Suppose $f \in End_{R}(M)$ is an $R$-module endomorphism. If $f$ is surjective then $f$ is not a right divisor of zero in $End_{R}(M)$ I am having trouble thinking about a practice problem that I feel should be pretty easy because it is part 1 of a 4 part problem.   I am probably overthinking it because I keep trying to construct maps of induced sequences of Hom since this comes from a section where we have introduced the notion of projectivity and injectivity for modules.
Let $R$ be a commutative ring with identity

Suppose $f \in End_{R}(M)$ is an $R$-module endomorphism.  If $f$ is surjective then $f$ is not a right divisor of zero in $End_{R}(M)$.  Conversely if or every submodule $N \neq M$ with $N \subset M$ there exits a linear form $x^{*} \in E^{*}$ which is zero on $N$ and surjective, every element of $End_{R}(M)$ which is not a right divisor of zero is a surjective endomorphism.

Does the first half of the problem follow simply from the fact that $f$ surjective implies $f^*$ injective?  There is a similar statement for injective endomorphsisms in the problem set but I cannot seem to come up with the ideas for either I was also wondering if there where any good texts or lecture notes where they cover module endomorphisims in a way that expalins these  type of problems clearly.

In the category of modules you can prove that a morphism is an epimorphism iff is surjective, where epimorphism means right cancellable. So consider another morphism $g$ that $gf=0$ and the fact that $0f=0$ gives you $gf=0f$ now it is right cancellable so $g=0$.So it isn't a right divisor of zero.
