Timeline for Homotopy groups of spheres in a $(\infty, 1)$-topos
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2016 at 22:38 | answer | added | Mike Shulman | timeline score: 10 | |
| Feb 15, 2016 at 10:30 | answer | added | Neil Strickland | timeline score: 2 | |
| Feb 14, 2016 at 14:56 | answer | added | user2529 | timeline score: 1 | |
| Oct 27, 2011 at 21:39 | vote | accept | Guillaume Brunerie | ||
| Oct 27, 2011 at 21:03 | history | edited | Guillaume Brunerie | CC BY-SA 3.0 |
Corrected bug in links
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| Oct 27, 2011 at 20:49 | history | edited | Guillaume Brunerie | CC BY-SA 3.0 |
Added the addition.
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| Oct 25, 2011 at 15:54 | answer | added | Hiro Lee Tanaka | timeline score: 12 | |
| Oct 25, 2011 at 14:33 | answer | added | Charles Rezk | timeline score: 28 | |
| Oct 24, 2011 at 23:35 | comment | added | Sam Gunningham | If the (ordinary) category of sets counts as an $\infty$-topos, then the spheres are all points, and the homotopy groups of those are all known. | |
| Oct 24, 2011 at 23:30 | comment | added | Sam Gunningham | I think the fact that $\pi _k (S^n)$ is a group should follow from general nonsense: $\Omega^k X$ is an $E_k$-object in $H$, so $\pi _0$ of it will be a group if $k>0$ (and abelian if $k>1$). I also think there would be a counterexample to your second point if you take $H$ to be sheaves of spaces over a sphere $S^k$. There will be a non-trivial global section of the constant sheaf with fibres $S^k$ (which should be the k-sphere object $\mathbb S^k$ in this category), so that $\pi _0(\mathbb S^k )=\mathbb Z$. | |
| Oct 24, 2011 at 23:11 | history | edited | Guillaume Brunerie | CC BY-SA 3.0 |
Changed "global elements" to "connected components"
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| Oct 24, 2011 at 23:08 | comment | added | Guillaume Brunerie | Thanks, I guess I wanted to take the collection of global elements up to homotopy, but for some reason I did not write the "up to homotopy" part. | |
| Oct 24, 2011 at 22:32 | comment | added | David Roberts♦ | A brief comment: if you take the collection of global elements of $\Omega^k\mathbb{S}^n$, then this naturally wants to form the 0-cells of an $\infty$-groupoid. Unless you take connected components it will not be a group, compare the case of $\Omega^k S^n$ in $Top$ - it is only an $A_\infty$-space, not a group. | |
| Oct 24, 2011 at 18:00 | history | asked | Guillaume Brunerie | CC BY-SA 3.0 |