How is an onto map implies $N+mM=M$ in Commutative Algebra?

Question

I am having hard time understanding some details in Proposition 2.8 which is on page 22 of Atiyah and Macdonald's book: Introduction to Commutative Algebra.

How the writers are claiming that being the composite map $$ N\rightarrow{M}\rightarrow{M/mM} $$ onto implies that $N+mM=M$?

Answer 1

Let $\iota$ be the map $N \to M$, let $\pi:M \to M/\mathfrak{m}M$ be the quotient map, and suppose $\pi \circ \iota$ is onto. In this problem $N$ is a submodule of $M$, so $\iota(n)$ is just $n$ thought of as an element of $M$. Let $x \in M$, then $x + \mathfrak{m}M \in \operatorname{im}(\pi\circ\iota)$, so $x + \mathfrak{m}M = \pi(\iota(n)) = n + \mathfrak{m}M$ for some $n \in N$. Therefore $x-n \in \mathfrak{m}M$, and $x = n + y$ for some $y \in \mathfrak{m}M$. We have written an arbitrary element of $M$ as a sum of something in $N$ and something in $\mathfrak{m}M$, therefore $M = N+\mathfrak{m}M$.

Answer 2

Given $x/mM\in M/mM$ you have $n\in N$ such that :

$\pi(i(n))=x+mM=n+mM$ for $x\in M $ and $n\in N$

i.e., $x=n+mM+mM=n+mM$ this mean $M\subseteq N+mM$

It is upto you to look for the other way.... (Which is trivial)