it preserves arbitrary suprema whenever they exist in $B$).

Does the converse always hold?

The converse holds in case $\pi$ induces an order preserving map $\pi^*:C\to B$ defined by $N_{\pi^*(c)}=\pi[N_c]$ for all $c\in C$, which occurs for example if $B$ is a complete boolean algebra. What if $B$ is not complete and $i$ is complete? In this case I'm stuck. Here is a sketch of the easy implication and of the partial converse:

(1) Assume $i$ is not complete. Let $D$ be a predense subset of $B$ such that $i[D]$ is not predense in $C$. Fix $c\in C$ such that $c\wedge i(d)=0$ for all $d\in D$. Then $A=\bigcup_{d\in D} N_d$ is a dense open subset of $St(B)$ and $\pi[N_c]\cap A=\emptyset$. Since $St(B)\setminus A$ is closed nowhere dense, this gives that $\pi[N_c]$ is closed (since $St(B)$ is compact Hausdorff) and nowhere dense, hence it is not open.

(2) It suffices to show that $\pi[N_c]$ is clopen for all $c\in C$. Let $A=\{b\in B: i(b)\wedge c=0\}$. By completeness of $B$, $\bigvee A=d$ exists, and by completeness of $i$, $i(d)\wedge c=0$. It is not hard to check that $\pi[N_c]=N_{\neg d}$ (since $G\not\in \pi[N_c]$ if and only if $G\cap A$ is non-empty).

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An example in which $i$ is a complete homomorphism but $\pi$ is not an open map: Notation: For any Boolean algebra $A$ and any $a$ in $A$, $S(a)$ is the set of all ultrafilters $F$ such that $a$ is in $F$. If $f:A\to B$ is a homomorphism, then $f^d: \operatorname{ult}(B)\to\operatorname{ult}(A)$ is its dual, assigning to each ultrafilter $F$ on $B$ the inverse image $f^{-1}[F]$. finco($\omega$) is the Boolean algebra of finite and cofinite subsets of $\omega$.

Let $f$ be the inclusion map from finco($\omega$) into P($\omega$).Then f is a complete homomorphism; see the Handbook of Boolean Algebras, pp. 21, 59. Suppose that $f^d$ is an open map. Let $E$ be the set of even integers.Then there is an a in finco($\omega$) such that $f^d[S(E)]=S(a)$. We claim that $E$ is a subset of $a$. Otherwise let $G$ be an ultrafilter on $B$ containing $E$ and the complement of $a$. Then $G$ is a member of $S(E)$, so with $F=f^d(G)$ we have $F$ a member of $S(a)$. So a is a member of $F$, and hence also of $G$, contradiction.

This proves the claim. But $a$ is not equal to $E$, so let $b$ be a nonzero member of finco($\omega$) below both $a$ and the complement of $E$. Let $H$ be an ultrafilter on $A$ such that $b$ is in $H$. Then also a is in H and so there is a $K$ in $S(E)$ such that $f^d(K)=H$. Then $b$ is in $K$, so the complement of $E$ is in $K$, contradiction.

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  • $\begingroup$ many thanks, nice reply. I suppose you mean the claim states that E is a subset of a, rather than E is not a subset of a. $\endgroup$ – matteo viale Nov 30 '17 at 9:21
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By Corollary 6.3 of [1], the morphism $i$ of Boolean algebras is complete if and only if the continuous function $\pi$ is quasi-open, which means that $int(\pi(U))$ is nonempty for every nonempty open subset $U$ of the domain. The example given by Don Monk shows that this is a strictly weaker property than being an open continous map, as one would expect.

[1] Bezhanishvili, Guram, Stone duality and Gleason covers through de Vries duality, Topology Appl. 157, No. 6, 1064-1080 (2010). ZBL1190.54015.

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