Renaming Mapping is Well-Defined/Proof 2

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

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

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

if and only if:

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

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: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.6$