Order-Preserving Mapping Not Always Semilattice Homomorphism

Theorem
Let $\struct {S, \circ}$ and $\struct {T, *}$ be semilattices.

Let $\preceq_1$ be the ordering on $S$ defined by:
 * $a \preceq_1 b \iff \paren {a \circ b} = b$

Let $\preceq_2$ be the ordering on $T$ defined by:
 * $x \preceq_2 y \iff \paren {x * y} = y$

Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be an order-preserving mapping.

Then:
 * $\phi: \struct {S, \circ} \to \struct {T, *}$ may not be a semilattice homomorphism

Proof
Let $S = \set{a, b, c}$.

Let $\circ : S \times S \to S$ be defined by:
 * $a \circ b = b \circ a = c$
 * $a \circ c = c \circ a = c$
 * $b \circ c = c \circ b = c$
 * $a \circ a = a$
 * $b \circ b = b$
 * $c \circ c = c$

Then a manual check shows that $\circ$ is:
 * closed
 * associative
 * commutative
 * idempotent

So $\struct{S, \circ}$ is a semilattice with ordering
 * $\preceq_1 = \set{\tuple{a, a}, \tuple{b, b}, \tuple{c, c}, \tuple{a, c}, \tuple{b,c}}$

Let $T = \set{p, q, r, s}$.

Let $* : S \times S \to S$ be defined by:
 * $p * q = q * p = r$
 * $p * r = r * p = r$
 * $q * r = r * q = r$
 * $p * s = s * p = s$
 * $q * s = s * q = s$
 * $r * s = s * r = s$
 * $p * p = p$
 * $q * q = q$
 * $r * r = r$
 * $s * s = s$

Then a manual check shows that $*$ is:
 * closed
 * associative
 * commutative
 * idempotent

So $\struct{T, *}$ is a semilattice with ordering
 * $\preceq_2 = \set{\tuple{p, p}, \tuple{q, q}, \tuple{r, r}, \tuple{s, s}, \tuple{p, r}, \tuple{q,r}, \tuple{p, s}, \tuple{q,s}, \tuple{r, s}}$

Let $\phi : S \to T$ be defined by:
 * $\map \phi a = p$
 * $\map \phi b = q$
 * $\map \phi c = s$

Then a manual check shows that $\phi$ is order-preserving.

Now:

Hence $\phi$ is not a semilattice homomorphism by definition.