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Its easy to find boolean lattices $A$ and $B$ together with a function $f : A \rightarrow B$ such that $f$ preserves both top and bottom elements, as well as binary meets, but not complements.

Example. Let $A$ and $B$ denote the four and two element boolean lattices respectively, and let $f$ map every element of $A$ to $\bot_B,$ except for $\top_A$ which is mapped to $\top_B$.

Note however that the function $f : A \rightarrow B$ defined above does not preserve joins; in particular, letting $a$ and $a'$ denote the two "middle" elements of $A$, we have the following.

  • $f(a \vee a') = f(\top_A) = \top_B$
  • $f(a) \vee f(a') = \bot_B \vee \bot_B = \bot_B$

Question. What's an example of boolean lattices $A$ and $B$ together with a function $f : A \rightarrow B$ subject to the following constraints?

  1. $f$ preserves top and bottom elements
  2. $f$ preserves binary meets and joins
  3. $f$ does not preserve all complements
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There is no such function. Given a boolean algebra, observe that $\lnot a$ is the unique element satisfying $a \land \lnot a = \bot$ and $a \lor \lnot a = \top$; thus, any function that preserves $\bot$, $\top$, $\lor$, and $\land$ must also preserve $\lnot$.

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Zhen Lin already answered, but I thought I'd augment. If $A$ and $B$ are two Boolean algebras and $f:A\rightarrow B$ is a lattice homomorphism then the following are equivalent:

  1. $f(0)=0$ and $f(1)=1$
  2. $f(\neg a)=\neg f(a)$ for all $a\in A$

The proof is straight-forward. Assuming 1. we get: $$0=f(0)=f(a\wedge\neg a)=f(a)\wedge f(\neg a)$$ $$1=f(1)=f(a\vee\neg a)=f(a)\vee f(\neg a)$$

Thus $f(\neg a)$ is a complement for $f(a)$ for all $a\in A$. Since complements in a Boolean lattice are unique $f(\neg a)=\neg f(a)$ for all $a\in A$.

Assuming 2. we have for any $a\in A$ that $$0=f(a)\wedge\neg f(a)=f(a\wedge\neg a)=f(0)$$ $$1=f(a)\vee\neg f(a)=f(a\vee\neg a)=f(1)$$

So in order to find a lattice homomorphism which doesn't preserve complements, it would necessarily (and sufficiently) not fix one of $0$ or $1$.

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