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I want to compute the fundamental group of a unit sphere $S^2$ placed at origin with three diameters along the three axes in $\mathbb{R}^3$. I understand that a sphere with one diameter is equivalent to the wedge sum of a sphere and a circle and thus, its fundamental group is $\mathbb{Z}$. But what happens in the case of more than one diameter? Also, I want to compute the fundamental group based at point $(1,0,0)$. Does the base point matter for the fundamental group of this space or it would be same for any point in my space? Can someone give me any hint on how to proceed?

Thanks

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But what happens in the case of more than one diameter?

A sphere with two diameters is homotopy equivalent to the wedge sum of a sphere with a bouquet (wedge sum) of three circles. To see this:

  1. Start with your sphere with two diameters in its interior
  2. Redraw the two diameters exterior to the sphere, which is homeomorphic.
  3. Slide the points of contact of the diameters until they coincide, which is homotopy equivalent (so now you have the wedge sum of the sphere with two circles intersecting at opposite points)
  4. And finally then collapse one of the edges to a point.

As shown in the sketch below, you now have $S^2\vee S^1\vee S^1\vee S^1$. Hence the fundamental group is free group on three generators.

sketch

Similarly, three diameters will give you the wedge sum of a sphere with a bouquet of five circles. Fundamental group is free group on five generators.

Also, I want to compute the fundamental group based at point $(1,0,0)$. Does the base point matter for the fundamental group of this space or it would be same for any point in my space?

Basepoint does not matter if the space is path connected, which this is.

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  • $\begingroup$ This is perfect answer. Thanks a lot! $\endgroup$ – Khushboo Nov 25 '17 at 18:08

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