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.

Edit: Ignore this answer. See Loronegro's.
Apparently I haven't done enough on Math.SE before, so I have to post this as an answer instead of a comment.
Hopefully this is helpful: first notice that $\pi c \le c$ for every $c\in C$ (the supremum of a bunch of things that are $\le c$ is also $\le c$). From this, you can deduce that, for every $u\in B$, $u\le c$ iff $u\le \pi c$. Therefore the sets $\{u\in B : u\le c\}$ and $\{u\in B : u\le \pi c\}$ are identical, so $\pi^{-1}\pi = 1$. By starting the argument with $\pi^{-1}c \le c$ instead of $\pi c \le c$, you can also conclude that $\pi\pi^{-1} = 1$.
You should feel cheated now, as I have avoided any mention of density. I guess proving that $\sum^C \{u\in B: u\le c\} = c$ requires the density of $B$.
