Exact Sequences of R-Modules

Question

In "A Course in Ring Theory by Passman" it is mentioned, "But the kernel of the combined epimorphism $P\rightarrow B\rightarrow C$ is clearly equal to $E$". I don't understand this part. How can the kernel of that combined epimorphism equal to $E$. I hope someone can shed light on this matter.

Answer 1

Let $\psi : B\to C$ and $\pi : P \to B$ be the relevant maps, then $$ \ker(\psi\circ\pi) = \{x \in P : \psi(\pi(x)) = 0\} = \{x\in P : \pi(x) \in \ker(\psi)\} $$ Since $0 \to A\hookrightarrow B\xrightarrow{\psi} C \to 0$ is exact, $\ker(\psi) = A$, hence $$\ker(\psi\circ\pi) = \pi^{-1}(A)$$ which is $E$ by definition.