Lower Adjoint Preserves All Suprema

Theorem
Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be {{Definition:Ordered Set|ordered sets]].

Let $d: T \to S$ be an lower adjoint of Galois connection.

Then $g$ preserves all suprema.

Proof
By definition of lower adjoint
 * $\exists g: S \to T: \left({g, d}\right)$ is a Galois connection

Let $X$ be a subset of $T$ such that
 * $X$ admits a supremum.

We will prove as lemma 1 that
 * $\forall s \in S: s$ is upper bound for $d^\to\left({X}\right) \implies d\left({\sup X}\right) \preceq s$

Let $s \in S$ such that
 * $s$ is upper bound for $d^\to\left({X}\right)$

We will prove as sublemma that
 * $g\left({s}\right)$ is upper bound for $X$

Let $t \in X$.

By definition of image of set:
 * $d\left({t})\right) \in d^\to\left({X}\right)$

By definition of upper bound:
 * $d\left({t}\right) \preceq s$

Thus by definition of Galois connection:
 * $t \precsim g\left({s}\right)$

This ends the proof of sublemma.

By definition of supremum:
 * $\sup X \precsim g\left({t}\right)$

Thus by definition of Galois connection:
 * $d\left({\sup X}\right) \preceq t$

This ends the proof of lemma 1.

We will prove as lemma 2 that
 * $d\left({\sup X}\right)$ is upper bound for $d^\to\left({X}\right)$

Let $s \in d^\to\left({X}\right)$.

By definition of image of set:
 * $\exists t \in T: t \in X \land d\left({t}\right) = s$

By definition of supremum:
 * $\sup X$ is upper bound for $X$

By definition of upper bound:
 * $t \precsim \sup X$

By definition of Galois connection:
 * $d$ is increasing mapping.

Thus by definition of increasing mapping:
 * $s \preceq d\left({\sup X}\right)$

This ends the proof of lemma 2.

Thus by definition of supremum:
 * $d^\to\left({X}\right)$ admits a supremum

and
 * $\sup\left({d^\to\left({X}\right)}\right) = d\left({\sup X}\right)$

Thus by definition:
 * $g$ preserves supremum on $X$

Thus by definition:
 * $g$ preserves all suprema.