When does a surjective module map admit a section such that the diagram commutes? Let $A$ be a polynomial ring over $\mathbb{R}$. Let $M, N$ be free $A$-modules.
Let $L$ be a finitely generated $A$-module.
Suppose we have $A$-module maps $f\colon M\to N$, $g\colon M\to L$ and $h\colon L\twoheadrightarrow N$ such that the following diagram commutes.

Do we have an $A$-linear section $s\colon N\to L$ such that the diagram commutes (i.e., $h\circ s = id$ and $s\circ f = g$)?
 A: Assumption: $h \circ g = f$. If $L$ is a projective $A$-module, there is always an $A$-linear section $s:N \rightarrow L$ with $h \circ s = id_N$. Hence
$$ h \circ ( s \circ f- g)=h \circ s \circ f - h \circ g = f-f =0. $$
Hence
$$P1. \text{   } s(f(x))-g(x) \in ker(h)$$
for all $x \in M$. Hence $s(f(x))=g(x) + m $ with $m \in ker(h)$ and
$$ s(f(x))\equiv g(x) mod(ker(h)).$$
Hence if $L$ is projective a section $s$ with property P1 exist. If $L$ is finitely generated this is not the case.
Question: "Do we have an A-linear section $s:N\rightarrow L$ such that the diagram commutes?"
Answer: Not in general.
A: Take $M=A^2,N=A,L=A^2$.
For all $a_1,a_2\in A$, set  $f(a_1,a_2)=a_1$, $h(a_1,a_2)=a_2$ and $g(a_1,a_2)=(a_2,a_1)$.
Then for all $a_1,a_2\ni A$, $h(g(a_1,a_2))=h(a_2,a_1)=a_1=f(a_1,a_2)$, so $h\circ g=f$.
Since $s$ is $A$-linear, $s(a)=a(u_1,u_2)=(au_1,au_2)$ for some fixed $(u_1,u_2)\in A^2$ (in fact $(u_1,u_2)=s(1)$)
Now $h(s(a))=au_2$ for all $a\in A$. Since we want $h(s(a))=a$ for all $a\in A$, we have $u_2=1$. Now $s(f(a_1,a_2))=s(a_1)=(a_1u_1,a_1u_2)=(a_1u_1,a_1)$
Hence we want $(a_1u_1,a_1)=g(a_1,a_2)=(a_2,a_1)$ for all $a_1,a_2\in A$,that is $a_1u_1=a_2$ for all $a_1,a_2\in A$, which is not possible (take $a_1=0$).
Hence $s$ does not exist in this case.
