prove that $M$ is Noetherian if and only if $M'$ and $M''$ are Noetherian.

Question

Let $0\to M'\xrightarrow{f} M\xrightarrow{g}M''\to0$ be an exact sequence in $A$-module. Then prove that $M$ is Noetherian if and only if $M'$ and $M''$ are Noetherian.

I've a proof from a book as follows :

I'm unable to understand only two line of this proof :

(1) $g^{-1}(N'')$ and $N''$ are not isomorphic. So why $g^{-1}(N'')$ is finitely generated implies $N''$ is finitely generated ?

(2) In the converse part of the proof why $f^{-1}(N)$ exists ? As the sequence is exact so $f$ is one-one only.

Answer 1

1. That is because $g$ is onto, hence the restriction of $g$ to $g^{-1}(N'')$ and $N''$ is too.
2. Being one-to-one doesn't preclude the existence of $f^{-1}(N)$: it's not empty, as it contains $0$.