Renaming Mapping is Bijection

Theorem
The Renaming Mapping is a bijection.

Proof

 * 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


 * 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 Range.

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.

Comment
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 $$r$$.