I have finite boolean algebras $B_1, B_2$ and an injective homomorphism $e : B_1 \hookrightarrow B_2$ between them. I'd like to know whether the following fact is true: Does there exist for every $y \in B_2$ some $a, x \in B_1$ such that $a$ is an atom (i.e. $0<a$ and there is no $b \in B_1$ with $0<b<a$. Equivalently, if $B_1$ is a powerset algebra, then $a$ is a singleton set) and $e(x) \le y \le e(x \lor a)$?
I know that $\bigvee_{B_2} {\downarrow\! y} \cap e[B_1]$ and $\bigwedge_{B_2} {\uparrow\! y} \cap e[B_1]$ exist and lie in $e[B_1]$. And $y$ lies between them. But if $y \notin e[B_1]$ does an atom $a \in B_1$ exist s.t. $e(a) \lor \left( \bigvee_{B_2} {\downarrow\! y} \cap e[B_1] \right) = \bigwedge_{B_2} {\uparrow\! y} \cap e[B_1]$? Or is there a better candidate for $x$?
(Note: I write $e[B_1]:=\{ e(x) \mid x \in B_1 \}$ and ${\downarrow\! y} := \{ y' \in B_2 \mid y' \le y \}$ and ${\uparrow\! y}$ similarly with the other inequality.)
Every finite Boolean algebra is isomorphic to a power set and every homomorphism of finite Boolean algebras is induced by a map of the corresponding sets. So your question is equivalently: given two finite sets $S_1$ and $S_2$ and a surjection $f:S_2\to S_1$, for each $y\subseteq S_2$ does there exist $x\subseteq S_1$ and $a\in S_1$ such that $f^{-1}(x)\subseteq y\subseteq f^{-1}(x\cup\{a\})$? Now it should be easy to see the answer is no: for instance, if $y$ contains some but not all of the points of $f^{-1}(\{s\})$ and $f^{-1}(\{t\})$ for two different $s,t\in S_1$, then $x$ would need to exclude both $s$ and $t$ but $x\cup\{a\}$ would need to include both $s$ and $t$.