Let $A$ and $B$ be commutative rings with $1 \neq 0$. Let $\varphi$ be a ring homomorphism with $\varphi(1) = 1$. We consider the extension and contraction of $\varphi$. Let $P \subset A$ be a prime ideal that satisfies that $\varphi^{-1}(\varphi(P)B) = P$. We want to prove that there exists such a prime ideal $P' \subset B$ that $\varphi^{-1}(P') = P$.
I am considering that we can use localization of $B$. But I have not come up with a good submonoid $D$ of $B$ at which we localize $B$. For example, if we set $D = \varphi(A \setminus P)$, I am not convinced that we can prove that $D \cap \varphi(P)B = \emptyset$ since, in general, it is not necessarily true that $f(X \cap Y) = f(X) \cap f(Y)$.
I appreciate it if you could tell me a correct solution.
P.S. I am working on the same question here. But my question above has not been resolved. In addition, seeing this, I fixed our main claim.
The answer for my concern is here. For a more detailed memo for me, here I write a solution in my word.
Since $A \setminus P = A \setminus \varphi^{-1}(P^{e}) = \varphi^{-1}(B \setminus P^{e})$, we have $\varphi(A \setminus P) = \varphi(\varphi^{-1}(B \setminus P^{e})) \subset B \setminus P^{e}$. Hence, $\varphi(A \setminus P) \cap P^{e} = \emptyset$. Localizing $B$ at $D = \varphi(A \setminus P)$, we obtain a proper ideal $D^{-1}(P^{e})$ in $D^{-1}B$.
Let $M$ be a maximal ideal of $B$ containing $D^{-1}(P^{e})$ and $P'$ be a pre-image of $M$. Since $M$ is a prime ideal of $D^{-1}B$, we see that $P'$ is a prime ideal of $B$ with $P' \cap D = \emptyset$, which implies that $\varphi^{-1}(P') \subset P$. In addition, we also have $\varphi^{-1}(P') \supset P$ since $D^{-1}(P^e) \subset M$, completing the proof.