Two projections of a ring which define the same set-theoretic map of the the Spec

Question

Suppose I have a commutative ring $A$ and an ideal $I$. And I have the projection $\pi: A \to A/I$ and suppose I have another ring morphism $\tau: A \to A/I$ and that $\pi^{-1} (P) = \tau^{-1}(P)$ for all primes $P$ of $A/I$. I'm wondering does this imply that $\pi = \tau$ or not? Either a proof or a counter example would be appreciated. Thank you!

Answer 1

This is false. For instance, if $A$ is a field and $I=0$, then $\pi$ is the identity map $A\to A/0=A$ but any other homomorphism $\tau:A\to A$ will also send the unique prime ideal $0$ to itself.