# Extend an automorphism from a partially ordered set $P$ to the complete Boolean algebra $B(P)$.

Cf. Jech's Set Theory p.221.

A partially ordered set $P$ is separative if for any $p,q$ with $p\nleq q$, there is $r\leq p$ such that $r$ and $q$ are incompatible, i.e., there is no $s\leq r,q$.

Let $P$ be a separative partially ordered set. If $\pi$ is an automorphism of $P$, then $\pi$ extends to an automorphism of the complete Boolean algebra $B(P)$ by $$\pi(u)=\sum\{\pi(p)\mid p\in P\wedge p\leq u\}$$ where $B(P)$ contains $P$ is given by the following:

Corollary 14.12. For every partially ordered set $(P,<)$ there is a complete Boolean algebra $B=B(P)$ and a mapping $e:P\to B^+$ such that:

(i) if $q\leq p$ then $e(q)\leq e(p)$;

(ii) $p$ and $q$ are compatible if and only if $e(p)\cdot e(q)\neq 0$;

(iii) $\{e(p)\mid p\in P\}$ is dense in $B$.

$B$ is unique up to isomorphism.

I want to check that $\pi$ is an automorphism of $B$. It is easy to see that

(i) $\pi(0)=0$ and $\pi(1)=1$;

(ii) $\pi(u\cdot v)\leq\pi(u)\cdot\pi(v)$;

(iii) $\pi(u+v)\geq\pi(u)+\pi(v)$,

but I cannot finish the nontrivial directions in (ii) and (iii); once this has been done, I can prove the bijectivity.

The idea for both is about the same: Work with $e(p)$ which are below/above the elements you are comparing, since $\pi(e(p))=e(\pi p)$. For simplicity of notation, let me just assume $e(p)=p$.
So, $\pi p\leq \pi(u\cdot v)\iff p\leq u\cdot v$, by the definition of $\pi$ on $B(P)$. But now, $p\leq u\cdot v$ if and only if $p\leq u$ and $p\leq v$, which holds if and only if $\pi p\leq\pi u$ and $\pi p\leq\pi v$, which is literally the definition of $\pi p\leq\pi(u)\cdot\pi(v)$.
So we get that $\pi p\leq\pi(u\cdot v)$ if and only if $\pi p\leq\pi(u)\cdot\pi(v)$.