This question is a follow up to this post: Question about an endomorphsim of modules $N \subset M$ given $\exists f \neq 0$ an endomorphism such that $f(M) \subset N$
I have been working through a bunch of similar exercises and after I had the proof to the above problem explained to me I thought I was in good shape to attack the problem below but I keep getting stuck on choosing the correct submodule.
Let $A$ be a commutative ring with identity and let $M$ be an $A$ module. Suppose for every submodule $N \neq M$ with $N \subset M$ there exits a linear form $x^{*} \in M^{*}$ which is zero on $N$ and surjective,
How do we show that if $ f \in End_A(M)$ is not a right divisor of zero then f is a surjective endomorphism?
I thought the proof would be very similar to the previous problem cited above. But when I consider $N = \operatorname{Im}(f)$ and assume $\operatorname{Im}(f) \neq M$ then I am having trouble getting a contradiction out of the behavior of the linear form on $N$. The next step in the argument I thought would give us a contradiction by considering the fact that $\operatorname{Im}(f) \subset \operatorname{ker}(x^*)$.