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Apr 21, 2017 at 4:54 history edited მამუკა ჯიბლაძე CC BY-SA 3.0
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Jun 20, 2012 at 23:14 vote accept Vidit Nanda
Jun 20, 2012 at 19:10 answer added Tyler Lawson timeline score: 14
Jun 20, 2012 at 18:29 comment added Donu Arapura Well, a morphism of filtered complexes which induces an isomorphism at the $E_1$ level (which implies it all the way down) is the same thing as a filtered quasi-isomorphism = isomorphism in the filtered derived category. Although this may not answer your question.
Jun 20, 2012 at 18:09 comment added Vidit Nanda Donu, I agree that it seems strong. But the question is natural: in the filtered derived category, does the equivalence $C \sim C'$ iff $E(C)$ is isomorphic to $E(C')$ go by another name?
Jun 20, 2012 at 17:58 comment added Donu Arapura I realized you're only asking for isomomorphisms, nevertheless, it feels too strong.
Jun 20, 2012 at 17:56 comment added Donu Arapura In other words, is the map from the filtered derived category to the category of spectral sequences full? I doubt it, although I don't have a counterexample.
Jun 20, 2012 at 17:52 answer added Ralph timeline score: 7
Jun 20, 2012 at 17:34 history asked Vidit Nanda CC BY-SA 3.0