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!



Question on injective hulls
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. $\Box$








