I was reading "Serial Rings" by Gennadi Puninski. There it is written that , "Since every module is a homomorphic image of a free module, every projective module is a direct summand of free module".(ie. if $P$ is a projective module, there exists a free module F such that, $ F=P \oplus T$ for some module $T$.)
But I can't understand how "Every module is a homomorphic image of a free module" implies that "Every projective module is a direct summand of free module".
(I have found a proof for "Every projective module is a direct summand of free module" but the first part of the above mentioned sentence wasn't used there.)
Let $P$ be a projective module. We wish to show that it is the direct summand of a free module. Let $\varphi : F \to P$ be a surjection, where $F$ is a free module.
Consider the following diagram below.
Since $P$ is projective, there exists a map $\psi : P \to F$ making the following diagram commute.
It now follows that $F = \operatorname{im} \psi \oplus \ker \varphi$. But $\varphi \circ \psi = \operatorname{id}_P$ implies that $\psi$ is injective and thus, $\operatorname{im} \psi \cong P$, as desired.