Increasing and Ordering on Mappings implies Mapping is Composition

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

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

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

where $\preceq, \precsim$ denotes the orderings on mappings.

Then $d \circ \left({g \circ d}\right)$ and $g = \left({g \circ d}\right) \circ g$

Proof
Let $t \in T$.

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

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

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

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

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

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

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

By Composition of Mappings is Associative:
 * $\left({d \circ \left({g \circ d}\right)}\right)\left({t}\right) \preceq d\left({t}\right)$

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

Thus $g = \left({g \circ d}\right) \circ g$ holds by mutatis mutandis.