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:

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:

Therefore $\psi$ is an isomorphism.

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