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).

  • $\begingroup$ 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 $P_{n+1}$. $\endgroup$ – Dean Young Apr 27 at 22:37
  • $\begingroup$ 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 $F_n = B_n \oplus T_{n-1}$. $\endgroup$ – co-co-nuts May 9 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.