Quotient Theorem for Sets

Theorem
Any mapping $$f: S \to T$$ can be factored uniquely into a surjection, followed by a bijection, followed by an injection.

Thus:
 * $$f = i \circ r \circ q_{\mathcal{R}_f}$$

where:


 * $$q_{\mathcal{R}_f}: S \to S / \mathcal{R}_f : q_{\mathcal{R}_f} \left({s}\right) = \left[\!\left[{s}\right]\!\right]_{\mathcal{R}_f}$$
 * $$r: S / \mathcal{R}_f \to \operatorname{Im} \left({f}\right) : r \left({\left[\!\left[{s}\right]\!\right]_{\mathcal{R}_f}}\right) = f \left({s}\right)$$
 * $$i: \operatorname{Im} \left({f}\right) \to T : i \left({t}\right) = t$$

Otherwise known as the factoring theorem or factor theorem.

Proof
From Factoring Mapping into Surjection and Inclusion, $$f$$ can be factored uniquely into:


 * A surjection $$g: S \to \operatorname{Im} \left({f}\right)$$, followed by:
 * The inclusion mapping $$i: \operatorname{Im} \left({f}\right) \to T$$ (an injection).

From the Quotient Theorem for Surjections, the surjection $$g$$ can be factored uniquely into:
 * The quotient mapping $$q_{\mathcal{R}_f}: S \to S / \mathcal{R}_f$$ (a surjection), followed by:
 * The renaming mapping $$r: S / \mathcal{R}_f \to \operatorname{Im} \left({f}\right)$$ (a bijection).

Thus:
 * $$f = i \circ \left({r \circ q_{\mathcal{R}_f}}\right)$$

By associativity of composition of mappings it can be seen that $$f = i \circ r \circ q_{\mathcal{R}_f}$$.