Relation Induced by Mapping is Equivalence Relation

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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 $f$
\(\ds \leadsto \ \ \) \(\ds \map f x\) \(=\) \(\ds \map f y\) 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.

$\blacksquare$


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