# Relation Induced by Mapping is Equivalence Relation

## Theorem

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

Let $\RR_f \subseteq S \times S$ be the relation induced by $f$:

$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$

Then $\RR_f$ is an equivalence relation.

## Proof

We need to show that $\RR_f$ is an equivalence relation.

Checking in turn each of the criteria for equivalence:

### Reflexive

$\RR_f$ is reflexive:

$\forall x \in S: \map f x = \map f x \implies x \mathrel {\RR_f} x$

$\Box$

### Symmetric

$\RR_f$ is symmetric:

 $\ds x$ $\RR_f$ $\ds y$ by definition $\ds \leadsto \ \$ $\ds \map f x$ $=$ $\ds \map f y$ Definition of $f$ $\ds \leadsto \ \$ $\ds \map f y$ $=$ $\ds \map f x$ Equality is Equivalence Relation, and so Symmetric $\ds \leadsto \ \$ $\ds y$ $\RR_f$ $\ds x$ Definition of $f$

$\Box$

### Transitive

$\RR_f$ is transitive:

 $\ds x$ $\RR_f$ $\ds y$ $\, \ds \land \,$ $\ds y$ $\RR_f$ $\ds z$ $\ds \leadsto \ \$ $\ds \map f x$ $=$ $\ds \map f y$ Definition of $f$ $\, \ds \land \,$ $\ds \map f y$ $=$ $\ds \map f z$ Definition of $f$ $\ds \leadsto \ \$ $\ds \map f x$ $=$ $\ds \map f z$ Equality is Equivalence Relation, and so Transitive $\ds \leadsto \ \$ $\ds x$ $\RR_f$ $\ds z$ Definition of $f$

$\Box$

Thus $\RR_f$ is reflexive, symmetric and transitive, and is therefore an equivalence relation.

$\blacksquare$