Upper Adjoint Preserves All Infima

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

Let $g: S \to T$ be an upper adjoint of Galois connection.

Then $g$ preserves all infima.

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

Let $X$ be a subset of $S$ such that
 * $X$ admits an infimum.

We will prove as lemma 1 that
 * $\forall t \in T: t$ is lower bound for $g^\to\left({X}\right) \implies t \precsim g\left({\inf X}\right)$

Let $t \in T$ such that
 * $t$ is lower bound for $g^\to\left({X}\right)$

We will prove as sublemma that
 * $d\left({t}\right)$ is lower bound for $X$

Let $s \in X$.

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

By definition of lower bound:
 * $t \precsim g\left({s}\right)$

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

This ends the proof of sublemma.

By definition of infimum:
 * $d\left({t}\right) \preceq \inf X$

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

This ends the proof of lemma 1.

We will prove as lemma 2 that
 * $g\left({\inf X}\right)$ is lower bound for $g^\to\left({X}\right)$

Let $t \in g^\to\left({X}\right)$.

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

By definition of infimum:
 * $\inf X$ is lower bound for $X$

By definition of lower bound:
 * $\inf X \preceq s$

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

Thus by definition of increasing mapping:
 * $g\left({\inf X}\right) \precsim t$

This ends the proof of lemma 2.

Thus by definition of infimum:
 * $g^\to\left({X}\right)$ admits an infimum

and
 * $\inf\left({g^\to\left({X}\right)}\right) = g\left({\inf X}\right)$

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

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