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Or rather a couple of questions.

Let $X$ be some topological space, $R$ be a (unital) PID and $G$ be an $R$-module. Am I correct in understanding that the singular cochain complexes $\mathrm{Hom}_\mathbb{Z}(C_*(X;\mathbb{Z}),G)$ and $\mathrm{Hom}_R(C_*(X;R),G)$ coincide (as complexes of abelian groups) and thus the cohomology groups $H^*(X;G)$ may be found from either one of them? ($C_*$ denotes the singular chain complex.)

Under the above impression, I'm trying to see how this relates to the universal coefficient theorems. As I see it, consequently one would have $$H^n(X;G)\cong \mathrm{Hom}_\mathbb{Z}(H_n(X;\mathbb{Z}),G)\oplus \mathrm{Ext}^1_\mathbb{Z}(H_{n-1}(X;\mathbb{Z}),G)\cong\\ \mathrm{Hom}_R(H_n(X;R),G)\oplus \mathrm{Ext}^1_R(H_{n-1}(X;R),G).$$ 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.

So, where is the nonsense lurking?

Update.(@Olivier Bégassat, @Drew. Way to long for a comment.)
Well, I tried thinking about it carefully... Consider the part of the singular chain complex $$\ldots\overset{\partial_{n+1}}{\longrightarrow}C_n(X;\mathbb{Z})\overset{\partial_n}{\longrightarrow}\ldots$$ One has $C_n(X;\mathbb{Z})=\ker\partial_n\oplus M$, for some subgroup $M\subset C_n(X;\mathbb{Z})$. Thus, $C_n(X;R)=\ker\partial_n\otimes R\oplus M\otimes R$ (a direct sum of $R$-modules, obviously). Now, denoting the boundary map of $C_*(X;R)$ by $\partial^R_*$, one has $$\ker\partial_n\otimes R\subset\ker\partial_n^R\subset C_n(X;R)$$ (embeddings of $R$-modules). This implies $$\ker\partial_n^R=\ker\partial_n\otimes R\oplus(\ker\partial_n^R\cap M\otimes R).$$ Finally, consider $\mathrm{im}\partial_{n+1}^R\subset\ker\partial_n\otimes R$ to obtain $$H_n(X;R)=\ker\partial_n^R/\mathrm{im}\partial_{n+1}^R=(\ker\partial_n\otimes R/\mathrm{im}\partial_{n+1}^R)\oplus(\ker\partial_n^R\cap M\otimes R)=\\H_n(X;\mathbb{Z})\otimes R\oplus\ldots$$ as $R$-modules indeed.

I know, some of this looks pretty suspicious (the $R$-module $H_n(X;R)$ is nearly always decomposable?!), but I just can't put my finger on a mistake in the above.

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I don't know if there is a mistake or not (your first assertion about the two ways of calculating the cohomology groups with coefficients in $G$ is correct), but are you sure the splitting $$H_n(X;R)\simeq H_n(X;R)\otimes_\Bbb{Z} R\oplus \mathrm{Tor}_1^\Bbb{Z}(H_{n-1}(X;\Bbb Z),R)$$ you want to use is automatically one of $R$-modules? You need this if you want to use $\mathrm{Hom}_R$ and $\mathrm{Ext}_R^1$. I'm doubly skeptical since there is no canonical way to split the universal coefficient theorem short exact sequences, and the sequence itself is one of $\Bbb Z$-modules. – Olivier Bégassat Feb 18 '14 at 1:11
Yes, several such questions do arise... Apparently, however, the embedding $H_n(X;\mathbb{Z})\otimes_\mathbb{Z}R\hookrightarrow H_n(X;R)$ is almost tautologically one of $R$-modules. – Igor Makhlin Feb 18 '14 at 12:05
I think if you're careful about it, you'll find that the maps can be chosen to be $R$-module maps, but that the splitting is not one of $R$-modules – Drew Feb 19 '14 at 9:52
Here's the answer I've posted on MO:…. – Igor Makhlin Feb 21 '14 at 16:29

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