Order isomorphic but not Boolean isomorphic

Question

Is there an example of Boolean lattices $(X,\le)$ and $(Y,\le)$ and a function $$f:X\to Y$$ such that for all $x,z\in X$ $$x\le z ~~~~ \leftrightarrow ~~~~f(x)\le f(z) $$

so that for some $a\in X$, $f(a')$ not a complement of $f(a)$?

( $a'$ is the complement of $a$ ).

Answer 1

If $f$ is bijective, then necessarily $f(0)=0$ and $f(1)=1$. Indeed, from $0=f(a)$ and $0\le a$ you get $f(0)\le f(a)=0$, so $f(0)=0$. Similarly, $f(1)=1$.

Let $a\in X$; set $f(b)=f(a)\land f(a')$. Then $f(b)\le f(a)$, so $b\le a$; also $f(b)\le f(a')$, so $b\le a'$. Therefore $b\le a\land a'=0$ and $b=0$; it follows that $f(a)\land f(a')=0$. Similarly, $f(a)\lor f(a')=1$.

In conclusion, $f(a')=f(a)'$.