# Renaming Mapping is Well-Defined/Proof 1

## Theorem

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

Let $r: S / \RR_f \to \Img f$ be the renaming mapping, defined as:

$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$

where:

$\RR_f$ is the equivalence induced by the mapping $f$
$S / \RR_f$ is the quotient set of $S$ determined by $\RR_f$
$\eqclass x {\RR_f}$ is the equivalence class of $x$ under $\RR_f$.

The renaming mapping is always well-defined.

## Proof

By Relation Induced by Mapping is Equivalence Relation, we have that $\RR_f$ is an equivalence relation.

To determine whether $r$ is well-defined, we have to determine whether $r: S / \RR_f \to \Img f$ actually defines a mapping at all.

Consider a typical element $\eqclass x {\RR_f}$ of $S / \RR_f$.

Suppose we were to choose another name for the class $\eqclass x {\RR_f}$.

Assume that $\eqclass x {\RR_f}$ is not a singleton.

For example, let us choose $y \in \eqclass x {\RR_f}, y \ne x$ such that:

$\eqclass x {\RR_f} = \eqclass y {\RR_f}$

then:

$\map r {\eqclass x {\RR_f} } = \map r {\eqclass y {\RR_f} }$

Hence from the definition, we have:

 $\ds y$ $\in$ $\ds \eqclass x {\RR_f}$ $\ds \leadsto \ \$ $\ds \map f y$ $=$ $\ds \map f x$ $\ds$ $=$ $\ds \map r {\eqclass x {\RR_f} }$ $\ds$ $=$ $\ds \map r {\eqclass y {\RR_f} }$

Thus $r$ is well-defined.

$\blacksquare$