Renaming Mapping is Well-Defined/Proof 1

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:

Thus $r$ is well-defined.