Increasing and Ordering on Mappings implies Mapping is Composition

Theorem

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

Let $g: S \to T, d: T \to S$ be mappings such that

$g$ and $d$ are increasing mappings

and

$d \circ g \preceq I_S$ and $I_T \preccurlyeq g \circ d$

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

Then $d = d \circ \paren {g \circ d}$ and $g = \paren {g \circ d} \circ g$

Proof

Let $t \in T$.

By definition of ordering on mappings:

$\map {I_T} t \preccurlyeq \map {\paren {g \circ d} } t$

By definition of identity mapping:

$t \preccurlyeq \map {\paren {g \circ d} } t$

By definition of increasing mapping:

$\map d t \preceq \map d {\map {\paren {g \circ d} } t}$

By definition of composition of mappings:

$\map d t \preceq \map {\paren {d \circ \paren {g \circ d} } } t$

By definition of ordering on mappings:

$\map {\paren {d \circ g} } {\map d t} \preceq \map {I_S} {\map d t}$

By definition of identity mapping:

$\map {\paren {d \circ g} } {\map d t} \preceq \map d t$

By definition of composition of mappings:

$\map {\paren {\paren {d \circ g} \circ d} } t \preceq \map d t$

By Composition of Mappings is Associative:

$\map {\paren {d \circ \paren {g \circ d} } } t \preceq \map d t$

Thus by definition of antisymmetry:

$\map d t = \map {\paren {d \circ \paren {g \circ d} } } t$

$\Box$

Thus $g = \paren {g \circ d} \circ g$ holds by mutatis mutandis.

$\blacksquare$

Sources

• Mizar article WAYBEL_1:20