Image of Preimage of Subring under Ring Epimorphism

Theorem
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.

Let $S_2$ be a subring of $R_2$.

Then:
 * $\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$

Proof
As $\phi$ is an epimorphism, it is a surjection, and so $\Img \phi = R_2$.

So $S_2 \subseteq \Img {R_1}$.

The result then follows from Image of Preimage under Mapping.