Image of Preimage of Subring under Ring Epimorphism

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

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

Then:
 * $\phi \left({\phi^{-1} \left({S_2}\right)}\right) = S_2$

Proof
As $\phi$ is an epimorphism, it is a surjection, and so $\operatorname{Im} \left({\phi}\right) = R_2$.

So $S_2 \subseteq \operatorname{Im} \left({R_1}\right)$.

The result then follows from Image of Preimage of Mapping.