I was told that it was $\mathbb{Z}$, and I can imagine a subgroup isomorphic to $\mathbb{Z}$ of 'wrappings' of the sphere around the point, but I am still convinced there are more homotopy classes. I don't see how the picture below is in one of the classes described above. Essentially the top and bottom of the sphere have been partially pushed through each other, and the red dot is the hole, which lies inside the middle pocket.
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2Well it deformation retracts onto $S^2$ – Seth Apr 20 '14 at 02:13
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1+1 I love this kind of question, where you're trying to reconcile something you've heard with what you know. – Ben Blum-Smith Apr 20 '14 at 02:25
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$X=\mathbb{R}^3\setminus \{(0,0,0)\}$ is homotopy equivalent to a 2-sphere $S^2$ since it deformation retracts onto the sphere by sending each open ray from the origin to the point where it hits the unit sphere. Therefore $\pi_2(X)=\pi_2(S^2)$. Thus the elements are exactly the homotopy classes of maps from $S^2$ to itself. These in turn are classified by the Brouwer degree of the map. Thus there is one for each integer and this identifies $\pi_2(S^2)$ with $\mathbb{Z}$.
In the example you're struggling with, the map is actually homotopic to the map that wraps the sphere around the origin once, with an orientation reversal (so corresponds to $-1\in\mathbb{Z}$). One way to see this is just by imagining inflating the middle pocket till it squeezes out the "outer pocket" and becomes the whole thing. (This is not a smooth homotopy because the "outer pocket" will become singular before disappearing, but it is a topological homotopy.)
Ben Blum-Smith
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For what it's worth: There's not an issue with making that homotopy smooth, only an issue with making it regular (a homotopy through immersions). – Nov 15 '18 at 09:53
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Right, I was just saying that the particular homotopy I described (on account of it being comparatively easy to visualize) is not smooth. ... But wait, I assumed the smooth homotopy you were thinking of was (the last 2/3rd of) this one https://www.youtube.com/watch?v=-6g3ZcmjJ7k and it's not obv. to me how it can avoid the origin? – Ben Blum-Smith Nov 16 '18 at 04:00
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I know this is a very old post, but here's a quick answer to your last question. Instead of thinking of this as a circle pulled through itself with a hole inside the center part, look at your picture left to right and think of it as a circle in the plane, not surrounding the hole, wedged with a circle surrounding the hole, then wedged with a circle not surrounding the hole.
Recall the plane is contractible. Meaning those loops on the left and right that are in the plane, not surrounding the hole, are contractible. So your picture is homotopy equivalent to a single circle surrounding the hole. So not only is your picture an element of the fundamental group, it's a generator!
Fun Fact: The above argument works perfectly well for $\pi_n(S^n)$ too!
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I think "homotopy equivalent" in the second paragraph should be "homotopic" -- a circle is certainly not homotopy equivalent to the wedge of three circles. – Ben Blum-Smith Apr 01 '21 at 19:11