Surjection on Total Ordering reflects Preordering

Theorem
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.

Let $f: S \to T$ be a mapping to an arbitrary set $T$.

Let $\RR$ be a relation on $T$ defined such that:
 * $\RR: = \set {\tuple {\map f x, \map f y}: x \preccurlyeq y}$

That is, $a$ is related to $b$ in $T$ they have preimages $x$ and $y$ under $f$ such that $x$ precedes $y$.

Then:
 * $f$ is a surjection onto set $T$


 * $\RR$ is a preordering on $T$
 * $\RR$ is a preordering on $T$

Sufficient Condition
Let $f: S \to T$ be a surjection.

By definition, a preordering is a relation which is both transitive and reflexive.

From Mapping on Total Ordering reflects Transitivity, $\RR$ is a transitive relation.

It remains to be shown that $\RR$ is a reflexive relation.

Let $a \in T$.

Because $f$ is a surjection, it follows that:
 * $\exists x \in S: a = \map f x$

Because $\preccurlyeq$ is a total ordering on $S$, it follows that:
 * $x \preccurlyeq x$

Hence by definition of $\RR$:
 * $\map f x \mathrel \RR \map f x$

That is:
 * $a \mathrel \RR a$

As $a$ was arbitrary, it follows that $\RR$ is reflexive.

Hence it follows that $\RR$ is a preordering.

Necessary Condition
Let $\RR$ be a preordering on $T$.

Then $\RR$ is a relation which is both transitive and reflexive.

Let $a \in T$.

As $\RR$ is reflexive, it follows that:
 * $\forall a \in T: a \mathrel \RR a$

That is:
 * $\forall a \in T: \exists x \in S: \map f x \mathrel \RR \map f x$

where $a = \map f x$

That means:
 * $\forall a \in T: \exists x \in S: a = \map f x$

That is:
 * $f$ is a surjection.