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I want to show that the loop $(\cos 2 \pi s, \sin 2 \pi s,0) \in S^2$ for $s \in [0,1]$ is homotopic to the constant loop with base point $(1,0,0)$. I can contract the loop $(\cos 2 \pi s, \sin 2 \pi s,0)$ to the constant loop using

$H(s,t)=(1-t)(\cos 2 \pi s, \sin 2 \pi s,0)+t(1,0,0)$

However, such a contraction does not stay on the sphere. I want to use some type of projection onto the northern hemisphere of $S^2$ but am not sure how to go about this.

Is projection the right way to go about this?

MovingUp
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  • Just divide by the norm of the vector $H(s,t)$ at each point. – Mikhail Katz Apr 28 '14 at 18:37
  • @user72649 but then this would still lie on the equator since dividing by the norm does not give us any height. – MovingUp Apr 28 '14 at 18:44
  • If $n$ is the norm of the vector then the vertical coordinate will be $\frac{t}{n}$ after division. Oh, I see. You should change $(1,0,0)$ to $(0,0,1)$ in your formula. – Mikhail Katz Apr 28 '14 at 18:46

2 Answers2

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Yes, take the stereographic projection of $S^2-N$ to the complex plane where $N$ is the north pole.(Riemann's projection).

This projection is a homeomorphism(stronger than homotopy). Then you can work on $\Bbb C$ or $\Bbb R^2$ and there you can shrink the circle to a point easily.

In general you can work these types of problems by working with the fundamental group. Here $S^2$ is simplyh connected and thus it's fundamental group is trivial.

$$p(x,y,z)=\frac {x+yi}{1-z}$$ for $z\neq 1$.

Haha
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Start with $H(s,t)=(1-t)(\cos 2 \pi s, \sin 2 \pi s,0)+t(0,0,1)$ and modify it by setting $K(s,t)=\frac{1}{|H(s,t)|} H(s,t)$. Then K homotopes the equator to the constant loop at the north pole. If you prefer the constant loop at $(1,0,0)$, follow the first homotopy by another one moving the north pole to $(1,0,0)$. There is really no need for exploiting fundamental groups, and moreover this type of homotopy is the basis for determining the fundamental group in the first place.

Mikhail Katz
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