What is the second homotopy group of $R^3 \setminus \{ (0,0,0) \}$

Question

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.

https://i.stack.imgur.com/x2yLj.jpg

Answer 1

$X=\mathbb{R}^3\setminus \{(0,0,0)\}$ 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.)