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I found a version of the universal coefficient theorem for cohomology here, which seems to be different to what is commonly found in textbooks. I report it here for convenience. It says that the following sequence is split exact

$$ 0 \longrightarrow H^n(X;\mathbb{Z}) \otimes G \longrightarrow H^n(X;G) \longrightarrow \mathrm{Tor}(H^{n+1}(X;\mathbb{Z}),G) \longrightarrow 0 $$

for a "coefficient" abelian group $G$. This seems to be a dualized version of the theorem for homology, that is, the theorem applied to the dual of the original chain complex. However, that proof relies on the fact that a chain with coefficients means $C_n(G)=C_n\otimes G$, whereas defining $C^n(G)=\mathrm{Hom}(C_n,G)$, it is not true that $C^n(G)=C^n\otimes G$.

So is that theorem true?

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  • $\begingroup$ A similar question is addressed here where it is claimed that the result is true if $G$ is also a ring, but that is not justified $\endgroup$ Apr 16 at 17:19

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You critically omit the condition that the homology of $X$ be of finite type listed under your link; without that assumption I doubt the formula is correct.

With this assumption, on the other hand, you can replace $C_*(X)$ with a free and levelwise finite-dimensional complex up to chain homotopy equivalence, as a consequence of which you have $C^*(X) \cong C_*(X)$ and $C^*(X; G) \cong C^*(X) \otimes G$, whence result follows from the (co)chain level universal coefficient theorem.

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