Timeline for Chain homotopy: Why du+ud and not du+vd?
Current License: CC BY-SA 2.5
12 events
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| Dec 15, 2009 at 19:30 | comment | added | Greg Kuperberg | Sorry, I meant WELL-BEHAVED complexes are generally bounded on one side. I dropped part of what I wanted to say. | |
| Dec 15, 2009 at 19:07 | comment | added | Tyler Lawson | @Greg: It's been my understanding that this isn't convention, it's usually specified if necessary? There is a perfectly good derived category of unbounded complexes. | |
| Dec 15, 2009 at 18:53 | comment | added | David E Speyer | Thanks Greg! For completeness, I'm going to add one more counter-example to the other answer. | |
| Dec 15, 2009 at 18:23 | comment | added | Greg Kuperberg | @Tyler: complexes are bounded on one side. @David: The top half of this solution together with the counterexample in the other answer are just a really nice answer. I learned stuff from it. | |
| Dec 15, 2009 at 17:19 | comment | added | Tyler Lawson | Minor comment: You need to assume something like the domain complex being bounded below (or having an exhaustive filtration with subquotients being bounded-below complexes of projectives) in order to be able to show that a map f which is zero in the derived category implies the existence of a homotopy from f to 0. | |
| Dec 15, 2009 at 15:12 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Dec 15, 2009 at 15:00 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Dec 15, 2009 at 14:52 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Dec 15, 2009 at 14:36 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Dec 15, 2009 at 13:25 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Dec 15, 2009 at 12:59 | history | answered | David E Speyer | CC BY-SA 2.5 |