Why does faithful module induce injection between Hom?

Question

This is from a paper of Matsumura:

The line that I don't understand is the faithfulness of $E$ implies injection $\text{Hom}(D,D) \rightarrow \text{Hom}(E,D)$. Why is this true?

Thanks in advance.

Answer 1

Presumably, $\operatorname{Hom}_A(D,D)=A$ means that the obvious map $A\to \operatorname{Hom}_A(D,D)$ is an isomorphism, so that every homomorphism $D\to D$ is given by multiplication by some element of $A$. Since the restriction map $\operatorname{Hom}_A(D,D)\to\operatorname{Hom}_A(E,D)$ is surjective, this means the same is true for homomorphisms $E\to D$: all of them are given by multiplication by some element of $A$. Since $E$ is faithful, multiplication by any nonzero element of $A$ is nonzero on $E$, and so the map $\operatorname{Hom}_A(D,D)\to\operatorname{Hom}_A(E,D)$ is injective (since all that map does is take multiplication by an element $a\in A$ on $D$ and restrict it to $E$).