Prove that every lattice homomorphism is order preserving but converse is not true.

Question

If $(L,*,+)$ and $(S,\cdot,\vee)$ are two lattices, a mapping $g\colon L\to S$ is called a lattice homomorphism from $L$ to $S$ if for any $a,b \in L$ we have $g(a*b) = g(a) \cdot g(b)$ and $g(a+b) = g(a) \vee g(b)$. How to prove that every lattice homomorphism is order preserving but converse is not true?

Answer 1

Hint: Given $a,b\in L,$ we have $a\le_L b$ iff $a*b=a$ iff $a+b=b$. The order relation on $S$ is similarly defined.

For a counterexample to the converse, see this picture.