# Quotient Theorem for Surjections

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## Contents

## Theorem

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

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

- $r \circ q_{\mathcal R_f} = f$

where:

- $\mathcal R_f$ is the equivalence induced by $f$
- $r: S / \mathcal R_f \to T$ is the renaming mapping
- $q_{\mathcal R_f}: S \to S / \mathcal R_f$ is the quotient mapping induced by $\mathcal R_f$.

This can be illustrated using a commutative diagram as follows:

- $\begin {xy} \[email protected] + [email protected] + 1em { S \[email protected]{-->}[rr]^*{f = r \circ q_{\mathcal R_f} } \ar[dd]_*{q_{\mathcal R_f} } && T \\ \\ S / \mathcal R_f \ar[urur]_*{r} } \end {xy}$

## Proof

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

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

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

Hence:

- $r \circ q_{\mathcal R_f} = f$
- $r$ is the only mapping $r: S / \mathcal R_f \to T$ that satisfies this equality.

$\blacksquare$

## Also known as

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

## Also see

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 10$: Theorem $10.5$ - 2000: James R. Munkres:
*Topology*(2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.4 \ \text{(b)}$