I was reading Keith Conrad's notes here and was wondering if there is any way to only prove (1) $\iff$ (2) which comes out as Let $0 → N → M→ P → 0$ be a short exact sequence of R-modules. The following are equivalent: (1) There is an $R$-linear map $f': M → N$ such that $f ' (f(n)) = n$ for all $n ∈ N.$
(2) There is an R-linear map $g ' : P → M$ such that $g(g ' (p)) = p $for all $p ∈ P.$
Here $f$ is from $N$ to $M$ and $g$ is from $M$ to $P$
The easiest way to show this equivalence is slightly indirect: statements (1) and (2) are both equivalent to:
(3): $M \simeq N \oplus P$ and under this isomorphism $f: N \rightarrow M$ is the map $\iota: n \mapsto (n, 0)$ and $g: M \rightarrow P$ is the map $\pi: (n, p) \mapsto p$.
It's easy to see that with these maps, we get the canonical split exact sequence (*):
$$ 0 \rightarrow N \rightarrow N \oplus P \rightarrow P \rightarrow 0 $$
Assuming (3), we can define the splitting maps required for (1) and (2) easily: Let $f': N \oplus P \rightarrow N$ be the map given by $(n,p) \mapsto n$. Then clearly $f'(f(n)) = f'( (n,0) ) = n$ for all $n \in N$. Similarly, we can define $g': P \rightarrow N \oplus P$ by $p \mapsto (0, p)$. Then we have $g(g'(p)) = g( (p,0)) = p$ for all $p \in P$.
Thus, (3) implies (1) and (2). Now it suffices to see that both (1) and (2) imply (3).
Assume (1). We can define a map $\varphi: M \rightarrow N \oplus P$ by $m \mapsto (f'(m), g(m))$. It's easy to check that $g = \pi \circ \varphi$ and that $\varphi \circ f = \iota$ (since $\varphi (f(n)) = (f'(f(n)), g(f(n)) = (n, 0)$) and that $\varphi$ is an isomorphism.
Similarly, assuming (2) we can define a map $\varphi: N \oplus P \rightarrow M$ by $(n,p) \mapsto (f(n), g'(p))$ and verify that this map turns the exact sequence into the canonical split exact sequence (*).
