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I thought about the fundamental group of a sphere where we remove two arbirary poins. Then I thought thad geometrically one can mabye find a deformation retract to the torus and thus the fundamental group would be $\Bbb{Z} \times \Bbb{Z}$ but I‘m not sure uf this is true.

Could maybe someone help me?

Thanks a lot

José Carlos Santos
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user1294729
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    If you remove one, you get a (topological) disk. Removing two, you get a punctured disk, which retracts to a circle, eh? – paul garrett Jan 30 '22 at 22:19
  • Sorry I don‘t see why removing two points from a sphere guves s retraction to a circle is there a possibility that you can draw it ? – user1294729 Jan 30 '22 at 22:22
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    Well, the first puncture makes a disk. Then a disk with its center gone can be "thinned" (topologically) to be the thinnest possible annulus, a circle. No? Shrink the outer edge's radius to 1, and increase the inner radius to 1 ... a circle. – paul garrett Jan 30 '22 at 22:23
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    Paul dissected this problem into two steps for you. Which step was unclear? "Both" is a fine answer, but just reiterating that you don't understand is not helpful. – diracdeltafunk Jan 30 '22 at 22:24
  • Aha so when I take away the first point from the sphere to get a disk I need to enlarge the given whole and wrap the surface around? – user1294729 Jan 30 '22 at 22:29

2 Answers2

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A different way to see it is that if you take the standard sphere in $\mathbb R^3$ and remove the north and south pole, you can take a projection (that's also a homeomorphism) onto the cylinder $\{(x,y,z) : x^2+y^2=1, |z| \leq 1\}$.

a formula would be to literally send $$(x,y,z) \mapsto \left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},z\right).$$

Note that on the North and South Pole, this formula is not well-defined! It's good that we removed it.

Geometrically this is kind of like "widening" a two holes in a sphere until you get to the cylinder.

This is actually a pretty common homeomorphism!

Once you have a cylinder, the circle $S^1$ is a deformation retract of your space.

Andres Mejia
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  • Very nice argument, +1. One convenient aspect of the usual stereographic projection argument given in José Carlos Santos's answer is that the same argument can be adapted to the case when one removes $n > 1$ points from the sphere. Is there a way to see what's going on in this answer too? It's slightly less clear to me geometrically how to see that this deformation retracts to a wedge of $n-1$ circles. – Alex Wertheim Jan 31 '22 at 03:24
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    @AlexWertheim Truthfully, I think the way that one proceeds will be very similar, since topologically a cylinder is just an annulus, so you would once again resort to having a disk with $(n-1)$ punctures :). I'm not sure about something more clever! – Andres Mejia Jan 31 '22 at 03:28
  • sorry are $(1,0,0)$ and $(-1,0,0)$ your north and south pole? – user1294729 Jan 31 '22 at 09:22
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    @aprozz yep that’s exactly right – Andres Mejia Jan 31 '22 at 21:25
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The sphere minus one point is homeomorphic to $\Bbb R^2$. Therefore, the sphere minus two points is homeomorphic to $\Bbb R^2\setminus\{(0,0)\}$, whose fundamental group is $(\Bbb Z,+)$.

José Carlos Santos
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