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).