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

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.