I am trying to show the right-exactness of a sequence of a sequence of modules, but I am stuck at the last step.
Let $R$ be a ring, and $L, M,N$ be $R$-modules (In the exercise I am trying to solve the ring and the modules are given explicitly, but it wont be important for my question). I want to show that the sequence $L \xrightarrow{f} M \xrightarrow{g}N \rightarrow 0$ is right-exact. I already managed to prove that $g$ is surjective and that $im(F) \subset ker(g)$. There is a hint in the exercise that I am supposed to prove $M / im(f) \cong N$, which I have done, but I don't see how this is enough. Of course from the homomorphism theorem I could get $M / im(f) \cong M / ker(g)$ but even this would not imply $ker(g) = im(f)$ in my opinion.
I am pretty sure that I am already done with the hard part and stuck with a triviality, but I don't get how to formally prove that $M / im(f) \cong N$ implies $ker(g) \subset im(f)$.
I assume that you have proved that $g$ induces an isomorphism $\bar g:M/Im f\rightarrow N$.
Denote by $p:M\rightarrow M/Im f$ the quotient map. Let $x\in Ker g$, $\bar g(p(x))=g(x)=0$ implies that $p(x)=0$ since $\bar g$ is an isomorphism, we deduce that $x\in Im f$ since the kernel of $p$ is $Im f$.