Question about isomorphisms between boolean algebras from Jech On page 79, Jech states the following:
"Note that if $h: B \rightarrow C$ (where B and C are Boolean Algebras) is a one to one mapping such that $u\leq v$ if and only if $h(u)\leq h(v)$, then $h$ is an isomorphism"
I am assuming that by "isomorphism" he means embedding. I have been able to show that $h(u+v) = h(u) + h(v)$ and $h(u\cdot v) = h(u)\cdot h(v)$, but I am struggling to prove that $h(0) = 0, h(1) = 1, \text{and}\ h(-u) = -h(u)$. Whenever I try to prove one of these, I run into a wall where I need one of the other results. 
Any help will be greatly appreciated!
 A: I don't have Jech on hand to check this, but I think "one-to-one mapping" here means bijection, confusingly enough. The phrase "one-to-one correspondence," for example, is pretty commonly used to refer to bijections. (I personally hate this since it clashes with the use of "one-to-one" for injections - or more precisely, I hate the use of "one-to-one" for injections, but that's much more entrenched.)
Under this interpretation the statement is true: any order-preserving bijection between Boolean algebras is an isomorphism of Boolean algebras. (This follows for a general model-theoretic reason: the Boolean algebra structure is definable from the ordering alone, so preserving-and-reflecting the ordering amounts to preserving-and-reflecting everything.)

The statement in question meanwhile is false if we read "one-to-one mapping" as just meaning injection (even if we read "isomorphism" as "embedding," or "isomorphism onto its image"). Indeed, contra your claim an order-preserving injection of Boolean algebras need not even preserve $\wedge$ and $\vee$: let $B$ be the four-element Boolean algebra $\{0^B,x,y,1^B\}$, let $C$ be the usual Boolean algebra of subsets of $\mathbb{N}$, and consider the map $$h: 0^B\mapsto \{3\}, x\mapsto \{3,4,5\}, y\mapsto \{3,4,6\}, 1^B\mapsto\{3,4,5,6,17\}.$$ As a map between posets $h$ is order-preserving, but $h$ is terrible in every other respect.
