TITLE: Module spectrum maps up to stable homotopy
QUESTION [7 upvotes]: Let $R$ be a commutative ring spectrum, $M$ and $N$ be a $R$-module spectra.
Let us consider $R$-module maps from $M$ to $N$ up to stable homotopy, that is maps $M \to N$ such that the composites $R \wedge M \to M \xrightarrow{f} N$ and $R \wedge M \xrightarrow{1 \wedge f} R \wedge N \to N$ are equal in the stable homotopy category.
Now suppose that $M = R \wedge X$ is a free $R$-module.
Is it true that the set of stable homotopy classes of $R$-module maps up to homotopy in the previous sense, $[M, N]_R = [R \wedge X, N]_R$, is in the natural bijection with stable homotopy classes of all maps $[X, N]$?
I believe that something like that (or even stronger) holds for genuine $R$-module maps, but what about homotopy $R$-module maps if we are interested in the stable homotopy classes of maps only?

REPLY [4 votes]: In any symmetric monoidal category $C$, the free $R$-module functor $R\otimes -: C\to Mod_R(C)$ is left adjoint to the forgetful functor $U:Mod_R(C)\to C$. Hence, $Hom_{Mod_R(C)}(R\otimes X,N) \cong Hom_C(X,U(N))$. Since the stable homotopy category is symmetric monoidal, the result you asked for follows.