Quotient Theorem for Surjections

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

Then there is one and only one bijection $r: S / \RR_f \to T$ such that:


 * $r \circ q_{\RR_f} = f$

where:
 * $\RR_f$ is the equivalence induced by $f$
 * $r: S / \RR_f \to T$ is the renaming mapping
 * $q_{\RR_f}: S \to S / \RR_f$ is the quotient mapping induced by $\RR_f$.

This can be illustrated using a commutative diagram as follows:


 * $\begin {xy} \xymatrix@L + 2mu@ + 1em {

S \ar@{-->}[rr]^*{f = r \circ q_{\RR_f} } \ar[dd]_*{q_{\RR_f} } && T \\ \\ S / \RR_f \ar[urur]_*{r} } \end {xy}$

Proof
From the definition of Induced Equivalence, the mapping $f: S \to T$ induces an equivalence $\RR_f$ on its domain.

As $f: S \to T$ is a surjection, $T = \Img f$ by definition.

From Renaming Mapping is Bijection, the renaming mapping $r: S / \RR_f \to T$ is a bijection, where $S / \RR_f$ is the quotient set of $S$ by $\RR_f$.

Hence:
 * $r \circ q_{\RR_f} = f$
 * $r$ is the only mapping $r: S / \RR_f \to T$ that satisfies this equality.

Also known as
Also known as the factor theorem for surjections.

Also see

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


 * Quotient Theorem for Sets