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:
 * $$x \mathcal{R}_\phi x' \land y \mathcal{R}_\phi y' \Longrightarrow \phi \left({x}\right) = \phi \left({x'}\right) \land \phi \left({y}\right) = \phi \left({y'}\right)$$

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 $$\phi$$ is an isomorphism.