A ring homomorphism $f:A\to B$ is said to have going-down property if the going-down theorem hold for $f(A)$ as a subring of $B$. In the book Introduction to Commutative Algebra by Atiyah and Macdonald, there is an exercise to prove that if $f$ has going-down property, then $\forall q\in\text{Spec}(B)$ the map $f^*:\text{Spec}(B_q)\to\text{Spec}(A_p)$ is surjective for $p=q^c.$
Since $\text{Spec}(A_p)$ are the prime ideals that are contained in $p$, my question is given an arbitrary $p'\subseteq p$, how can we relate it to a prime ideal of $f(A)$ in order to use the going down property? I've tried to use the extension of ideals and the image of ideals, but neither of these ideas help me to get the wanted result. My question is how to prove the surjectivity of $f^*$? Any help is appreciated.
