Please don't give an alternative proof. I know how to prove this another way, but this argument intrigues me, despite that it (perhaps because it?) makes no sense to me.
I don't understand this proof in Jonathan Rosenbergs K-theory text. The following was proved in one direction (easily):
The following are equivalent definitions of a projective module:
- $P$ is projective if whenever $A\to P$ is a surjection, there is a right inverse $P\to A$.
- $P$ satisfies the standard diagram completion property (without tikzcd, too much of a bother to render, but yall know it)
They are proving the first implies the second:
Please explain the 'we may suppose $\varphi$ is one-to-one'. I am not sure if this is an implication from the replaced diagram, and if it is how it is so. Otherwise I cannot see how it is fair to do this, since it isn't WLOG - and additionally why we stop caring about the diagram immediately afterwards.
What happens here is that we have a diagram $$\require{AMScd} \begin{CD} & & P \\ & & @VV{(\varphi,id_P)}V \\ M\times P@>{\psi\times id_P}>> N\times P \\ @V{\pi_M}VV @VV{\pi_N}V\\ M@>>{\psi}> N\end{CD}$$where the lower square commute. Now the existence of a map $\alpha :P\to M$ such that $\psi\alpha=\varphi$ is equivalent to the existence of a map $\alpha':P\to M\times P$ such that $(\psi\times id_P)\alpha'=(\varphi,id_P)$; indeed, given $\alpha$ it suffices to define $\alpha'=(\alpha,id_P)$, and given $\alpha'$ it suffices to define $\alpha=\pi_M\alpha'$.
So in order to prove that $P$ is projective, it is enough to complete the upper part of the diagram above, where now $(\varphi,id_P)$ is one-to-one; so if we know that the diagram completion can always be done under this condition the proof is complete.