Let's say I have two commutative rings $A, B$ and a ring epimorphism $\phi : A \to B$. this induces a continuous map $\phi^* : \operatorname{Spec}(B) \to \operatorname{Spec}(A)$. Let $f \in B$ and choose $x \in \phi^{-1}(f)$, is it true that for any $\mathfrak{q} \in \overline{V(\{x\})}$ there exists a $\mathfrak{p} \in \overline{V(\{f\})}$ such that $\mathfrak{q} = \phi^{-1}(\mathfrak{p})$? Note that $\overline{V(\{x\})}$ and $\overline{V(\{f\})}$ denote the complements of $V(\{x\})$ and $V(\{f\})$ respectively.
If this statement is true, then how can I prove this? I don't immediately see any way to prove this since the only things we know about $\mathfrak{q}$ is that it is a prime ideal of $A$ that doesn't contain $x$, so I don't see any way I could tell what $\mathfrak{p}$ would even be given so little information.
This is not true, the morphism $\phi^*$ is injective, but not necessarily surjective. Take $f=0$ and $x=0$, $V(0)=Spec(A)$. Then if $\phi^*$ is not surjective take any $q$ which is not in its image.
Example $\phi:\mathbb{Z}\rightarrow\mathbb{Z}/p$ where $p$ is a prime.