Timeline for Uniqueness of loop spaces
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| May 18, 2011 at 14:55 | history | edited | Dr Shello | CC BY-SA 3.0 |
added 24 characters in body
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| May 17, 2011 at 6:06 | history | edited | Dr Shello | CC BY-SA 3.0 |
added 165 characters in body
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| May 16, 2011 at 14:54 | history | edited | gowers | CC BY-SA 3.0 |
Corrected spelling
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| May 16, 2011 at 8:31 | comment | added | S. Carnahan♦ | You might also ask a slightly different question: "Given a homotopy equivalence of based loop spaces, how do I tell if it is a loop map?" | |
| May 16, 2011 at 7:47 | answer | added | Tilman | timeline score: 12 | |
| May 16, 2011 at 6:25 | comment | added | Dr Shello | @Mariano: yes, right. The question is what assumptions are needed to make that true. | |
| May 16, 2011 at 6:23 | comment | added | Dr Shello | @Tyler: this is interesting - what are references for this sort of thing? | |
| May 16, 2011 at 6:19 | comment | added | Dr Shello | Yes, of course you would want $Y$ to be connected to avoid trivialities. | |
| May 16, 2011 at 3:57 | comment | added | Tyler Lawson |
Don't have time to leave a full answer, but if the homotopy groups of X are concentrated in a narrow "stable" range (roughly the top nonzero group below about twice the dimension of the bottom nonzero group) then there is a unique homotopy type. Similarly if Y has very small dimension relative to its connectivity. One can compare the obstruction theory for $Map(Y_1, Y_2)$ with that for $Map(\Omega Y_1, \Omega Y_2)$, together with a comparison of the cohomology of $Y_1$ and $\Omega Y_1$, to get a fuller statement.
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| May 16, 2011 at 3:27 | answer | added | Mike Shulman | timeline score: 25 | |
| May 16, 2011 at 3:14 | comment | added | Ryan Budney | Clearly the answer to that is no. | |
| May 16, 2011 at 3:12 | comment | added | Mariano Suárez-Álvarez | The question is, probably: «if $\Omega X_1$ and $\Omega X_2$ are homotopy equivalent, are $X_1$ and $X_2$ homotopy equivalent?» | |
| May 16, 2011 at 3:01 | comment | added | Ryan Budney | You could of course make it unique by choosing $Y$ to be $BX$ but it's not clear (to me) what you're after. | |
| May 16, 2011 at 2:54 | comment | added | Ryan Budney | It's never unique. Given any pointed space $Y$ with $\Omega Y \simeq X$, you can take the disjoint union of $Y$ with another (unpointed) space $W$, and $\Omega (Y \cup W) \simeq X$. This is keeping the base-point from $Y$. | |
| May 16, 2011 at 2:50 | history | asked | Dr Shello | CC BY-SA 3.0 |