Given a separative poset $P$, we can form it's Boolean completion $B(P)$. This is a Boolean algebra whose elements are regular cuts on $P$, as defined here. Also, $P$ embeds densely into $B(P)$ by mapping $p$ to the principal ideal $U_p$.
Let's now take a poset $Q$ and a $P\subseteq Q$, both of which are separative. The inclusion $P\to Q$ is monotone. What I want to know is, can we somehow induce a complete embedding $f\colon B(P)\to B(Q)$ from this that makes the obvious diagram commute?
It seems to me there is only one thing that should work. Since $P$ is densely embedded in $B(P)$ and we know $f$ on $P$, the only reasonable choice seems to be $$f(U)=\bigvee_{p\in U}U_p$$
But this doesn't even seem to be a Boolean algebra homomorphism. Let's look at the meets: $$f(U\cap V)=\bigvee_{p\in U\cap V} U_p \stackrel{?}{=} \bigvee_{p\in U} U_p\cap \bigvee_{q\in V} U_q= f(U)\cap f(V)$$
Obviously, the left hand set is contained in the right hand set, but do we have equality? I just don't see it.
Have I missed something obvious? Is this even the right way of defining the map $f$?
Added: Apostolos dealt with the meets in the comments, just using distributivity. But I'm having trouble with joins now. We have $$f(U\vee V)=\bigvee_{p\in U\vee V} U_p^Q \stackrel{?}{=} \bigvee_{p\in U\cup V} U_p^Q= f(U)\vee f(V)$$ (I'll be using $U_p^P$ and $U_p^Q$ to distinguish principal ideals in $P$ and $Q$ respectively)
To focus on the nontrivial inclusion, take $x\in f(U\vee V)$ and $y\leq x$. We wish to find a $p_0\in U\cup V$ such that $U_y^Q\cap U_{p_0}^Q\neq\emptyset$. According to the definition of the join, there exists a $p\in U\vee V$ such that $U_y^Q\cap U_p^Q\neq\emptyset$. Taking an element $q$ of this set, we get $q\leq y$ and $q\leq p$. Now, if we could find an element of $P$ below this $q$, we would be done (since $p\in U\vee V$), but I don't see any reason for such an element to exist. In particular, in the application I have in mind, $P$ is very far from being dense in $Q$. How do I proceed here?
One trivial counterexample comes to mind:
Take $Q$ to be the power set of $\mathbb R$ and $P$ the power set of $\mathbb N$, both without the zero element, i.e. only non-empty sets. The order on both is inclusion.
Trivially $P$ and $Q$ are separative and $P\subseteq Q$. However since both are the positive elements of a complete Boolean algebra, and the completion is unique [up to isomorphism], the completion is both is simply adding the zero element.
However an embedding of a Boolean algebra into another must respect the $0,1$ elements, but $1_{B(P)}=\mathbb N\neq\mathbb R=1_{B(Q)}$. So there is no obvious way of closing the diagram.