All Suprema Preserving Mapping is Lower Adjoint of Galois Connection

Theorem
Let $\struct {S, \preceq}$ be an ordered set.

Let $\struct {T, \precsim}$ be a complete lattice.

Let $d: T \to S$ be all suprema preserving mapping.

Then there exists a mapping $g: S \to T$ such that $\tuple {g, d}$ is Galois connection and:
 * $\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$

where
 * $\min$ 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 $t$

Proof
Define a mapping $d: T \to S$:
 * $\forall s \in S: \map g s := \map \sup {d^{-1} \sqbrk {s^\preceq} }$

We will prove as lemma 1 that:
 * $d$ is an increasing mapping.

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

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

By Supremum of Lower Closure of Element:
 * $\map \sup {x^\precsim} = x$ and $\map \sup {y^\precsim} = y$

By definition of mapping preserves all suprema:
 * $d$ preserves the supremum on $x^\precsim$ and $d$ preserves the supremum on $y^\precsim$

By definition of mapping preserves the supremum on set:
 * $\map \sup {\map {d^\to} {x^\precsim} } = \map d x$ and $\map \sup {\map {d^\to} {y^\precsim} } = \map d y$

By Image of Subset under Mapping is Subset of Image:
 * $\map {d^\to} {x^\precsim} \subseteq \map {d^\to} {y^\precsim}$

Thus by Supremum of Subset:
 * $\map d x \preceq \map d y$

This ends the proof of lemma 1.

We will prove as lemma 2 that
 * $\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$

Let $s \in S$.

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

By Image of Inverse Image:
 * $d \sqbrk {d^{-1} \sqbrk {s^\preceq} } \subseteq s^\preceq$

By Supremum of Subset and Supremum of Lower Closure of Element:
 * $\map \sup {d \sqbrk {d^{-1} \sqbrk {s^\preceq} } } \preceq \map \sup {s^\preceq} = s$

By definition of lower closure of element:
 * $\map \sup {d \sqbrk {d^{-1} \sqbrk {s^\preceq} } } \in s^\preceq$

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

By definition of mapping preserves the supremum:
 * $\map \sup {d \sqbrk {d^{-1} \sqbrk {s^\preceq} } } = \map d {\map g s}$

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

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

This ends the proof of lemma 2.

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

Thus by lemma 2:
 * $\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$