Let $h: B\to B'$ be an onto map, where $\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ and $\mathcal{B'}:=(B',\leq,\lor,\land,^c,0,1)$ are Boolean algebras. Show that $h$ is an isomorphism from $\mathcal{B}$ to $\mathcal{B'}$ iff for all $x,y\in B$, $x\leq y$ iff $h(x)\leq h(y)$.
To prove one side of the implication, let us start by assuming that $h$ is an isomorphism. So, $h$ is one-one and onto both. We can use the homomorphism axioms as stated here. Now, $x \leq y$ iff $x \land y = x$. So $h(x\land y) = h(x) \land h(y) = h(x)$. This implies $h(x)\leq h(y)$. All the arguments are reversible, and the double implication is proved.
I'm unable to show the other side of the implication, i.e. if for all $x,y\in B$, $x \leq y$ iff $h(x)\leq h(y)$ then $h$ is an isomorphism. We already know that $h$ is onto, so all that remains to be shown is that $h$ is one-one. Starting with $h(x) = h(y)$, it should suffice to prove $x=y$. How do I go about this? I have thought of writing $h(x)=h(y)$ as $h(x)\leq h(y)$ and $h(y)\leq h(x)$ (do we need to prove this as well?), but I didn't reach the conclusion after this.
Thanks!
If $h(x) = h(y)$, then both $$h(x) \leq h(y)$$ and $$h(y) \leq h(x)$$ hold.
Now, from the equivalence it follows that $$x \leq y$$ and $$y \leq x,$$ and therefore, $x = y$, by anti-symmetry.