Sections and homomorphisms of modules Here's a problem I already solved.
Prove that a homomorphism $r: M \rightarrow N$ of right A-modules admits a section $v: N \rightarrow M$ if and only if $r$ is surjective and $M = L \oplus \operatorname{ker}(r)$ where $L$ is a submodule of $M$. Call this (*)
Now here's the problem.
Suppose that the sequence $0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0$ of right $A$-modules is exact, where $u: L \rightarrow M$ and $r: M \rightarrow N$.
Prove that the homomorphism $u$ admits a retraction $p: M \rightarrow L$ if and only if $r$
admits a section $v: N \rightarrow M$.
My try:
=>) This is the one that bothers me. Do we really need to assume $u$ is a retraction? by exactness of the sequence we have that $r$ is surjective. So define a section $v: N \rightarrow M$ as follows: for each $n \in N$ pick $m \in M$ such that $r(m) = n$. Set $v(n) = m$ then $r \circ v = 1_{N}$, so that $r$ admits the section $v$.
<=) Suppose $r: N \rightarrow M$ admits a section, then by (*) $M = \operatorname{ker}(r) \oplus \operatorname{Im}(v)$ and by exactness we have $M=\operatorname{Im}(u) \oplus \operatorname{Im}(v)$. Notice that by exactness we also have that $u$ is an injection. Define a retraction $p: M \rightarrow L$ as follows: given $m \in M$ write $m$ as $u(l)+v(n)$ where $l \in L$ and $n \in N$. Then set $p(m)=l$.
Since $M$ is a direct sum of $\operatorname{Im}(u)$ and $\operatorname{Im}(v)$ and $u$ is an injection we have that $p$ is well-defined and by construction $p \circ u = 1_{L}$ so that $p$ is a retraction of $u$.
Where is the mistake?
 A: Your map $v$ is not well defined.  For example, look at the short exact sequence $0\to\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\to 0$.  Then by your reasoning, we can define a map $v\colon\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}$ by $v(0)=6$ and $v(1)=17$.  This does satisfy $rv=1$, but $v$ is most certainly not a homomorphism.
A: So the only thing that remains is to show that if
$$0\to L\stackrel{u}{\to} M\stackrel{r}{\to} N\to 0$$
is an exact sequence, and $u$ admits a retraction $p\colon M\to L$, then $r$ admits a section $v\colon N\to M$.
Intuition: The retraction is just the projection onto the direct summand. You can isolate the "other factor" of the direct sum by looking at the composition going "the other way"; that is, by considering the submodule of $M$ given by $u\circ p$. Then every element can be written as something in $u\circ p(M)$ and something in a submodule that will turn out to be isomorphic to $N$; the obvious way to do this is to start with $m\in M$, and write it as $m= u(p(m)) + (m - u(p(m)))$. So that's what guides us below. 
Let $p$ be a retraction of $u$, so that $p\circ u = \mathrm{id}_N$.
Given $n\in N$, we define $v(n)$ as follows: there exists $m\in M$ such that $r(m)=n$. Define $v(n) = m-u(p(m))$. (This "twist" to your idea is what makes it work). 
First, I claim that this is well-defined: if $n=r(m)=r(m')$, then $m-m'\in \mathrm{ker}(r)=\mathrm{Im}(u)$, so there exists $t\in N$ such that $m-m'=u(t)$. Then 
$$u(p(m)) - u(p(m')) = u(p(m-m')) = u(p(u(t)) = u(t) =m-m',$$ 
hence $m-u(p(m)) = m'-u(p(m'))$. Thus, $v(n)$ depends only on $n$, and not the chosen preimage. 
Second, I claim that $v$ is a homomorphism: if $n,n'\in N$, let $m,m'\in M$ with $r(m)=n$, $r(m')=n'$. Then $r(m+m')=n+n'$, so
$$\begin{align*}
v(n+n') &= (m+m') - u(p(m+m')) \\
&= m+m' + u(p(m)) + u(p(m')) \\
&= (m+u(p(m))) + (m'+u(p(m')))\\
 &= v(n)+v(n');
\end{align*}$$
and if $n\in N$ and $\alpha\in R$, then letting $m\in M$ be such that $r(m)=n$, then $r(\alpha m) = \alpha n$, so
$$v(\alpha n) = (\alpha m) - u(p(\alpha m)) = \alpha m - \alpha u(p(m)) = \alpha(m-u(p(m))) = \alpha v(n).$$
Third, I claim that $v$ is a section of $r$: indeed, let $n\in N$. If $m\in M$ is such that $r(m)=n$, then:
$$r(v(n)) = r(m+u(p(m))) = r(m) + r(u(p(m))) = r(m)+0 = r(m) = n.$$
Thus, $r\circ v = \mathrm{id}_N$, as desired.
