algebraic-topologyfundamental-groupsspheres
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
The approach is correct, but you can use a simpler method. The space you care about is homotopic to the wedge sum of two spheres and one circle. Thus the fundamental group is Z. To see why this space is homotopic to the space I said, you just check first that you can contract the upper half-sphere, thus the lower half-sphere becomes to a sphere, and the torus becomes a sphere with its south and north poles glued together. A sphere with its south and north poles glued together is homotopic to a sphere with a line connect its south and north poles. After these deformations you have a space which is what I said above.
We have two representations of the Klein bottle as a fundamental polygon. The first is:
And the second is formed by cutting this polygon across the diagonal, flipping one piece and reattaching it to give essentially two real projective planes glued together:
You should be able to see that as CW complexes and a 2-cell attached according to the diagram, these both form the Klein bottle with non-orientable genus 2.
Removing a point from this 2-cell produces a space that deformation rectacts onto the 1-skeleton, which in both cases obviously forms the wedge sum of two circles and the fundamental group is $\Bbb{Z} * \Bbb{Z}$.
Let's see if we can develop some sort of physical intuition for this. If the point (or by deformation, hole) we remove is in the right place, we can embed this in $\Bbb{R}^3$ to get a physical intuition.
Which then forms 
And you can see rather easily that this deforms to:
Which obviously has the fundamental group of $\Bbb{Z} * \Bbb{Z}$, as this deformation retracts onto $S^1 \vee S^1$.
Best Answer
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.