Mapping reflects Preordering

Theorem

Let $S$ and $T$ be sets.

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

Let ${\precsim} \subseteq T \times T$ be a preordering on $T$.

Let $\mathcal R$ be the relation defined on $S$ by the rule:

$x \mathrel {\mathcal R} y \iff \map f x \precsim \map f y$

Then $\mathcal R$ is a preordering on $S$.

Proof

Reflexivity

 $\displaystyle x$ $\in$ $\displaystyle S$ $\displaystyle \leadsto \ \$ $\displaystyle \map f x$ $\in$ $\displaystyle T$ Definition of $f$ $\displaystyle \leadsto \ \$ $\displaystyle \map f x$ $\precsim$ $\displaystyle \map f x$ as $\precsim$ is a preordering on $T$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $\mathcal R$ $\displaystyle x$ by definition of $\mathcal R$

Thus $\mathcal R$ is reflexive.

$\Box$

Transitivity

 $\displaystyle x$ $\mathcal R$ $\displaystyle y$ $\, \displaystyle \land \,$ $\displaystyle y$ $\mathcal R$ $\displaystyle z$ $\displaystyle \leadsto \ \$ $\displaystyle \map f x$ $\precsim$ $\displaystyle \map f y$ $\, \displaystyle \land \,$ $\displaystyle \map f y$ $\precsim$ $\displaystyle \map f z$ by definition of $\mathcal R$ $\displaystyle \leadsto \ \$ $\displaystyle \map f x$ $\precsim$ $\displaystyle \map f z$ as $\precsim$ is a preordering on $T$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $\mathcal R$ $\displaystyle z$ by definition of $\mathcal R$

Thus $\mathcal R$ is transitive.

$\Box$

So, by definition, $\mathcal R$ is a preordering.

$\blacksquare$