Equivalent properties of projective modules

Question

For a left $R$ module $P$, the following are equivalent:

1. Given $M\xrightarrow{\psi} N \to 0$ exact and $\omega: P \to N$, there exists $\tilde{\omega}: P \to M$ such that $\psi \circ \tilde{\omega} = \omega$.
2. Every short exact sequence $0 \to A \to B \to P \to 0$ splits.
3. $P$ is a direct summand of a free module.

In proving this equivalence, we typically go $1 \implies 2 \implies 3 \implies 1$. I'm interested in more direct proofs of $3 \implies 2$ and $2 \implies 1$. Does anyone know any?

Update: I think I showed $3 \implies 2$...I'll post it.

Answer 1

For $2 \Rightarrow 1$ the pullback $M \times_N P \to P$ of a surjective map is surjective so by $(2)$ it splits and we have maps $P \to M \times_N P \to M$. Check that these define $\tilde\omega$ as desired.

For $3 \Rightarrow 2$ take $S$ such that $P \oplus S$ is free. Then there is a short exact sequence $0 \to A \to B \oplus S \to P \oplus S \to 0$ which splits because $P \oplus S$ has a basis. Show that the splitting maps $P$ into $B$ and that this restriction splits the original short exact sequence.