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Let $f:C\rightarrow D$ be a chain map between chain complexes over a PID $R$. It's given that the chain map $H_q(f)$ is an isomorphism for every $q$ and that for every $q$, $C_q,D_q$ are free $R$-modules. Must $f$ be a chain equivalence ? If I recall correctly, this will be true if $R$ is assumed to be a field. I don't know about the general PID case though.

Thank you,

Amr
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  • Do you know how to prove this for a field? You can utilise the same ideas here. Recall that submodules of free modules over a PID are free. – shubhankar Nov 06 '20 at 22:18
  • @ShubhankarSahai I did that a long time, I wasn't hard I guess. I thought to ask first before taking the risk of trying to prove a false statement. I ll try proving it myself, thank you!! – Amr Nov 06 '20 at 22:20
  • @Shubhankar can you give me a reference please ? Perhaps the way I proved it for fields wont be easilt generalized to PID s. Thank you – Amr Nov 08 '20 at 13:14
  • https://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence/10983#10983

    The answer is for chain complexes of free abelian groups. However the upshot is that there is a splitting $C_i=B_{i-1}\oplus Z_i$ because $B_{i-1}$ is free which holds in any pid. In fact if you should be able to prove the same for chain complexes of projective modules over a pid, since in this case submodules of projective modules are projective and you still have that splitting.

     – shubhankar Nov 09 '20 at 20:07
  • if you prove it for a field, then the proof is exactly the same and proceeds with examining the splitting. – shubhankar Nov 09 '20 at 20:08

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