I want to show that
$f$ has the going-down property $\Leftrightarrow$ For any prime ideal $\mathfrak{q}$ of $B$, if $\mathfrak{p}=\mathfrak{q}^c$, then $f^{*}:\textrm{Spec}(B_{\mathfrak{q}}) \rightarrow \textrm{Spec}(A_{\mathfrak{p}})$ is surjective.
I have proved ($\Leftarrow$), but there's something wrong in ($\Rightarrow$).
pf of ($\Rightarrow$): First, I understood \textrm{Spec}(A_{\mathfrak{p}}) = \{\mathfrak{p}' \in \textrm{Spec}(A) | \mathfrak{p}' \subset \mathfrak{p} \}. Let \mathfrak{p}' \subset \mathfrak{p}. Then f(\mathfrak{p}') \subset f(\mathfrak{p}) are prime ideals in $f(A)$. From $f(\mathfrak{p})=f(f^{-1}(\mathfrak{q}))=\mathfrak{q} \cap f(A)$, since f has the going-down property, there exists \mathfrak{q}' \subset \mathfrak{q} such that \mathfrak{q}' \cap f(A) = f(\mathfrak{p}'). Now f^{*}(\mathfrak{q}')=f^{-1}(\mathfrak{q}')=f^{-1}(\mathfrak{q}' \cap f(A))=f^{-1}(f(\mathfrak{p}'))
If \mathfrak{p}' \supset \ker f, then f^{-1}(f(\mathfrak{p}'))=\mathfrak{p}', so the proof is done. But isn't it possible that \mathfrak{p}' does not contain $\ker f$? But $f^{*}$ to be surjective, $\textrm{Spec}(A_{\mathfrak{p}})$ must consists of contracted ideal. Is the problem wrong?