Timeline for Homotopy groups of a Bouquet of n-spheres
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 15, 2012 at 12:12 | comment | added | Hugo Chapdelaine | @Ryan, yes I meant that the complement of the unknot in $\mathbf{R}^3$ deformation retracts to $S^2\vee S^1$. | |
| Mar 14, 2012 at 23:11 | comment | added | Ryan Budney | The $n$-component unlink in $\mathbb R^3$ has the homotopy-type of a wedge $n$ copies of $S^2$ and $n$ copies of $S^1$. | |
| Mar 14, 2012 at 22:42 | comment | added | Ryan Budney | I mean the unknot. The complement of the unknot in $\mathbb R^3$ has a geometrically natural deformation retract consisting of a 2-sphere that contains the unknot (in the Jordan-Schoenflies sense) together with an arc that connects two distinct points in $S^2$. This is the arc that runs from the north pole to the south pole, through the centre of the unknot (in the sense of centroid). But this 2-complex is homotopy-equivalent to $S^2\vee S^1$, just homotope the attaching map of the arc so that the two endpoints go to the same place | |
| Mar 14, 2012 at 22:36 | comment | added | Scott Carter | @Ryan Do you mean unlink in your lower comment? | |
| Mar 14, 2012 at 20:53 | comment | added | Ryan Budney | The complement of the unknot in $\mathbb R^3$ is homotopy-equivalent to $S^1 \vee S^2$. I suspect you're thinking of the universal cover of this space. The universal cover of $S^1 \vee S^2$ is homotopy-equivalent to $\vee_\infty S^2$. | |
| Mar 14, 2012 at 20:46 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Mar 14, 2012 at 20:45 | comment | added | Hugo Chapdelaine | You are right completely right Ryan, I forgot the point at $\infty$, it is the unknot complement in $\mathbf{R}^3$ which is homotopy equivalent to a countable wedge of $S^2$. I'll remove my motivation. | |
| Mar 14, 2012 at 19:10 | comment | added | Ryan Budney | To be precise, an unknot complement in $S^3$ is homeo/diffeomorphic to $S^1 \times \mathbb R^2$. A countable wedge of 2-spheres has $\pi_2 X$ isomorphic to a direct sum of countable-many copies of $\mathbb Z$. Knot complements in the 3-sphere are all $K(\pi,1)$ type spaces. Wedges of spheres have infinitely many non-trivial homotopy groups. | |
| Mar 14, 2012 at 19:05 | comment | added | Ryan Budney | This is called the Hilton-Milnor theorem. Rationally the homotopy groups are a free lie algebra with respect to the whitehead product. Google provides you with the appropriate references. The result you state about the countable wedge of 2-spheres related to knot complements is false. | |
| Mar 14, 2012 at 19:01 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |