I am trying to understand some variations of singular chain complex and their homology.

Suppose that $X$ is a topological space. We have the abelian groups $C_n(X)$ of simplicial $n$-chains and the corresponding chain complex $$ \dotsb \xrightarrow{\enspace \partial \enspace} C_n \xrightarrow{\enspace \partial \enspace} C_{n-1} \xrightarrow{\enspace \partial \enspace} \dotsb $$

Every abelian group can be identified with a $\mathbb Z$-module over the integers. So the above chain complex of abelian groups translates into a chain complex of $\mathbb Z$-modules. The homology groups of the chain complex above are abelian groups, too, and as $\mathbb Z$-modules they translate to the homology modules of the $\mathbb Z$-module complex.

Q1: Is this correct so far?

More akin to the textbooks, I can consider the tensor product of the original chain complex with an abelian group $G$, giving the chain complex with coefficients in $G$, $$ \dotsb \xrightarrow{\enspace \partial \enspace} C_n \otimes G \xrightarrow{\enspace \partial \enspace} C_{n-1} \otimes G \xrightarrow{\enspace \partial \enspace} \dotsb $$

Q2: Are the homology groups of this complex isomorphic to the homology groups of the first complex?

Above we have two variations of the original chain complex. Now suppose that I have a ring $R$.

Let $C_n^R$ be the free $R$-module generated by the singular chains of $X$. We then have a chain complex of $R$-modules $$ \dotsb \xrightarrow{\enspace \partial \enspace} C_n^R \xrightarrow{\enspace \partial \enspace} C_{n-1}^R \xrightarrow{\enspace \partial \enspace} \dotsb $$

We can also forget the ring structure of $R$ and consider it an abelian group. We then have the chain complex with coefficients in $R$: $$ \dotsb \xrightarrow{\enspace \partial \enspace} C_n \otimes R \xrightarrow{\enspace \partial \enspace} C_{n-1} \otimes R \xrightarrow{\enspace \partial \enspace} \dotsb $$

Q3: How do the homology groups / modules of these chain complex relate to each other?

  • $\begingroup$ For Q2, see the Universal Coefficient Theorem: en.wikipedia.org/wiki/Universal_coefficient_theorem $\endgroup$ – Lord Shark the Unknown Aug 27 '18 at 6:33
  • $\begingroup$ For a (commutative, with unit) ring $R$, $C_n^R \simeq C_n\otimes R$ $\endgroup$ – Max Aug 27 '18 at 7:57
  • $\begingroup$ Let me add to @Max's comment that we equip the $\mathbb Z$-module $C_n\otimes_{\mathbb Z} R$ with the structure of an $R$-module by multiplication into the right factor of simple tensors. $\endgroup$ – Christoph Aug 27 '18 at 10:33

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