Is the following true: If two chain complexes of free abelian groups have isomorphic homology modules then they are chain homotopy equivalent.
[Math] Does homology detect chain homotopy equivalence
ac.commutative-algebraat.algebraic-topologyhomological-algebrahomology
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Best Answer
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.