Is the following true: If two chain complexes of free abelian groups have isomorphic homology modules then they are chain homotopy equivalent.



Does homology detect chain homotopy equivalence?
Yes, this is true. Suppose $C\_\*$ is such a chain complex of free abelian groups.


For each $n$, choose a splitting of the boundary map $C\_n \to B\_{n-1}$, so that $C\_n \cong Z\_n \oplus B\_{n-1}$. (You can do this because $B\_{n-1}$, as a subgroup of a free group, is free.) For all $n$, you then have a sub-chain-complex $\cdots \to 0 \to B\_n \to Z\_n \to 0 \to \cdots$ concentrated in degrees $n$ and $n+1$, and $C\_\*$ is the direct sum of these chain complexes.


Given two such chain complexes $C\_\*$ and $D\_\*$, you get a direct sum decomposition of each, and so it suffices to show that any two complexes $\cdots \to 0 \to R\_i \to F\_i \to 0 \to \cdots$, concentrated in degrees $n$ and $n+1$, which are resolutions of the same module $M$ are chain homotopy equivalent; but this is some variant of the fundamental theorem of homological algebra.


This is special to abelian groups and is false for modules over a general ring.


