Timeline for Are the C(S^n, S^n)'s homeomorphic ?
Current License: CC BY-SA 2.5
11 events
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| Dec 13, 2009 at 16:51 | comment | added | Mariano Suárez-Álvarez | @Igor: my question "why is it enough..." was not addressed to you but to a comment that disappeared :P | |
| Dec 13, 2009 at 16:48 | comment | added | Reid Barton | Ah, and I was only speaking with reference to the question in the title, not the related questions. | |
| Dec 13, 2009 at 14:57 | comment | added | Reid Barton | @Igor: I agree with your point 3--specifically, the connecting maps in the homotopy LES might differ for various components; but you didn't use anything about them in your argument. | |
| Dec 13, 2009 at 13:43 | history | edited | Igor Belegradek | CC BY-SA 2.5 |
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| Dec 13, 2009 at 13:26 | history | edited | Igor Belegradek | CC BY-SA 2.5 |
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| Dec 13, 2009 at 7:21 | comment | added | Reid Barton | $SF_n = \Omega^n S^n$ is a loop space, so its components are all homotopy equivalent, and therefore your argument should apply to any component of $C(S^n, S^n)$. | |
| Dec 13, 2009 at 6:04 | comment | added | Andy Putman | By the way, I deleted my comment because after reading Igor's post more carefully, I realized that everything I said in it is basically contained in his post. | |
| Dec 13, 2009 at 5:32 | comment | added | Ady | Also, what about the second related question ? | |
| Dec 13, 2009 at 5:21 | comment | added | Mariano Suárez-Álvarez | Why is it enough to show the identity components are not homeomorphic? | |
| Dec 13, 2009 at 4:39 | comment | added | Ady | I can agree C(S^n, S^n) (as a space of pointed maps) has the same homotopy groups as S^n just shifted by n dimensions. But the original question was about C(S^n, S^n) as a whole, i.e., no basepoint, no pointed maps. Also, if two components are not homeo, does it follow the two spaces are not homeo ? | |
| Dec 13, 2009 at 4:08 | history | answered | Igor Belegradek | CC BY-SA 2.5 |