I don't understand a step in the statement of the structure theorem of finitely generated modules over a PID.
The source I'm using is lecture notes modules and categories (page 35).
Also, if you have improvements for my question, please put it. I find it difficult to pose this question since I don't get a step somewhere in the middle of a big proof.
I will first state the step I don't get. After that I explain where the terms come from.
They claim that the short exact sequence $0 \rightarrow F_2^{'} \rightarrow F_2 \overset{f}{\rightarrow} R$ splits by the section $R \rightarrow F_2$, $1 \mapsto x / f(x)$. I don't understand the definition of this map $R \rightarrow F_2$, $1 \mapsto x / f(x)$, it seems to me that it uses that $f(x)$ is a unit (since we divide by $f(x)$), and I don't see why or how.
Now I explain where the terms come from. (But it might be easier to simply download the lecture notes) They start with the short exact sequence $0 \rightarrow F_1 \rightarrow F_2 \rightarrow M \rightarrow 0$, with $F_i$ of finite rank, which they got from a lemma for finitely generated modules over a PID ($M$). Where they interpret $F_1$ as a submodule of $F_2$. They select an element $x \in F_1$ for which the content $c_{F_2}(x)$ is maximal (amongst all the ideals in $R$ of the form $c_{F_2}(x)$ with $x \in F_1$).
This maximality was not fully clear to me; I think the ideal is maximal with respect to the inclusion relation (and that this maximal ideal exists by Zorn's lemma). By another lemma they have that there exists a surjective $f: F_2 \rightarrow R$ such that $f(x)$ generated the ideal $c_{F_2}(x)$. The kernel of $f$ is denoted by $F_2^{'}$.
They only use that $f(x)$ divides $x$ which is a much less restrictive statement. They say that there exist a $y\in F_2$ such that $f(x)y=x$ holds for all $x\in F_2$.
By definition $f(x)$ generates $c_{F_2}(x)$ if for any $g\in Hom(F_2,R)$ there is a $r_g\in R$ such that $g(x)=r_g f(x)$
Now if $f(x)y=x$ holds then also $g(f(x)y)=g(x)$ therefore $f(x)g(y)=g(x)$. So basically we are looking for an element $y\in F_2$ such that for all $g\in Hom(F_2,R)$ we have that $r_g=g(y)$. But such a $y$ exists uniquely since $Hom(F_2,R)\cong F_2$ (because $F_2$ is free).