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I am reading the Hatcher's book and am a bit confused with (co)homologies with coefficients. I would really appreciate if somebody clarifies the following to me.

Let $C$ be a chain complex of $R$-modules for a ring $R$. Then I can take the associated cochain complex of $R$-modules $C^*=Hom_R(C,G)$, where $G$ is some fixed $R$-module. If $C$ comprises only free modules, under which conditions on $G$ (or $R$) does $C^*$ consist of free modules too ? When does the universal coefficient theorem for cohomologies hold ?

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Your question is more about ordinary algebra than homological algebra: you don't really need the chain complex structure on $C$ at all, except to induce a cochain complex structure on $C^*$. So henceforth I'll pretend $C$ is just an $R$-module.

Fact. If $C = R^{\oplus \kappa}$ is the free left $R$-module on $\kappa$-many generators and $G$ is a left $R$-module, then $$\textrm{Hom}_R(C, G) \cong G^{\times \kappa}$$ as left $R$-modules, naturally in $G$.

Fact. The product of finitely many left $R$-modules is also the direct sum, i.e. $$A \times B \cong A \oplus B$$ naturally in $A$ and $B$.

Proposition. If $\textrm{Hom}_R(C, G)$ is a free left $R$-module for all free left $R$-modules $C$, then $G$ must be a free left $R$-module.

Proof. Take $C = R$, then $\textrm{Hom}_R(C, G) \cong G$, so $G$ is a free left $R$-module.

Proposition. If $C$ is a finitely generated free left $R$-module, then $\textrm{Hom}_R(C, G)$ is a free left $R$-module for all free left $R$-modules $G$ (not necessarily finitely-generated!).

Proof. If $C = R^{\oplus n}$, then $\textrm{Hom}_R(C, G) \cong G^{\times n} \cong G^{\oplus n}$, and the direct sum of free left $R$-modules is a free left $R$-module.


There's not much more we can say in general. For example, if $R$ is a field, then all $R$-modules are free. So there's nothing to prove. On the other hand, for $R = \mathbb{Z}$, $C = \mathbb{Z}^{\oplus \aleph_0}$, $G = \mathbb{Z}$, we get $\textrm{Hom}_\mathbb{Z} (\mathbb{Z}^{\oplus \aleph_0}, \mathbb{Z}) \cong \mathbb{Z}^{\times \aleph_0}$, but the latter $\mathbb{Z}$-module is not free.

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