Timeline for Transversality in the proof of the Blakers-Massey Theorem. Is it necessary?
Current License: CC BY-SA 2.5
8 events
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| May 27, 2013 at 14:53 | comment | added | Ronnie Brown | @John Klein: As an algebraic topologist, I prefer to get the connectivity result from a determination of the critical group in the Blakers-Massey theorem as a tensor product, and then to note that a tensor product is zero if one of the terms is zero. This kind of explains the dimensions involved. It also suggests the problem of whether one can use the techniques of Puppe/Goodwillie/Munson to give a different, and hopefully more geometric, proof of the van Kampen theorem for $n$-cubes of spaces proved with J.-L. Loday, see my answer below. That theorem also has other consequences. | |
| May 31, 2012 at 18:47 | comment | added | John Klein | @Mark Grant: yes, it seems to do just that. In fact Brian wrote me an email the other day about this. | |
| May 31, 2012 at 8:31 | comment | added | Mark Grant | @John Klein: Does this preprint of Brian Munson arxiv.org/abs/1205.6668 answer your extended question? | |
| Feb 6, 2011 at 23:05 | comment | added | John Klein | I'd rather not at this point. | |
| Feb 4, 2011 at 9:35 | comment | added | j.p. | @John Klein: It might be better if you post the question in your comment as a separate question. | |
| Feb 4, 2011 at 5:02 | comment | added | John Klein | I just skimmed the argument. I think you're right. I now have another question related to this: Can one derive the $n$-ad connectivity theorem along similar lines? | |
| Feb 4, 2011 at 5:01 | vote | accept | John Klein | ||
| Feb 2, 2011 at 19:08 | history | answered | Charles Rezk | CC BY-SA 2.5 |