Constant loop homotopic to circle on the sphere

Question

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?

Answer 1

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 $z\neq 1$.

Answer 2

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.