Statement (False) : Let $\phi: A \rightarrow B$ be a ring homomorphsim , $\phi^*: \text{Spec}(B) \rightarrow \text{Spec}(A)$. Also, $\operatorname{Int}$ is a topological interior set and for $f \in A$ , $X_f:=\{ \mathfrak{p} \in \text{Spec} : f \notin \mathfrak{p} \} $. Then, is it true $\operatorname{Int}( \phi^*(Y_{\phi(f)}))=X_f$ ,where $X=\text{Spec}(A) , Y=\text{Spec}(B)$ ?
[Question] Is it true? If it is false, is there some true statements which is associated with image $\phi^*$ of open set?
[Motivation]
I know that three facts as followings
(i) For $f \in A$, $ {\phi^*}^{-1}(X_f)=Y_{\phi{(f)}}$ .
(ii) For an ideal $\mathfrak{a}$ of $A$, ${\phi^*}^{-1}(V(\mathfrak{a}))=V(\mathfrak{a}B)$ ,where $V(\mathfrak{a}) := \{ \mathfrak{p} \in \text{Spec} : \mathfrak{a} \subseteq \mathfrak{p} \} $
(iii) For an ideal $\mathfrak{b}$ of $B$, $\overline{{\phi^*}(V(\mathfrak{b}))}=V({\phi}^{-1}(\mathfrak{b}))$
Since for $f \in A$ , $X_f$ is a basic open set, (i) is associated with the inverse image of open sets.
Also, (ii) , (iii) are associated with the image and inverse image of closed sets.
I hope that there is a statement of image $\phi^*$ of open sets.
[EDIT]
By comment of KReiser, I see now above statement is false. In fact, take $A=k[x,y]$ , $B=^{k[x,y]}/ _{(y)}$ , $f=1$ , $\phi:A \rightarrow B$ is a natural projection map . Then, take natural mapping ${\phi}^* : \mathbb{A} \rightarrow \mathbb{A}^2$ and Im ${\phi}^*$= $V((y))$ , so $\operatorname{Int} (\text{Im} {\phi}^*) = \operatorname{Int} ({\phi}^*(Y_{\phi(1)})) = \emptyset $
However, I hope to find a true statement related to $\phi^*(Y_{\phi(f)})$