Galois Connection is Expressed by Minimum

Theorem
Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be ordered sets.

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

Then $\tuple {g, d}$ is a Galois connection :
 * $g$ is an increasing mapping and
 * $\forall t \in T: \map d t = \min \set {g^{-1} \sqbrk {t^\succsim} }$

where
 * $\min$ denotes the minimum
 * $g^{-1} \sqbrk {t^\succsim}$ denotes the image of $t^\succsim$ under relation $g^{-1}$
 * $t^\succsim$ denotes the upper closure of $t$

Sufficient Condition
Let $\tuple {g, d}$ be a Galois connection.

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

Let $t \in T$.

By definition of reflexivity:
 * $\map d t \preceq \map d t$

By definition of Galois connection:
 * $t \precsim \map g {\map d t}$

By definition of upper closure:
 * $\map g {\map d t} \in t^\succsim$

By definition of image of set under relation:
 * $\map d t \in g^{-1} \sqbrk {t^\succsim}$

By definition of lower bound:
 * $\forall s \in S: s$ is lower bound for $g^{-1} \sqbrk {t^\succsim} \implies s \preceq \map d t$

We will prove that:
 * $\map d t$ is lower bound for $g^{-1} \sqbrk {t^\succsim}$

Let $s \in g^{-1} \sqbrk {t^\succsim}$.

By definition of image of set:
 * $\map g s \in t^\succsim$

By definition of upper closure of element:
 * $t \precsim \map g s$

Thus by definition of Galois connection:
 * $\map d t \preceq s$

By definition of infimum:
 * $g^{-1} \sqbrk {t^\succsim}$ admits an infimum

and
 * $\map \inf {g^{-1} \sqbrk {t^\succsim} } = \map d t$

Thus:
 * $\map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$

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

Let $\forall t \in T: \map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$

Thus:
 * $g$ is increasing mapping.

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

Let $x, y \in T$ such that
 * $x \precsim y$

By Upper Closure is Decreasing:
 * $y^\succsim \subseteq x^\succsim$

By Image of Subset under Relation is Subset of Image/Corollary 3:
 * $g^{-1} \sqbrk {y^\succsim} \subseteq g^{-1} \sqbrk {x^\succsim}$

By assumption
 * $\map d x = \map \min {g^{-1} \sqbrk {x^\succsim} } = \map \inf {g^{-1} \sqbrk {x^\succsim} }$

and
 * $\map d y = \map \min {g^{-1} \sqbrk {y^\succsim} } = \map \inf {g^{-1} \sqbrk {y^\succsim} }$

By Infimum of Subset:
 * $\map d x \preceq \map d y$

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

We will prove that
 * $\forall s \in S, t \in T: t \precsim \map g s \iff \map d t \preceq s$

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

First implication:

Let $t \precsim \map g s$.

By definition of upper closure of element:
 * $\map g s \in t^\succsim$

By definition of image of set:
 * $s \in g^{-1} \sqbrk {t^\succsim}$

By assumption:
 * $\map d t = \map \min {g^{-1} \sqbrk {t^\succsim} } = \map \inf {g^{-1} \sqbrk {t^\succsim} }$

By definition of infimum:
 * $\map d t$ is lower bound for $g^{-1} \sqbrk {t^\succsim}$

Thus by definition of lower bound:
 * $\map d t \preceq s$

Second implication:

Let $\map d t \preceq s$

By assumption:
 * $\map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$

By definition of smallest element of set:
 * $\map d t \in g^{-1} \sqbrk {t^\succsim}$

By definition of image of set:
 * $\map g {\map d t} \in t^\succsim$

By definition of upper closure of element:
 * $t \precsim \map g {\map d t}$

By definition of increasing mapping:
 * $\map g {\map d t} \precsim \map g s$

Thus by definition of transitivity:
 * $t \precsim \map g s$

Thus by definition:
 * $\tuple {g, d}$ is Galois connection.