Why free abelian chain complexes in homology?

Question

I was wondering is there any specific reason why the chain complexes in homology are free abelian?

Thanks in advance

Answer 1

This question also occurred to me when first studying homology. To be a bit more precise, in singular homology, one can ask why the default construction of an $n$-chain, an element of $C_n(X)$, is defined to be the free abelian group with basis the $n$-simplices $\sigma_{\alpha}:\Delta^n\to X$ (the continuous mappings of the standard $n$-simplex into $X$).

One reason is that $n$-chains are then finite formal linear sums $\sum a_{\alpha}\sigma_{\alpha}$ with coefficients $a_{\alpha}\in\mathbb{Z}$, which can be used to build "$n$-surfaces in $X$." In particular, we can build an $n$-chain which is a sum of "$n$-spheres in $X$" where each sphere can "wrap around itself" any number of times in any direction (to be more precise, each term $a\sigma$ consists of $a$ copies of $\sigma$, with the orientation reversed for negative $a$). For homology using non-integer coefficients, we have to discard the "number of wrappings" idea.

We then build a chain complex, a sequence of homomorphisms of $n$-chains $\partial_{n}\colon C_{n}\left(X\right)\to C_{n-1}\left(X\right)$ with $\partial_{n}\partial_{n+1}=0$, which can be viewed in terms of these copies of spheres:

Finally, the $n$th homology group is defined as $H_{n}(X)\equiv\textrm{Ker}\partial_{n}/\textrm{Im}\partial_{n+1}$, i.e. the group of cosets consisting of homologous $n$-cycles that can all be obtained from each other by adding the boundary of some $(n+1)$-volume in $X$.

For simple spaces, we then expect that $H_n$ will be a direct product of $\mathbb{Z}$ components, one for each “$n$-dimensional hole” in $X$ that is not the boundary of a $(n+1)$-volume, since each such hole can be "wrapped around" any number of times in either direction, and none of these "wrappings" are homologous.

This viewpoint is worked out in more detail here.