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Let $C$ be a chain complex over a principal ideal domain $R$. How can I construct a chain complex $F$ of free $R$-modules which is quasi-isomorphic to $C$?

Edit: It is well known how to do this with a chain complex concentrated in degree $0$, and we could take that as motivation. In that case, a projective resolution allows us to construct a chain complex with the same homological properties, but which is more tractible (e.g. because a quasi-isomorphism of projectives is a homotopy equivalence).

Ronald J. Zallman
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co-co-nuts
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  • First suppose that the chain complex is bounded below. We inductively construct a chain complex of projectives $P_$ with a map $P_ \rightarrow C_*$. How to do this for the smallest nonzero degree is clear enough. Suppose we have constructed part of a chain complex $P_n \rightarrow \cdots \rightarrow P_0$ and a part of a map $P_i \rightarrow C_i$ for $i \leq n$. Let $B'n$ be the preimage of $B_n \subset C_n$ in $P_n$. Choose a free module $P{n+1}$ projecting onto $B'n$. This induces a map $P{n+1} \rightarrow B_{n}$ which lifts to a map $P_{n+1} \rightarrow C_{n+1}$ since $P_{n+1}$. – Ronald J. Zallman Apr 27 '19 at 22:37
  • Yes, the construction I finally have is obtained from the S.E.S $0 \to T_n \to B_n \to H_n(C) \to 0$ where $B_n$ is any free $R$ module. And define $F_n = B_n \oplus T_{n-1}$. – co-co-nuts May 09 '19 at 14:08

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