Isomorphism involving tensor products of homomorphism groups This post is related to the following question: A counterexample to the isomorphism $M^{*} \otimes M \rightarrow Hom_R( M,M)$.  I have been trying to isolate what hypothesis can be eliminated in these questions surrounding the tensor products of homomorphism groups and I was wondering if the following questions was formulated in such a way as to prevent a counterexample in the post quoted above.
Let $R$ be a commutative ring with identity.  Let $M,N,P$ be $R$ modules and let $\theta : Hom_R (M,N) \otimes_R P \to Hom_R (M,N \otimes_R P)$ be the canonical mapping given by $(f,y) \mapsto (x \mapsto f(x \otimes y))$ for $f \in Hom_R (M,N)$ and $ y \in P$.

If M is a finitely generated projective module is the canonical mapping an isomorphism?

 A: Suppose $F$ and $G$ are two contravariant additive functors from the category of $R$-modules to abelian groups, and that $\alpha:F\to G$ is a natural transformation. In your example, $F(M)=\hom(M,N)\otimes P$, $G(M)=\hom(M,N\otimes P)$ and $\alpha=\theta$.
One can check easily that


*

*if $M$, $M'$ are $R$-modules such that $\alpha_M:F(M)\to G(M)$ and $\alpha_{M'}:F(M')\to G(M')$ are isomorphisms, then $\alpha_{M\oplus M'}:F(M\oplus M')\to G(M\oplus M')$ is an isomorphism and, conversely,

*if $M$, $M'$ are $R$-modules such that $\alpha_{M\oplus M'}:F(M\oplus M')\to G(M\oplus M')$ is an isomorphism, then  $\alpha_M:F(M)\to G(M)$ and $\alpha_{M'}:F(M')\to G(M')$ are isomorphisms.
Using this fact, we can check that

if $\alpha_R$ is an isomorphism, then for all finitely generated projective $R$-modules $P$, the map $\alpha_P$ is an isomorphism.

Indeed, if $P$ is such a projective module, there exists another such projective module $Q$, and $n\geq1$ and an isomorphism $f:P\oplus Q\to R^n$. Since $\alpha_R$ is an isomorphism, the first point above implies by induction that $\alpha_{R^n}$ is an isomorphism. Since $\alpha$ is a natural transformation and $f$ is an isomorphism, this implies in turn that $\alpha_{P\oplus Q}$ is an isomorphism and, finally, the second point above implies now that $\alpha_P$ is an isomorphism.
