How can $A/P^c$ be viewed as a subring of $B/P$?

Question

Let $f : A \to B$ be a commutative ring hom. If for ideal $J$ in $B$ we define $J^c = f^{-1}(J)$ and we let $P$ be a prime ideal, then how is $A/P^c$ viewed as a subring of $B/P$? Especially since $A \to B$ is not necess. injective.

This is out of Matsumura's Commutative Algebra.

Answer 1

$A\to B$ isn't injective. $A/P^c\to B/P$, on the other hand, is.

Consider the composite homomorphism $A\to B\to B/P$, and what the kernel of this homomorphism is.