This seemingly simple question stands unanswered on math.stackexchange.com for a couple of days (https://math.stackexchange.com/questions/680211/a-question-about-the-universal-coefficient-theorem), so I decided to repost it here, although it is hardly research related.

Let $X$ be some topological space, $R$ be a (unital) PID and $G$ be an $R$-module. I'm trying to see how the fact that the cohomology groups $H^*(X;G)$ may be found as the cohomology of either $\mathrm{Hom}_\mathbb{Z}(C_*(X;\mathbb{Z}),G)$ or $\mathrm{Hom}_R(C_*(X;R),G)$ relates to the universal coeffients theorems.

One has \begin{align} H^n(X;G)&\cong \mathrm{Hom}_\mathbb{Z}(H_n(X;\mathbb{Z}),G)\oplus \mathrm{Ext}^1_\mathbb{Z}(H_{n-1}(X;\mathbb{Z}),G)\\ &\cong\mathrm{Hom}_R(H_n(X;R),G)\oplus \mathrm{Ext}^1_R(H_{n-1}(X;R),G). \end{align} Now let us try to derive the latter isomorphism from the universal coefficient theorem for homology. Replacing the homology groups in the second line with the corresponding expressions of the form $*\otimes_\mathbb{Z} R\oplus \mathrm{Tor}_1^\mathbb{Z}(*,R)$ we obtain quite the isomorphism. I can see that $\mathrm{Hom}_\mathbb{Z}(H_n(X;\mathbb{Z}),G)\cong \mathrm{Hom}_R(H_n(X;\mathbb{Z})\otimes_\mathbb{Z}R,G)$ via tensor-hom adjunction, i.e. the very first summands on both sides coincide. However, the isomorphism still seems dubious to me, since, if I'm not mistaken, any of the four summands on the right may actually be nonzero.

I do understand that the most suspicious part is the exact sequence in the UCT for homology being a split exact sequence of $R$-modules. However, in the stackexchange question I provide an argument seemingly showing that it actually is such.

I would be truly grateful for a detailed explanation of what is what here.

**MAJOR UPDATE.** Thanks to Neil Strickland's comment the question has boiled down to the following.

As I see it, the UCTs imply the following isomorphism of $R$-modules: $$\mathrm{Hom}_\mathbb{Z}(H_n(X;\mathbb{Z}),G)\oplus \mathrm{Ext}_\mathbb{Z}(H_{n-1}(X;\mathbb{Z}),G)=\\\mathrm{Hom}_R(H_n(X;\mathbb{Z})\otimes_\mathbb{Z} R,G)\oplus \mathrm{Hom}_R(\mathrm{Tor}^\mathbb{Z}(H_{n-1}(X;\mathbb{Z}),R),G)\oplus\\\mathrm{Ext}_R(H_{n-1}(X;\mathbb{Z})\otimes_\mathbb{Z} R,G)\oplus\mathrm{Ext}_R(\mathrm{Tor}^\mathbb{Z}(H_{n-2}(X;\mathbb{Z}),R),G).$$ I would like to see a direct proof.

Let us refer to the isomorphism as $A\oplus B=C\oplus D\oplus E\oplus F$. Tensor-hom adjunction shows that $A=C$. Thus we are to show $B=D\oplus E\oplus F$.

Now, as Neil pointed out, $R$ is either a free $\mathbb{Z}$-module or a $\mathbb{Z}/p\mathbb{Z}$-algebra for some prime $p$. In the first case it isn't hard to see that $D=F=0$ and $B=E$, achieving a proof. I'm now interested in the second case, what is the situation here?