Renaming Mapping is Bijection

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

Let $r: S / \mathcal R_f \to \operatorname{Im} \left({f}\right)$ be the renaming mapping, defined as:


 * $r: S / \mathcal R_f \to \operatorname{Im} \left({f}\right): r \left({\left[\!\left[{x}\right]\!\right]_{\mathcal R_f}}\right) = f \left({x}\right)$

where:
 * $\mathcal R_f$ is the equivalence induced by the mapping $f$
 * $S / \mathcal R_f$ is the quotient set of $S$ determined by $\mathcal R_f$
 * $\left[\!\left[{x}\right]\!\right]_{\mathcal R_f}$ is the equivalence class of $x$ under $\mathcal R_f$.

The renaming mapping is a bijection.

Proof of Injectivity
To show that $r: S / \mathcal R_f \to \operatorname{Im} \left({f}\right)$ is an injection:

Thus $r: S / \mathcal R_f \to \operatorname{Im} \left({f}\right)$ is an injection.

Proof of Surjectivity
To show that $r: S / \mathcal R_f \to \operatorname{Im} \left({f}\right)$ is a surjection:

Note that for all mappings $f: S \to T$, $f: S \to \operatorname{Im} \left({f}\right)$ is always a surjection from Surjection by Restriction of Codomain.

Thus by definition $\forall y \in \operatorname{Im} \left({f}\right): \exists x \in S: f \left({x}\right) = y$.

Thus $r: S / \mathcal R_f \to \operatorname{Im} \left({f}\right)$ is a surjection.

As $r: S / \mathcal R_f \to \operatorname{Im} \left({f}\right)$ is both an injection and a surjection, it is by definition a bijection.

Different approaches
considers the case where $r$ is an injection, but does not stress its bijective aspects from this particular perspective:


 * This type of factorization of mappings ... is particularly useful when the set of inverse images $\alpha^{-1} \left({a'}\right)$ coincides with $\overline S$; for, in this case, the mapping $\overline a$ is 1-1. Thus if $\overline a \overline \alpha = \overline b \overline \alpha$, then $a \alpha = b \alpha$ and $a \sim b$. Hence $\overline a = \overline b$. Thus we obtain here a factorization $\alpha = \nu \overline \alpha$ where $\overline \alpha$ is 1-1 onto $T$ and $\nu$ is the natural mapping.

Note that in the above, Jacobson uses:
 * $\alpha$ for $f$
 * $a'$ for the image of a representative element $a$ of $S$ under $\alpha$
 * $\overline S$ for $S / \mathcal R_f$
 * $\nu$ for the quotient mapping $q_{\mathcal R_f}: S \to S / \mathcal R_f$
 * $\overline a$ and $\overline b$ for representative elements of $\overline S$
 * $\overline \alpha$ for the renaming mapping $r$.

takes the approach of deducing the existence of the mapping $r$, and then determining under which conditions it is either injective or surjective. From there, the surjective restriction of $r$ is taken, and $\mathcal R$ is then identified with the equivalence induced by $f$.

Hence the bijective nature of $r$ is constructed rather than deduced.