Timeline for Torsion in homology or fundamental group of subsets of Euclidean 3-space
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 12 at 4:04 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| Jan 20, 2022 at 23:59 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
fixed arxiv front-end link and gave full published reference with doi link
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| Nov 19, 2010 at 20:24 | comment | added | Sergey Melikhov | Hi Ryan, sorry that I messed up dimensions in the original response. | |
| Nov 19, 2010 at 19:51 | history | edited | Sergey Melikhov | CC BY-SA 2.5 |
added 82 characters in body
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| Nov 19, 2010 at 19:43 | comment | added | Ryan Budney | Dear Sergey, sorry for taking so long to get back on this. Yes I believe you are correct and I stated too much in ruling out torsion. I'll correct that statement. | |
| Nov 19, 2010 at 19:35 | history | edited | Sergey Melikhov | CC BY-SA 2.5 |
added 129 characters in body; deleted 7 characters in body
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| Nov 19, 2010 at 19:26 | history | edited | Sergey Melikhov | CC BY-SA 2.5 |
minor correction
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| Nov 19, 2010 at 19:18 | history | edited | Sergey Melikhov | CC BY-SA 2.5 |
corrected a misprint
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| Nov 16, 2010 at 11:10 | comment | added | Sergey Melikhov | Perhaps I should clarify that the Barratt-Milnor example, as I see it, shows precisely that singular homology is pathological on compact metric spaces (because it doesn't feel dimension). I wanted to understand why this is so, but haven't signed up for all pathology. There are other ways in which singular homology is pathological on compact metric spaces: it fails the strong excision axiom and the Alexander duality; it doesn't understand in any way the operation of inverse limit; K(Z,n) doesn't represent singular cohomology. Steenrod homology is free of all these deficiencies. Hope this helps. | |
| Nov 15, 2010 at 19:38 | comment | added | Sergey Melikhov | I'm only saying that one who is interested in whether there is torsion in singular $H_4$ of the $2$-dimensional Barratt-Milnor example (which I think is conceivable) might find it helpful to look at my geometric proof of the Barratt-Milnor original result that its singular $H_3$ is nonzero (and in fact uncountable). Personally I am not interested at all, because I don't know of any single true application of singular homology/homotopy beyond the case where it coincides with Steenrod homology/homotopy. (They coincide on spaces that are homotopy equivalent to ANRs, including CW-complexes.) | |
| Nov 14, 2010 at 10:13 | comment | added | BS. | Are you saying that there is torsion in $H_4$ of Barratt-Milnor ? | |
| Nov 13, 2010 at 23:51 | history | edited | Sergey Melikhov | CC BY-SA 2.5 |
added 3 characters in body
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| Nov 13, 2010 at 23:43 | history | answered | Sergey Melikhov | CC BY-SA 2.5 |