Direct Sum of Projective and Free Modules In Kevin Ventullo's answer on Mathoverflow it is mentioned that if $F$ is a free module on infinitely many generators and $P$ is a projective module, then $P \oplus F$ is again free.
However, I cannot see why this should be true.
Could you please show me why this holds?
Thank you very much for your help!

I think the proof is not well written.
What the proof needs is a free module $F_0$ such that $P_i$ is a direct summand of $F_0$, for $i=0,\dots,n$: just take $Q_i$ with $P_i\oplus Q_i$ free and form $F_0=\bigoplus_{i=0}^n(P_i\oplus Q_i)$.
Then show that, if $F=F_0^{(\omega)}$ (a direct sum of a countable number of copies of $F_0$) and $P$ is a direct summand of $F_0$, then $P\oplus F$ is free: this is easy because, if $F_0=Q\oplus P$, we just need to formalize with suitable maps the idea that
$$
P\oplus (Q\oplus P)\oplus(Q\oplus P)\oplus\dotsb=
(P\oplus Q)\oplus(P\oplus Q)\oplus\dotsb
$$
Hence $P\oplus F\cong F$ is free.
