Quotient Epimorphism is Epimorphism/Ring

Theorem
Let $\left({R, +, \circ}\right)$ be a ring whose zero is $0_R$ and whose unity is $1_R$.

Let $J$ be an ideal of $R$.

Let $\left({R / J, +, \circ}\right)$ be the quotient ring defined by $J$.

Let $\phi: R \to R / J$ be the natural (ring) epimorphism from $R$ to $R / J$:
 * $x \in R: \phi \left({x}\right) = x + J$

Then $\phi$ is a ring epimorphism whose kernel is $J$.

Proof
The fact that $\phi$ is a homomorphism can be verified easily.

$\phi$ is surjective because $\forall x + J \in R / J: x + J = \phi \left({x}\right)$.

Therefore $\phi$ is an epimorphism.

Let $x \in \ker \left({\phi}\right)$.

Then:

Thus $\ker \left({\phi}\right) = J$.