Find a homomorphism that maps a point in a Boolean algebra into the image of a proper filter Let $\mathbb B$ be a Boolean algebra. Let $F$ be a proper filter in $\mathbb B$ (i.e. $0 \notin F$), and let $I$ be its dual ideal. Suppose there exists $a \in \mathbb B$ such that $a \notin F \cup I$.
Question: Does there exist a Boolean algebra $\mathbb B'$ and a homomorphism $h: \mathbb B \to \mathbb B'$ such that $h(a) \in h(F)$?
All I have been able to observe so far is that $h$ cannot be injective. 

Let $G = [F\cup\{a\})$ be the filter generated by $F \cup \{a\}$.
We have that $c \in G$ iff $a \wedge b \leq c$ for some $b \in F$.
Let us see that $G$ is proper.
If $0 \in G$ then there exist $b \in F$ such that $b \wedge a = 0$.
It follows that $b \leq a'$, whence $a' \in F$ and $a \in I$, a contradiction.
Now, let $b \in G \cap I$.
It follows that $b' \in F$, thence $b,b' \in G$ and $0 \in G$, a contradiction.
Therefore $G \cap I = \varnothing$.
Finally (and here you need (AC) or actually, something a little bit weaker, I think, but (AC) will do it), given that $G$ is an filter, $I$ is an ideal and $G \cap I = \varnothing$, there exists an ultrafilter $U$ such that $G \subseteq U$ and $U \cap I = \varnothing$.  
Let $\mathbb B'$ be the two-element Boolean algebra, and make $h(x) = 1$ iff $x \in U$.
It follows that $h(a) = 1 \in \{1\} = h(F)$.
