at.algebraic topology - 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.

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