Given $f$ a ring epimorphism and $\alpha$ an ideal then I need to proof that $f(\sqrt{\alpha}) = \sqrt{\alpha^e}$ where $\sqrt{}$ denotes the radical of the ideal and $\alpha^e = \langle f(\alpha) \rangle$ is the extended ideal. Here by surjectivity one does not need extension.
I proved in general the $\subseteq$ part, however, the other seems more tricky.
The statement is wrong. Consider the ring homomorphism $f:\Bbb Z \to \Bbb Z/4\Bbb Z$ and $\alpha=(0)$, then $\alpha^e=(\overline{0})$ and we get $f(\sqrt{(0)})=f((0))=(\overline{0})$, but $\sqrt{(\overline{0})}=(\overline{2})$.