Suppose $M$ and $N$ are simple modules over a commutative ring $R$, and $P_N$ is the projective cover of $N$. If there is a nonzero morphism $P_N\to M$, does this imply $N\cong M$?
I'm not too familiar with projective covers other than the definition I read on the stacks project. Since $M$ is simple, the morphism $P_N\to M$ is an epimorphism. Since $M$ and $N$ are simple, I know it's enough just to have some nonzero map between them.
Let $K\subset P_N$ be the kernel of the map $f:P_N\to M$. Since $p:P_N\to N$ is a projective cover, $p(K)$ must be a proper submodule of $N$, and hence $0$ since $N$ is simple. It follows that $p$ factors through $P_N/K=M$ to give a nonzero map $M\to N$.