Quotient Theorem for Epimorphisms

Theorem

Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ 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 $\left({S / \mathcal R_\phi}, {\circ_{\mathcal R_\phi}}\right)$ be the quotient structure defined by $\mathcal R_\phi$.

Then:

• The induced equivalence $\mathcal R_\phi$ is a congruence for $\circ$;
• There is one and only one isomorphism $\psi: \left({S / \mathcal R_\phi}, {\circ_{\mathcal R_\phi}}\right) \to \left({T, *}\right)$ which satisfies $\psi \bullet q_{\mathcal R_\phi} = \phi$.

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

Proof

• First we check that $\mathcal R_\phi$ is compatible with $\circ$.

We note that by definition of induced equivalence:

$x \mathcal R_\phi x' \land y \mathcal R_\phi y' \implies \phi \left({x}\right) = \phi \left({x'}\right) \land \phi \left({y}\right) = \phi \left({y'}\right)$

Then:

	\(\displaystyle \)	\(\displaystyle \phi \left({x \circ y}\right)\)	\(=\)	\(\displaystyle \phi \left({x}\right) * \phi \left({y}\right)\)	\(\displaystyle \)	Morphism Property
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({x'}\right) * \phi \left({y'}\right)\)	\(\displaystyle \)	equality defined above
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({x' \circ y'}\right)\)	\(\displaystyle \)	Morphism Property

Thus $\left({x \circ y}\right) \mathcal R_\phi \left({x' \circ y'}\right)$ by definition of induced equivalence.

So $\mathcal R_\phi$ is compatible with $\circ$.

• 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:

	\(\displaystyle \)	\(\displaystyle \forall x, y \in S: \psi \left({\left[\!\left[{x}\right]\!\right]_{\mathcal R_\phi} \circ_{\mathcal R_\phi} \left[\!\left[{y}\right]\!\right]_{\mathcal R_\phi} }\right)\)	\(=\)	\(\displaystyle \psi \left({\left[\!\left[{x \circ y}\right]\!\right]_{\mathcal R_\phi} }\right)\)	\(\displaystyle \)	Definition of Quotient Structure
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({x \circ y}\right)\)	\(\displaystyle \)	Definition of Epimorphism
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi \left({x}\right) * \phi \left({y}\right)\)	\(\displaystyle \)	by the Morphism Property
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \psi \left({\left[\!\left[{x}\right]\!\right]_{\mathcal R_\phi} }\right) * \psi \left({\left[\!\left[{y}\right]\!\right]_{\mathcal R_\phi} }\right)\)	\(\displaystyle \)	Definition of Quotient Mapping

Therefore $\psi$ is an isomorphism.

Moreover, on the strength of the Quotient Theorem for Surjections, such a $\psi$ is unique.

$\blacksquare$

Notes

Some authors call this the Factor Theorem for Epimorphisms.

Sources

• Seth Warner: Modern Algebra (1965): $\S 12$: Theorem $12.5$