MATH.18:5.1 - Recover an order from an operation, then compare allowed maps

One account starts with a set S and a binary operation a∨b satisfying associativity, commutativity and idempotence:

(a∨b)∨c=a∨(b∨c); a∨b=b∨a; a∨a=a.

Another starts with a partial order in which every pair has a least upper bound. An upper bound of a and b is an element z with a<=z and b<=z. The least upper bound lies below every such z. Antisymmetry makes it unique. We want to move constructions between these descriptions.

From the operation, define a<=b to mean a∨b=b. Idempotence gives reflexivity. If a∨b=b and b∨a=a, commutativity gives a=b, establishing antisymmetry. If a∨b=b and b∨c=c, associativity gives:

a∨c=a∨(b∨c)=(a∨b)∨c=b∨c=c.

Thus the relation is transitive. Also a∨b is an upper bound of a and b. If z is any upper bound, then (a∨b)∨z=a∨(b∨z)=a∨z=z. So a∨b<=z and the original operation supplies the least upper bound.

Conversely, define a∨b to be the least upper bound in the ordered account. Uniqueness makes the operation commutative and idempotent. Both (a∨b)∨c and a∨(b∨c) are the least upper bound of the same three elements, giving associativity.

The round trips recover both primitives. Starting from the operation returns it as the least upper bound just proved. Starting from the order returns it because a<=b holds exactly when b is the least upper bound of a and b.

Now the intended use expands: can every order-preserving map be used as a map preserving the binary operation? A join-preserving map is monotone: apply it to a∨b=b. The converse fails. Take subsets of {u,v} ordered by inclusion, with union as join, and target {0,1} with its usual order. Define h to return 1 only on the full set and 0 on all other subsets. It is monotone, yet:

h({u} union {v})=1, while max(h({u}),h({v}))=0.

For constructions that combine joins, select join-preserving maps in both accounts. If all monotone maps are needed, retain that broader ordered account and its different transformation class. The objects matched; the expanded map claim required and received its own answer.