I am studying a proof which shows that a particular $R$-module map $\pi$ is surjective onto a module $M$. The details of the map are complex, so I won't give them here, and will just sketch what I need to ask my question.
Denote the image of $\pi$ by $N$, and the radical of $R$ by $J(R)$, so the radical of $N$ is $J(R)N$ and the radical of $M$ is $J(R)M$. It is given that $J(R)M+N=M$.
The proof then asserts that $J(R)(M/N)\cong (J(R)M+N)/N$.
Once I have this isomorphism, I can finish the proof with Nakayama's Lemma, but I can't get this step to work.
What I have tried is, working from the right hand side:
$(J(R)M+N)/N\cong J(R)M/(J(R)M\cap N)$ by the second isomorphism theorem for modules. Now, $J(R)M\cap N=J(R)N$ so we can rewrite this as $(J(R)M+N)/N\cong (J(R)M)/(J(R)N)$.
So, if I could show that $J(R)(M/N)\cong (J(R)M)/(J(R)N)$, then I'd be done, but I don't know why this is true.
I think you're overthinking it. The fact is that $J(R)(M/N)= (J(R)M+N)/N$ as sets (even stronger than isomorphism.)
Clearly $J(R)M\subseteq J(R)M+N$, so the $\subseteq$ containment holds.
Now what does an element of the right side look like: $(jm+n)+N$ for $j\in J(R), m\in M $ and $n\in N$. But this is equal to $jm+N=j(m+N)\in J(R)(M/N)$. So the $\supseteq$ contaiment holds as well.