Timeline for Is there an obstruction which classifies "quasi-isomorphism but not chain equivalence"?
Current License: CC BY-SA 3.0
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Feb 26, 2016 at 13:38 | comment | added | Vidit Nanda | Thanks, Fernando and Dylan, for the references. I'll look carefully for signs of hope... | |
| Feb 25, 2016 at 19:04 | comment | added | Dylan Wilson | 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. | |
| Feb 25, 2016 at 19:03 | comment | added | Dylan Wilson | 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. | |
| Feb 25, 2016 at 12:59 | history | edited | Vidit Nanda |
Added tag
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| Feb 24, 2016 at 22:48 | comment | added | Fernando Muro | 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. | |
| Feb 24, 2016 at 17:09 | history | asked | Vidit Nanda | CC BY-SA 3.0 |