Let $M$ be a left $R-module$. Prove that the following are equivalent:
(a) M is simple;
(b) Every non zero homomorphism $M \to N$ is a monomorphism;
I know that a module $M$ is simple if and only if every cyclic submodule generated by a non-zero element of $M$ equals $M$.I assume (a) but i couldnt go longer. Please some hint.
Edit: (a)->(b): If $M$ is simple then $M$'s only submodules are ${0}$ and $M$. Let $M$ be simple and $\phi:M\to N$ be a non zero homomorphism. $\operatorname{Ker}\phi$ is a submodule of $M$. Then $\operatorname{Ker}\phi=0$ or $\operatorname{Ker}\phi=M$ but it cant be $M$
(b)->(a): Let $N\leq M$ and $\pi:M\to M/N$ be natural homomorphism. Then $\operatorname{Ker}\pi=N$, we know that $\pi$ is monomorphism because of our assumption so $\operatorname{Ker}\pi=\{0\}$. Thus $N=\{0\}$