Renaming Mapping is Well-Defined

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Theorem

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

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

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

where:

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


The renaming mapping is always well-defined.


Proof

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

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

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

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

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

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

$\eqclass x {\mathcal R_f} = \eqclass y {\mathcal R_f}$

then:

$\map r {\eqclass x {\mathcal R_f} } = \map r {\eqclass y {\mathcal R_f} }$


Hence from the definition, we have:

\(\displaystyle y\) \(\in\) \(\displaystyle \eqclass x {\mathcal R_f}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map f y\) \(=\) \(\displaystyle \map f x\)
\(\displaystyle \) \(=\) \(\displaystyle \map r {\eqclass x {\mathcal R_f} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map r {\eqclass y {\mathcal R_f} }\)

Thus $r$ is well-defined.

$\blacksquare$


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