How can I show the following:
Let $f: M \rightarrow N$ be a morphism in $\text{mod}(A)$, where $A$ is an Artin algebra. Suppose $f \neq 0$. Then there exists a simple module $S$ with its injective hull $I(S)$ and a morphism $q: N \rightarrow I(S)$ such that $qf \neq 0$.
Any help is appreciated!
Let me explain the underlined portion of Lemma 2.2.
Lemma. If $h\colon A\to B$ is a nonzero module homomorphism, then there are a simple module $S$, its injective hull $I_S$, and a map $q\colon B\to I_S$ such that $qh\neq 0$.
Apply this in the proof with $A=P_{S}$, $B=I_{S_r}/S_r$, and $h=pvf$.
Proof of Lemma. Since $h\neq 0$, there is some $a\in A$ such that $b:=h(a)$ is not zero. Let $C\leq B$ be a submodule of $B$ maximal for $b\notin C$. There is a least submodule of $B$ properly containing $C$, and it is $C^*=\langle C\cup \{b\}\rangle$. The natural map $\nu\colon B\to B/C$ is a surjective homomorphism onto a module $B/C$ that is an essential extension of a simple module $S:=C^*/C$. The module $B/C$ embeds into the injective hull $I_S$ of $S$; let this embedding be written $\iota\colon B/C\to I_S$. Now define $q:=\iota \nu\colon B\to I_S$. The composition $qh$ is not zero since $qh(a)=q(b)=\iota\nu(b)\neq 0$, since $\nu(b)\neq 0$ and $\iota$ is an embedding. $\Box$
This is false: Let $A=K[x]/(x^3)$ and $S$ the unique simple $A$-module. The injective hull is $\pi : S \rightarrow A$ with cokernel of dimension 2 and thus not simple.
Is there an assumption missing? (it would help if you cite the article) For example it is true when $I(S)$ is also projective and $A$ has Loewy length 2.
