A curiosity on complete homomorphisms of boolean algebras

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.

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.