# Renaming Mapping is Well-Defined/Proof 1

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## 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.

$\blacksquare$

## Sources

- 1951: Nathan Jacobson:
*Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts*... (previous) ... (next): Introduction $\S 3$: Equivalence relations - 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Factoring Functions: Theorem $10$