# Quotient Mapping on Structure is Epimorphism

## Theorem

Let $\RR$ be a congruence relation on an algebraic structure $\struct {S, \circ}$.

Then the quotient mapping from $\struct {S, \circ}$ to the quotient structure $\struct {S / \RR, \circ_\RR}$ is an epimorphism:

$q_\RR: \struct {S, \circ} \to \struct {S / \RR, \circ_\RR}: \forall x, y \in S: \map {q_\RR} {x \circ y} = \map {q_\RR} x \circ_\RR \map {q_\RR} y$

## Proof

The quotient mapping $q_\RR: S \to S / \RR$ is the canonical surjection from $S$ to $S / \RR$.

Next we show that this is a homomorphism:

 $\ds \map {q_\RR} x$ $=$ $\ds \eqclass x \RR$ Definition of Quotient Mapping $\ds \map {q_\RR} y$ $=$ $\ds \eqclass y \RR$ Definition of Quotient Mapping $\ds \map {q_\RR} {x \circ y}$ $=$ $\ds \eqclass {x \circ y} \RR$ Definition of Quotient Mapping $\ds \eqclass {x \circ y} \RR$ $=$ $\ds \eqclass x \RR \circ_\RR \eqclass y \RR$ Definition of Operation Induced on Quotient Set $\ds \leadsto \ \$ $\ds \map {q_\RR} {x \circ y}$ $=$ $\ds \map {q_\RR} x \circ_\RR \map {q_\RR} y$ Definition of Quotient Mapping

Thus the morphism property is shown to hold.

So the quotient mapping $q_\RR: \struct {S, \circ} \to \struct {S / \RR, \circ_\RR}$ has been shown to be a homomorphism which is a surjection, and is thus an epimorphism.

$\blacksquare$