Equivalent characterizations of projective modules

If $P$ is projective and $0\to K\to E\stackrel{q}{\to }P\to 0$ is short exact, lift ${\mathrm{id}}_P$ through $q$ using [F1]; the lift is a section, so the sequence splits by [L1].

If every such sequence splits, then for a surjection $q:E\to M$ and map $f:P\to M$, form the pullback module $T=\{(x,e)\in P\oplus E:f(x)=q(e)\}$. The projection $T\to P$ is surjective with kernel isomorphic to $\ker q$, so its short exact sequence splits; a section followed by the projection $T\to E$ is a lift of $f$. Thus $P$ is projective.

By [L2], applying ${\mathrm{Hom}}_R(P,-)$ to $0\to A\to B\stackrel{q}{\to }C\to 0$ is exact through ${\mathrm{Hom}}_R(P,B)$; its last map is surjective exactly when every $P\to C$ lifts through $q$. Hence [F1] makes assertions 1 and 3 equivalent.

If $P$ is projective, lift ${\mathrm{id}}_P$ through the canonical surjection $R^{(P)}\to P$ from [L3]. This section splits the free cover by [L1], so $P$ is a direct summand of $R^{(P)}$.

A direct summand of a projective module is projective: precompose a map from the summand with the projection, lift the resulting map, and restrict the lift along the inclusion. Under AC the free ambient module in assertion 4 is projective by [L4], so assertion 4 implies assertion 1.

Steps 1.1 and 1.2 prove $1\iff 2$, step 1.3 proves $1\iff 3$, and steps 1.4 and 1.5 prove $1\iff 4$ with the stated choice boundary.