# Renaming Mapping is Well-Defined

## 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 1

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.

$\blacksquare$

## Proof 2

From Condition for Mapping from Quotient Set to be Well-Defined:

- there exists a mapping $\phi: S / \RR \to T$ such that $\phi \circ q_\RR = f$

- $\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$

But by definition of the equivalence induced by the mapping $f$:

- $\forall x, y \in S: \tuple {x, y} \in \RR_f \implies \map f x = \map f y$

The result follows directly.

$\blacksquare$

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $2$