Timeline for Other Homology Theories still Count Holes?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 16, 2012 at 19:17 | vote | accept | Chris Gerig | ||
| Mar 16, 2012 at 3:31 | history | edited | Chris Gerig | CC BY-SA 3.0 |
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| Mar 16, 2012 at 2:23 | history | edited | Chris Gerig | CC BY-SA 3.0 |
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| Mar 15, 2012 at 12:07 | answer | added | Liviu Nicolaescu | timeline score: 4 | |
| Mar 15, 2012 at 10:33 | answer | added | David Farris | timeline score: 7 | |
| Mar 15, 2012 at 8:06 | answer | added | Jonny Evans | timeline score: 7 | |
| Mar 15, 2012 at 7:21 | comment | added | Chris Gerig | @Andy, is this better? | |
| Mar 15, 2012 at 6:59 | history | edited | Chris Gerig | CC BY-SA 3.0 |
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| Mar 15, 2012 at 6:52 | history | edited | Chris Gerig | CC BY-SA 3.0 |
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| Mar 15, 2012 at 6:19 | answer | added | Steven Landsburg | timeline score: 21 | |
| Mar 15, 2012 at 5:03 | comment | added | Andy Putman | This is a pretty vague question. | |
| Mar 15, 2012 at 4:23 | comment | added | Ryan Budney | have no solution. | |
| Mar 15, 2012 at 4:22 | comment | added | Ryan Budney | Technically, singular homology does not quite count holes. $H_0 X$ is free abelian on the path-components of $X$, so there's one more copy of $\mathbb Z$ than the number of $0$-dimensional holes. Said another way ,if you treat a contractible space as "having no holes", then $H_0$ can't be measuring holes as it's not trivial. There's a calibration issue -- you need to take the associated reduced homology. That way, the homology theory is trivial on a contractible space. So sure, it measures holes, in that you can describe non-trivial homology classes as extension problems that... | |
| Mar 15, 2012 at 3:54 | history | asked | Chris Gerig | CC BY-SA 3.0 |