In short, I have a projective module and a free module, and want to construct a module homomorphism between the two. Is this always possible, at least in some way?
Let me go into more detail. Suppose we have some commutative ring $S$ and a finite group $G$ so that we can construct the (most likely noncommutative) group ring $R=S[G]$. Now, suppose we have a projective $R$-module $P$ such that $rk_S(P)=rk_S(F)$ where $F$ is some finitely generated free $R$-module. I want to construct a $R$-module homomorphism from $P\to F$ or $F\to P$ (either suits my purpose). I was thinking of the following argument but am unsure if it's valid. Suppose $F=R^a$ for some $a\geq 1$. Since $P$ is projective, then there exists some $R$-mod $Q$ such that $P\oplus Q\cong R^\alpha$ where $\alpha\geq a$ (equal to $a$ when $P$ is free). I can inject $R^a\to R^\alpha$ by sending $(r_1,\ldots,\,r_a)\mapsto(r_1,\ldots,\,r_a,\,0,\ldots,\,0)$ (here we have $\alpha-a$ zeroes). Furtheremore, I can project $R^\alpha\to P$ by mapping $(p,\,q)\mapsto p$. Is the composition of these two maps not an $R$-mod homomorphism, or am I being silly?