Can we construct a homomorphism from a projective module into a free module?

Question

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?

Answer 1

Given any $R$-modules $M$ and $N,$ there is at least one $R$-module homomorphism $M\to N.$ In particular, there is always the trivial homomorphism $m\mapsto 0_N.$

Answer 2

If $M$ is a nontrivial projective module and $N=R^n$ for $n\geq 1$ and you want a nontrivial homomorphism $M\to N$, you could do the following.

Since every projective module is (isomorphic to) a direct summand of a free module $F$, you would have an injective homomorphism $g:M\to F$. If $F$ has lower rank than $M$, then it is easy to come up with another injective homomorphism $f:F\to N$, and you've got an injective homomorphism $fg:M\to N$.

If it happens that $F$ has higher rank, then you can certainly use a surjective homomorphism $f$ from $F$ onto $N$, whence you obtain a nonzero map $fg:M\to N$ again.