# Quotient Theorem for Epimorphisms

## Contents

## Theorem

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.

Let $\mathcal R_\phi$ be the equivalence induced by $\phi$.

Let $S / \mathcal R_\phi$ be the quotient of $S$ by $\mathcal R_\phi$.

Let $q_{\mathcal R_\phi}: S \to S / \mathcal R_\phi$ be the quotient mapping induced by $\mathcal R_\phi$.

Let $\struct {S / \mathcal R_\phi}, {\circ_{\mathcal R_\phi} }$ be the quotient structure defined by $\mathcal R_\phi$.

Then:

- The induced equivalence $\mathcal R_\phi$ is a congruence relation for $\circ$
- There is one and only one isomorphism $\psi: \struct {S / \mathcal R_\phi}, {\circ_{\mathcal R_\phi} } \to \struct {T, *}$ which satisfies $\psi \bullet q_{\mathcal R_\phi} = \phi$.

where, in order not to cause notational confusion, $\bullet$ is used as the symbol to denote composition of mappings.

## Proof

### Proof of Congruence Relation

Let $x, x', y, y' \in S$ such that:

- $x \mathop {\mathcal R_\phi} x' \land y \mathop {\mathcal R_\phi} y'$

By definition of induced equivalence:

\(\displaystyle x \mathop {\mathcal R_\phi} x'\) | \(\implies\) | \(\displaystyle \map \phi x = \map \phi {x'}\) | |||||||||||

\(\displaystyle y \mathop {\mathcal R_\phi} y'\) | \(\implies\) | \(\displaystyle \map \phi y = \map \phi {y'}\) |

Then:

\(\displaystyle \map \phi {x \circ y}\) | \(=\) | \(\displaystyle \map \phi x * \map \phi y\) | Definition of Epimorphism (Abstract Algebra) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi {x'} * \map \phi {y'}\) | equality shown above | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi {x' \circ y'}\) | Definition of Epimorphism (Abstract Algebra) |

Thus $\paren {x \circ y} \mathop {\mathcal R_\phi} \paren {x' \circ y'}$ by definition of induced equivalence.

So $\mathcal R_\phi$ is a congruence relation for $\circ$.

$\Box$

### Proof of Unique Isomorphism

From the Quotient Theorem for Surjections, there is a unique bijection from $S / \mathcal R_\phi$ onto $T$ satisfying $\psi \bullet q_{\mathcal R_\phi} = \phi$.

Also:

\(\displaystyle \forall x, y \in S: \map \psi {\eqclass x {\mathcal R_\phi} \circ_{\mathcal R_\phi} \eqclass y {\mathcal R_\phi} }\) | \(=\) | \(\displaystyle \map \psi {\eqclass {x \circ y} {\mathcal R_\phi} }\) | Definition of Quotient Structure | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi {x \circ y}\) | Definition of Epimorphism (Abstract Algebra) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi x * \map \phi y\) | Definition of Epimorphism (Abstract Algebra) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \psi {\eqclass x {\mathcal R_\phi} } * \map \psi {\eqclass y {\mathcal R_\phi} }\) | Definition of Quotient Mapping |

Therefore $\psi$ is an isomorphism.

$\blacksquare$

## Also known as

Some authors call this the **Factor Theorem for Epimorphisms**.

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 12$: Theorem $12.5$