# Mapping reflects Preordering

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## 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.

$\blacksquare$

## Sources

- 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: Further exercises: $6$