Let $A$ be a commutative unitary ring, $I$ an ideal contained in the Jacobson radical of $A$; let $M$ be an $A$-module and $N$ a finitely generated $A$-module, and let $u: M \to N$ be a homomorphism.
If the induced homomorphism $\bar u: M/ IM\to N/IN$ is surjective, then $u$ is surjective.
Some approaches that may be used:
1) If $M/ IM = 0$ then by Nakayama Lemma $M=0$. Now if we can show somehow that $(N/u(M))/I(N/u(M))=0$ then we are done. Since $N/u(M)=0 \implies u(M)=N$.
To use the Nakayama Lemma we need N to be finitely generated and I to be inside the Jacobson radical both of which are satisfied.
Now we need to use the surjectivity of the induced map and some sort of embedding to show that $(N/u(M))/I(N/u(M))=0$, but I cannot find a proper well-defined embedding using the surjectivity of the induced map. Any other approach would be really helpful too.
