Galois Connection Implies Order on Mappings

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

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

where
 * $\preceq, \precsim$ denote the orders on mappings,
 * $I_S$ denotes the identity mapping of $S$

Proof
Let $s \in S$.

By definition of reflexivity:
 * $g\left({s}\right) \precsim g\left({s}\right)$

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

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

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

Thus by definition of order on mappings:
 * $d \circ g \preceq I_S$

Let $t \in T$.

By definition of reflexivity:
 * $d\left({t}\right) \preceq d\left({t}\right)$

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

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

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

Thus by definition of order on mappings:
 * $I_T \precsim g \circ d$