Galois Connection is Expressed by Maximum

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
 * $d$ is an increasing mapping and:
 * $\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$

where:
 * $\max$ denotes the maximum
 * $d^{-1} \sqbrk {s^\preceq}$ denotes the image of $s^\preceq$ under relation $d^{-1}$
 * $s^\preceq$ denotes the lower closure of $t$

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

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

Let $s \in S$.

By definition of reflexivity:
 * $\map g s \preceq \map g s$

By definition of Galois connection:
 * $\map d {\map g s} \preceq s$

By definition of lower closure:
 * $\map d {\map g s} \in s^\preceq$

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

By definition of upper bound:
 * $\forall t \in T: t$ is an upper bound for $d^{-1} \sqbrk {s^\preceq} \implies \map g s \precsim t$

We will prove that:
 * $\map g s$ is an upper bound for $d^{-1} \sqbrk {s^\preceq}$

Let $t \in d^{-1} \sqbrk {s^\preceq}$.

By definition of image of set:
 * $\map d t \in s^\preceq$

By definition of lower closure of element:
 * $\map d t \preceq s$

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

By definition of supremum:
 * $d^{-1} \sqbrk {s^\preceq}$ admits a supremum

and
 * $\map \sup {d^{-1} \sqbrk {s^\preceq} } = \map g s$

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

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

Let:
 * $\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$

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

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

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

By Preimage of Subset is Subset of Preimage:
 * $d^{-1} \sqbrk {x^\preceq} \subseteq d^{-1} \sqbrk {y^\preceq}$

By assumption:
 * $\map g s = \map \max {d^{-1} \sqbrk {x^\preceq} } = \map \sup {d^{-1} \sqbrk {x^\preceq} }$

and:
 * $\map g y = \map \max {d^{-1} \sqbrk {y^\preceq} } = \map \sup {d^{-1} \sqbrk {y^\preceq} }$

By Supremum of Subset:
 * $\map g s \preceq \map g y$

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

Thus:
 * $d$ 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 $s \in S, t \in T$.

Second implication:

Let $\map d t \preceq s$.

By definition of lower closure of element:
 * $\map d t \in s^\preceq$

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

By assumption:
 * $\map g s = \map \max {d^{-1} \sqbrk {s^\preceq} } = \map \sup {d^{-1} \sqbrk {s^\preceq} }$

By definition of supremum:
 * $\map g s$ is an upper bound for $d^{-1} \sqbrk {s^\preceq}$

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

First implication:

Let $t \precsim \map g s$

By assumption:
 * $\map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$

By greatest element of set:
 * $\map g s \in d^{-1} \sqbrk {s^\preceq}$

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

By definition of lower closure of element:
 * $\map d {\map g s} \preceq s$

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

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

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