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Fix a ring $R$ and let $C_\bullet$, $D_\bullet$ be (possibly unbounded) chain complexes of $R$-modules. Assume that $f_\bullet:C_\bullet \to D_\bullet$ is a quasi-isomorphism: that is to say, $f$ is a chain map whose induced maps on homology $H_\bullet f: H_\bullet C \to H_\bullet D$ are isomorphisms. Here's the question:

Is there some (possibly cohomological or $K$-theoretic) obstruction to $f$ forming one-half of a chain homotopy equivalence between $C_\bullet$ and $D_\bullet$?

Note that the question is explicitly not asking whether given two quasi-isomorphic chain complexes, there exists a chain homotopy equivalence. I'm interested in the prescribed quasi-isomorphism $f$ taking part in such an equivalence.

What I know so far is that quasi isomorphism and chain equivalence coincide for bounded complexes of vector spaces, I have gone through this related MO question and hence Weibel's Chapter 10 (particularly Theorem 10.4.8). But neither of those appear to address this question directly.

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    $\begingroup$ A nice reference dealing with this kind of problem in great generality is Weiss's "Hammock localization in Waldhausen categories", but your setting doesn't fit because the class of quasi-isomorphisms is not the saturation of the class homotopy equivalences. I would say that the answer is 'no' in your case, but don't expect a counterexample (at least from me). You can still define an obstruction in relative $K$-theory whose vanishing is necessary, but sufficiency seems to need Weiss hypotheses or something similar. $\endgroup$
    – Fernando Muro
    Commented Feb 24, 2016 at 22:48
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    $\begingroup$ It's worth pointing out that $f$ is automatically a homotopy equivalence if $C_*$ is homotopy equivalent to something built out of bounded complexes of projectives (explicitly: cofibrant objects in the model structure on chain complexes where fibrations are levelwise epimorphisms and weak equivalences are quasi-isomorphisms), this is the analog of Whitehead's theorem. Actually, you can get away with a lot less: I think it's enough to find some suitably nice summand of your map that has this property. $\endgroup$
    – Dylan Wilson
    Commented Feb 25, 2016 at 19:03
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    $\begingroup$ The explicit conditions can be found by tying together various results in May and Ponto's "More Concise..." book: See section 18.6 and 17.3.10. $\endgroup$
    – Dylan Wilson
    Commented Feb 25, 2016 at 19:04
  • $\begingroup$ Thanks, Fernando and Dylan, for the references. I'll look carefully for signs of hope... $\endgroup$
    – Vidit Nanda
    Commented Feb 26, 2016 at 13:38

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