Ordering on Mappings Implies Galois Connection

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

Let $g: S \to T$ and $d: T \to S$ be mappings such that:
 * $g$ and $d$ are increasing mappings

and
 * $d \circ g \preceq I_S$ and $I_T \precsim g \circ d$

Then
 * $\left({g, d}\right)$ is Galois connection.

where
 * $\preceq, \precsim$ denote the orderings on mappings
 * $I_S$ denotes the identity mapping of $S$
 * $\circ$ denotes the composition of mappings.

Proof
We will prove that
 * $\forall s \in S, t \in T: t \precsim g\left({s}\right) \iff d\left({t}\right) \preceq s$

Let $s \in S, t \in T$.

First implication:

Let
 * $t \precsim g\left({s}\right)$

By definition of increasing mapping:
 * $d\left({t}\right) \preceq d\left({g\left({s}\right)}\right)$

By definition of ordering on mappings:
 * $\left({d \circ g}\right)\left({s}\right) \preceq I_S\left({s}\right)$

By definition of composition:
 * $d\left({g\left({s}\right)}\right) \preceq I_S\left({s}\right)$

By definition of identity mapping:
 * $d\left({g\left({s}\right)}\right) \preceq s$

Thus by definition of transitivity:
 * $d\left({t}\right) \preceq s$

Second implication:

Let
 * $d\left({t}\right) \preceq s$

By definition of increasing mapping:
 * $g\left({d\left({t}\right)}\right) \precsim g\left({s}\right)$

By definition of ordering on mappings:
 * $I_T\left({t}\right) \precsim \left({g \circ d}\right)\left({t}\right)$

By definition of composition:
 * $I_T\left({t}\right) \precsim g\left({d\left({t}\right)}\right)$

By definition of identity mapping:
 * $t \precsim g \left({d\left({t}\right)}\right)$

Thus by definition of transitivity:
 * $t \precsim g \left({s}\right)$

Thus by definition:
 * $\left({g, d}\right)$ is Galois connection.