Given a chain complex $A_\bullet\in\mathrm{Ch(\mathbf{Ab})}$, are there exist some chain complex $A'_\bullet\in\mathrm{Ch(\mathbf{Ab})}$ whose homology groups are all isomorphic to that of $A_\bullet$ such that $A'_p$ are all free abelian groups?
I first thought of stronger condition: Are chain complexes chain equivalent to free ones? but that was false, so I'm wondering if this weaker condition holds.
I think this boils down to "given a sequence of Abelian groups, is there a free chain complex having these as homology groups?".
The answer to this is yes. Let $A_n$ be your $n$-th Abelian group. Then it has a free presentation $$0\to F_n'\to F_n\to A_n\to0$$ which is an exact sequence with $F_n$ and $F_n'$ free. Then the homology of the complex $$\cdots\to0\to F_n'\to F_n\to0\to\cdots\tag{*}$$ is just $A_n$ in one position. Now take the direct sum of the (*) over all $n$.
