$\operatorname{Hom}(C,R) \otimes G \simeq\operatorname{Hom}(C,G)$

Question

Given $C,G$ free $R-$modules, I was interesed in proving that $\operatorname{Hom}C,R) \otimes G \simeq\operatorname{Hom}(C,G)$ if $G$ is finitely generated.

I thought $\psi : \varphi \otimes g \longmapsto \varphi \cdot g$ could be the isomorphism I was looking for but I wasn't sure about the inverse, I didn't came up with a direct proof of an isomorphism.

Any help or hint would be appreciated.

Edit : FiMePr answer was given before adding the word $free$ to the question.

Answer 1

I think the result is false if you don't add more hypotheses :

Let $R = \mathbb{Z}$, $C = G = \mathbb{Z}/2$.

Then $Hom(C, R) = Hom(\mathbb{Z}/2, \mathbb{Z}) = 0$, so $Hom(C, R) \otimes G = 0$

but $Hom(C,G) = Hom(\mathbb{Z}/2, \mathbb{Z}/2) = \mathbb{Z}/2$.

However, the result is true if $C,G$ are free and finitely generated, i.e., if both of them are isomorphic to some $R^n$, for $n$ an integer (not necessarily the same integer for $C$ and $G$).

EDIT : With the edited hypotheses :

$Hom(\oplus_{i \in I} R, R) \otimes R^n = (\prod_{i \in I} Hom(R,R))\otimes R^n = R^I \otimes R^n = \oplus_{k=1}^n R^I \otimes R = R^{n \times I} = \prod_{i \in I} Hom(R, R^n) = Hom(\oplus_{i \in I} R, R^n)$.

EDIT^2 : If we only assume $G$ to be free and finitely generated :

$Hom(C,G) = Hom(C, R^n) = Hom(C,R)^n = \oplus_{i=1}^n Hom(C, R) = (\oplus_{i=1}^n Hom(C, R))\otimes R = \oplus_{i=1}^n (Hom(C, R)\otimes R) = Hom(C, R)\otimes (\oplus_{i=1}^n R) = Hom(C, R) \otimes R^n = Hom(C,R) \otimes G$.