If $\psi(m+IM)=\phi(m)+IN$ for an R-module homomorphism $\phi$, why is $\psi(M/IM)=(\phi(M)+IN)/IN$?

Question

Let $I$ be a nilpotent ideal in a commutative ring $R$, let $M$ and $N$ be $R$-modules and let $ϕ:M→N$ be an $R$-module homomorphism. Show that if the induced map $\psi :M/IM→N/IN$ is surjective, then $ϕ$ is surjective.

On every solution I've found for this exercise, it is stated that $\psi(M/IM)=(\phi(M)+IN)/IN$.

I don't understand why this is true. Every element of $M/IM$ is of the form $m+IM$, and $\psi(m+IM)=\phi(m)+IN$. So why is $\psi(M/IM)$ not just $\phi(M)+IN\subset N/IN$?

Answer 1

The induced map is defined by $$ \psi(x+IM)=\phi(x)+IN $$ The image of $\psi$ is some submodule of $N/IN$, so it has the form $K/IN$, for a unique submodule $K$ of $N$, with $IN\subseteq K$.

Let's show that $K=\phi(M)+IN$. If $y\in K$, then $y+IN\in\psi(M/IM)$, which is equivalent to saying that $y+IN=\phi(x)+IN$, for some $x\in M$. Therefore $y-\phi(x)\in IN$, so $y\in\phi(M)+IN$. Conversely, if $y\in\phi(M)+IN$, then $y=\phi(x)+z$, with $z\in IN$ and therefore $$ y+IN=\phi(x)+z+IN=\phi(x)+IN\in\psi(M/IM) $$ so $y\in K$.

Answer 2

That is because $\phi(M)+IN$ is a submodule of $N$, not of $N/IN$. Therefore, you have to determine the canonical image of $\phi(M)+IN$ in $N/IN$, which obviously is $(\phi(M)+IN)/IN$.

More generally, if $N'$ is a submodule of $N$, its image in $N/IN$ is simply $(N'+IN)/IN$.