This is an exercise from Assem's book "Algèbre et module" (more precisely, III.50). I couldn't solve it after thinking for a while. Let
$\require{AMScd}$ \begin{CD} 0 @>>> L @>i>> P @>p>> V @>>> 0\\ @. @| @VuVV @VVvV \\ 0 @>>> L @>>i'> M @>>p'> N @>>> 0 \end{CD} be a commutative diagram with exact lines. If $v$ factors through $M$, then the first line splits. Call that morphism say $h: V \longrightarrow M$ and assume that it commutes inside the two respective triangles in the right square (if you can do it only assuming commutativity of the lower triangle, that's even better).
You can assume that everything is a module over a ring. Still, I would prefer a proof inside abelian categories or something more general like pseudo-abelian with possibly more assumptions.
By factoring through the images of $u$ and $v$, I can reduce to the case where $u$ and $v$ are epi. I could find some relations between $\ker (u)$, $\ker (v)$ and $\ker (h)$. Nothing very illuminating though. But maybe I'm just overseeing something trivial or maybe there's a nasty trick around.
Thanks in advance.
Take the fiber product $P':=M\times_NV$. Note that the map $P'\to V$, as a fiber product of a surjective map, is also surjective. Letting $L':=\ker(P'\to V)$, we obtain two maps of exact sequences:
\begin{CD} 0@>>>L@>>>P@>>>V@>>>0\\ @.@VVV@VVV@|@.\\ 0@>>>L'@>>>P'@>>>V@>>>0\\ @.@VVV@VVV@VVV@.\\ 0@>>>L@>>>M@>>>N@>>>0 \end{CD}
Note that the vertical map $L'\to L$ is an isomorphism by definition of $P'$, thus the leftest vertical arrows are all isomorphisms. By the five lemma, the middle vertical map $P\to P'$ is also an isomorphism. Then the lift $V\to M$ of $V\to N$ along with the identity map $V\to V$ induces, by definition of $P'$, a map $V\to P'$ which splits $P'\to V$.