Galois Connection is Expressed by Maximum

Theorem
Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be ordered sets.

Let $g: S \to T$, $d: T \to S$ be mappings.

Then $\left({g, d}\right)$ is Galois connection
 * $d$ is increasing mapping and
 * $\forall s \in S: g\left({s}\right) = \max\left({d^{-1}\left[{s^\preceq}\right]}\right)$

where
 * $\max$ denotes the maximum
 * $d^{-1}\left[{s^\preceq}\right]$ denotes the image of $s^\preceq$ under relation $d^{-1}$
 * $s^\preceq$ denotes the lower closure of $t$

Sufficient Condition
Let $\left({g, d}\right)$ be a Galois connection.

Thus by definition of Galois connection:
 * $d$ is an increasing mapping

Let $s \in S$.

By definition of reflexivity:
 * $g\left({s}\right) \preceq g\left({s}\right)$

By definition of Galois connection:
 * $d\left({g\left({s}\right)}\right) \preceq s$

By definition of lower closure:
 * $d\left({g\left({s}\right)}\right) \in s^\preceq$

By definition of image of set under relation:
 * $g\left({s}\right) \in d^{-1}\left[{s^\preceq}\right]$

By definition of upper bound:
 * $\forall t \in T: t$ is upper bound for $d^{-1}\left[{s^\preceq}\right] \implies g\left({s}\right) \precsim t$

We will prove that
 * $g\left({s}\right)$ is upper bound for $d^{-1}\left[{s^\preceq}\right]$

Let $t \in d^{-1}\left[{s^\preceq}\right]$.

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

By definition of lower closure of element:
 * $d\left({t}\right) \preceq s$

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

By definition of supremum:
 * $d^{-1}\left[{s^\preceq}\right]$ admits a supremum

and
 * $\sup\left({d^{-1}\left[{s^\preceq}\right]}\right) = g\left({s}\right)$

Thus:
 * $g\left({s}\right) = \min\left({d^{-1}\left[{s^\preceq}\right]}\right)$

Necessary Condition
Let $d: T \to S$ be an increasing mapping.

Let $\forall s \in S: g\left({s}\right) = \max\left({d^{-1}\left[{s^\preceq}\right]}\right)$

We will prove that
 * $g$ is increasing mapping.

Let $x, y \in S$ such that
 * $x \preceq y$

By Lower Closure is Increasing:
 * $x^\preceq \subseteq y^\preceq$

By Image of Subset under Relation is Subset of Image/Corollary 3:
 * $d^{-1}\left[{x^\preceq}\right] \subseteq d^{-1}\left[{y^\preceq}\right]$

By assumption
 * $g\left({x}\right) = \max\left({d^{-1}\left[{x^\preceq}\right]}\right) = \sup\left({d^{-1}\left[{x^\preceq}\right]}\right)$

and
 * $g\left({y}\right) = \max\left({d^{-1}\left[{y^\preceq}\right]}\right) = \sup\left({d^{-1}\left[{y^\preceq}\right]}\right)$

By Supremum of Subset:
 * $g\left({x}\right) \preceq g\left({y}\right)$

Thus by definition:
 * $g$ is an increasing mapping.

Thus:
 * $d$ is increasing mapping.

We will prove that
 * $\forall s \in S, t \in T: t \precsim g\left({s}\right) \iff d\left({t}\right) \preceq s$

Let $s \in S, t \in T$.

Second implication:

Let $d\left({t}\right) \preceq s$.

By definition of lower closure of element:
 * $d\left({t}\right) \in s^\preceq$

By definition of image of set:
 * $t \in d^{-1}\left[{s^\preceq}\right]$

By assumption
 * $g\left({s}\right) = \max\left({d^{-1}\left[{s^\preceq}\right]}\right) = \sup\left({d^{-1}\left[{s^\preceq}\right]}\right)$

By definition of supremum:
 * $g\left({s}\right)$ is upper bound for $d^{-1}\left[{s^\preceq}\right]$

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

First implication:

Let $t \precsim g\left({s}\right)$

By assumption
 * $g\left({s}\right) = \max\left({d^{-1}\left[{s^\preceq}\right]}\right)$

By greatest element of set:
 * $g\left({s}\right) \in d^{-1}\left[{s^\preceq}\right]$

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

By definition of lower closure of element:
 * $d\left({g\left({s}\right)}\right) \preceq s$

By definition of increasing mapping:
 * $d\left({t}\right) \preceq d\left({g\left({s}\right)}\right)$

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

Thus by definition:
 * $\left({g, d}\right)$ is Galois connection.