The exercise is
Let $A$ is a ring, $S$ a multiplicatively closed subset of $A$, and $\phi:A \rightarrow S^{-1}A $ the canonical homomorphism. Show $\phi^{*}:\operatorname{Spec}(S^{-1}A)\rightarrow \operatorname{Spec}(A)$ is a homeomorphism from $\operatorname{Spec}(S^{-1}A)$ onto its image in $X=Spec(A)$
I have proved that $\phi^{*}$ is injective ,and since $\phi^{*}$ is continuous ,then it is enough to prove $\phi^{*}$ is closed map or open map. And I want to prove it is a closed map. Suppose $\alpha $ is an ideal of $S^{-1}A$ ,then we know $\overline{\phi^{*}(V(\alpha))}=V(\alpha^{\mathrm c})$. So it is enough to prove ${\phi^{*}(V(\alpha))}=V(\alpha^{\mathrm c})$. And
$p\in \phi^{*}(V(\alpha))\Leftrightarrow \ \exists q\in V(\alpha)$,such that $p=\phi^{*}(q)$ $\Leftrightarrow \exists\, q,\alpha \subseteq q,p=q^{\mathrm c}\Rightarrow \alpha^{\mathrm c}\subseteq p$
But I cannot prove that $ \exists\, q,\alpha \subseteq q,\,p=q^{\mathrm c}\Leftarrow \alpha^{\mathrm c}\subseteq p$. And even I think it is may be not true.
Are there mistakes in my proof ?