All Suprema Preserving Mapping is Lower Adjoint of Galois Connection

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

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

where
 * $\min$ 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 $t$

Proof
Define a mapping $d: T \to S$:
 * $\forall s \in S: g\left({s}\right) := \sup\left({d^{-1}\left[{s^\preceq}\right]}\right)$

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:
 * $\sup\left({x^\precsim}\right) = x$ and $\sup\left({y^\precsim}\right) = y$

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:
 * $\sup\left({d^\to\left({x^\precsim}\right)}\right) = d\left({x}\right)$ and $\sup\left({d^\to\left({y^\precsim}\right)}\right) = d\left({y}\right)$

By Image of Subset under Relation is Subset of Image/Corollary 2:
 * $d^\to\left({x^\precsim}\right) \subseteq d^\to\left({y^\precsim}\right)$

Thus by Supremum of Subset:
 * $d\left({x}\right) \preceq d\left({y}\right)$

This ends the proof of lemma 1.

We will prove as lemma 2 that
 * $\forall s \in S: g\left({s}\right) = \max\left({d^{-1}\left[{s^\preceq}\right]}\right)$

Let $s \in S$.

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

By Image of Inverse Image:
 * $d\left[{d^{-1}\left[{s^\preceq}\right]}\right] \subseteq s^\preceq$

By Supremum of Subset and Supremum of Lower Closure of Element:
 * $\sup\left({d\left[{d^{-1}\left[{s^\preceq}\right]}\right]}\right) \preceq \sup\left({s^\preceq}\right) = s$

By definition of lower closure of element:
 * $\sup\left({d\left[{d^{-1}\left[{s^\preceq}\right]}\right]}\right) \in s^\preceq$

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

By definitions mapping preserves the supremum:
 * $\sup\left({d\left[{d^{-1}\left[{s^\preceq}\right]}\right]}\right) = d\left({g\left({s}\right)}\right)$

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

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

This ends the proof of lemma 2.

Thus by Galois Connection is Expressed by Maximum:
 * $\left({g, d}\right)$ is a Galois connection.

Thus by lemma 2:
 * $\forall s \in S: g\left({s}\right) = \max\left({d^{-1}\left[{s^\preceq}\right]}\right)$