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.