Let's say that we have two abelian groups, $H$ and $G$. We can view them as $\mathbb{Z}$-modules, so it makes sense to tensor them over $\mathbb{Z}$. We also get a somehow obvious map:
$$\mathrm{Hom}(H,\mathbb{Z}) \otimes G \longrightarrow \mathrm{Hom}(H,G); f \otimes g \mapsto (h \mapsto f(h) \cdot g)$$
It is well-known and relatively easy to prove, that if both $H$ and $G$ are free and finitely generated abelian, then the above map is an isomorphism:
$$\mathrm{Hom}(H,\mathbb{Z}) \otimes G \cong \mathrm{Hom}(H,G)$$
This isomorphism also works for arbitrary free modules with finite rank over a ring. But if we take the ring to be $\mathbb{Z}$, it is often the case that stronger statements hold. Explicitely I am wondering, if we can loosen some constraints on $H$ or $G$ if we take the ring to be $\mathbb{Z}$.
My suspicion is that this isomorphism still holds, if we only require $H$ to be free and finitely generated, and $G$ to be any arbitrary abelian group. Because in this case we have an isomorphism $H \cong \mathrm{Hom}(H, \mathbb{Z})$.
Does this work, and if yes, is it possible to even require less?
For some context, I am interested in some minimal conditions on the homology groups of spaces or generally chain complexes, such that I can formulate a statement of the form $H^n(X;R) \cong H^n(X;\mathbb{Z})\otimes R$ for some ring of coefficients $R$.