The completion of a Boolean algebra is unique up to isomorphism Jech defines a completion of a Boolean algebra $B$ to be a complete Boolean algebra $C$ such that $B$ is a dense subalgebra of $C$.
I am trying to prove that given two completions $C$ and $D$ of $B$, then the mapping $\pi: C \rightarrow D$ given by $\pi(c) = \sum^D \{ u \in B \ \vert \ u \le c \} $ is an isomorphism. I understand that $c \not= 0 \implies \pi(c) \not= 0$, and that given any $d \in D$, I can write it as $d = \sum^D \{ u \in B \ \vert \ u \le d \} $, but I am unsure how to proceed. Any help would be appreciated.

I am afraid that it doesn't make sense that $\pi (c)\leq c$ because both elements live in different lattices $C$ and $D$, so they are not comparable. So it's better to denote differently both orders as $\leq_C$ and $\leq_D$.
So let $\pi (c)= \sum^D${$u\in B:u\leq_C c$}, and let $\rho: D\to C$ be such that $\rho(d)=\sum^C${$u\in B:u\leq_D d$}.
For any $c\in C$ and $u\in B$ we have $u\leq_C c\Rightarrow u\leq_D\pi(c)$, by the property of the supremum.  Conversely, let $u\in B$ be such that $u\leq_D\pi(c)$. Then $u=u\wedge\pi(c)=u\wedge\sum^D${$b\in B:b\leq_C c$}$=\sum^D${$u\wedge b:b\leq_C c$}, since $D$ is a complete Boolean algebra, and so it has the general distributivity. Now, if $b\leq_C c$ then $u\wedge b\leq_Cu\wedge c\leq_C c$. Therefore $u\leq_C c$. We have shown that $u\leq_C c\Leftrightarrow u\leq_D\pi(c)$.
Therefore $c=\sum^C${$u\in B:u\leq_C c$}$=\sum^C${$u\in B:u\leq_D \pi(c)$} $=\rho(\pi(c))$.
Similarly $\pi(\rho(d))=d$ for each $d\in D$. So $\rho=\pi^{-1}$, and $\pi$ is an isomorphism.
