Projective Modules, Definition and Etymology

Question

Question that I have after reading some of the answers to this question about the etymology of projective modules. It goes as follows:

Let $P$ be an $R$-module, $R$ being some arbitrary ring. Is it the case that if there exists a free $R$-module $M$ such that $P$ is isomorphic to a submodule $P'$ of $M$, and a surjective map $f: M \rightarrow P'$ such that $f^2 = f$, then $P$ is necessarily a projective $R$-module?

EDIT: I have come to learn that this indeed is the case, though I still could very much use a formal proof.

Presumably you also assume that $P$ is a submodule of $M$ (since otherwise the condition does not make sense). — Tobias Kildetoft, Feb 17 '20 at 18:17

score 1 · Accepted Answer · answered Feb 17 '20 at 18:29

Yes, although you need to phrase it differently. Note that in order for "$f^2=f$" to even make sense, $f$ has to be a self-map. So, rephrasing your question, is it the case that given $P$ as submodule of a free module $M$ and a map $f:M\to M$ such that $f(M)=P$ and $f^2=f$, then $P$ is a projective module?

The only way we can have $f^2=f$ is if $f$ is the identity on $P$, since suppose $z=f(y)$, then $f(z)=f^2(y)=f(y)=z$. So $f$ is the identity on $P$, so the inclusion map splits the exact sequence $0\to \ker(f) \to M \to P \to 0$

Projective Modules, Definition and Etymology

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