Lower Adjoint at Element is Minimum of Preimage of Singleton of Element implies Composition is Identity

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

Let $g: S \to T, d: T \to S$ be mappings such that
 * $\forall t \in T: \map d t = \min \set {g^{-1} \sqbrk {\set t} }$

Then $g \circ d = I_T$

where $I_T$ denotes the identity mapping of $T$.

Proof
Let $t \in T$.

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

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

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

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

Thus by definition of composition of mappings:
 * $\map {\paren {g \circ d} } t = t$