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

Theorem
Let $L = \left({S, \preceq}\right), R = \left({T, \precsim}\right)$ be ordered sets.

Let $g:S \to T, d:T \to S$ be mappings such that
 * $\forall t \in T: d\left({t}\right) = \min\left({g^{-1}\left[{\left\{ {t}\right\} }\right]}\right)$

Then $g \circ d = I_T$

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

Proof
Let $t \in T$.

By assumption:
 * $d\left({t}\right) = \min\left({g^{-1}\left[{\left\{ {t}\right\} }\right]}\right)$

By definition of minimum:
 * $d\left({t}\right) = \inf\left({g^{-1}\left[{\left\{ {t}\right\} }\right]}\right)$ and $d\left({t}\right) \in g^{-1}\left[{\left\{ {t}\right\} }\right]$

By definition of image of set:
 * $g\left({d\left({t}\right)}\right) \in \left\{ {t}\right\}$

By definition of singleton:
 * $g\left({d\left({t}\right)}\right) = t$

Thus by definition of composition of mappings:
 * $\left({g \circ d}\right)\left({t}\right) = t$