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.

$\blacksquare$

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

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Theorem $10.5$
• 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.4 \ \text{(b)}$