I have the following question:



> 
> Let $A$ be an Artin algebra. Let $S\_1$ and $S\_2$ be simple modules in $\text{mod}(A)$ and let $P(S\_1)$ be the projective cover of $S\_1$. Let $f: P(S\_1) \rightarrow S\_2$ be module homomorphism with $f \neq 0$. Then $S\_1 \cong S\_2$.
> 
> 
> 


Any help is highly appreciated!



Question on simple modules and projective covers
Here, $P\_S$ is a projective cover of a simple module $S$. This means that there is a surjection $\sigma\colon P\_S\to S$ from projective $P\_S$ onto $S$, which has superfluous kernel $K$. The fact that $P\_S/K\cong S$ is simple implies that $K$ is a maximal submodule of $P\_S$, and the fact that $K$ is superfluous then implies that every proper submodule of $P\_S$ is contained in $K$. In particular, the only simple quotient of $P\_S$ up to isomorphism is $P\_S/K\cong S$.
In the underlined part of Lemma 2.2, $g\colon P\_s\to S\_r$ is a nonzero homomorphism onto a simple module. The kernel must be $K$ and we must have $S\_r\cong P\_S/K\cong S$.




