Upper Adjoint of Galois Connection is Surjection implies Lower Adjoint at Element is Minimum of Preimage of Singleton of Element

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

Let $g: S \to T, d:T \to S$ be mappings such that:
 * $\tuple {g, d}$ is a Galois connection

and
 * $g$ is a surjection.

Then
 * $\forall t \in T: \map d t = \min \set {g^{-1} \sqbrk {\set t} }$

Proof
By definition of Galois connection:
 * $g$ is an increasing mapping.

Let $t \in T$.

By definition of surjection:
 * $\Img g = T$

By Image of Preimage under Mapping/Corollary:
 * $g \sqbrk {g^{-1} \sqbrk {t^\succeq} } = t^\succeq$

By Galois Connection is Expressed by Minimum:
 * $\map d t = \min \set {g^{-1} \sqbrk {t^\succeq} }$

By definition of min operation:
 * $\map d t = \inf \set {g^{-1} \sqbrk {t^\succeq} }$ and $\map d t \in g^{-1} \sqbrk {t^\succeq}$

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

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

By definition of minimum element:
 * $g^{-1} \sqbrk {t^\succeq}$ admits an infimum.

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

By definition of surjection:
 * $\exists s \in S: t = \map g s$

By definition of singleton:
 * $t \in \set t$

By Set is Subset of Upper Closure
 * $\set t \subseteq \set t^\succeq$

By Upper Closure of Singleton:
 * $\set t^\succeq = t^\succeq$

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

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

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

By definition of antisymmetry:
 * $\map g {\map d t} = t$

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

By Image of Subset under Relation is Subset of Image/Corollary 3:
 * $g^{-1} \sqbrk {\set t} \subseteq g^{-1} \sqbrk {t^\succeq}$

We will prove that
 * $\map d t$ is an infimum of $g^{-1} \sqbrk {\set t}$

Thus by Lower Bound is Lower Bound for Subset:
 * $\map d t$ is lower bound for $g^{-1} \sqbrk {\set t}$

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

Thus by definition of min operation:
 * $\map d t = \min \set {g^{-1} \sqbrk {\set t} }$