Quotient Theorem for Sets

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

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

where:


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

This can be illustrated using a commutative diagram as follows:


 * QuotientTheoremForSets.png

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

As Composition of Mappings is Associative it can be seen that $f = i \circ r \circ q_{\mathcal R_f}$.

Comment
This construction is known as the canonical decomposition of $f$.

Also see

 * Factoring Mapping into Quotient and Injection
 * Factoring Mapping into Surjection and Inclusion


 * Quotient Theorem for Surjections