Can one use only two coordinate patches to piece up a circle?

Question

Let $S := \{ (x,y) \in \mathbb{R}^{2} \mid x^{2} + y^{2} = 1\}$. Then $S$ is a smooth 1-manifold, for we can simply cover $S$ by the diffeomorphisms $\alpha_{1}: x \mapsto (x, \sqrt{1-x^{2}}) :\ ]-1,1[ \to \mathbb{R}^{2}$, $\alpha_{2}: x \mapsto (x, -\sqrt{1-x^{2}}):\ ]-1,1[ \to \mathbb{R}^{2},$ $\alpha_{3}: y \mapsto ( -\sqrt{1-y^{2}},y):\ ]-1,1[ \to \mathbb{R}^{2},$ and $\alpha_{4}: y \mapsto (\sqrt{1-y^{2}},y):\ ]-1,1[ \to \mathbb{R}^{2}.$

But is it possible to piece up a circle by only two coordinate patches?

Answer 1

Sure. Define $\beta_1(t)=(\cos t, \sin t)$ and $\beta_2(t)=(-\cos t, \sin t)$, both with domain $]-\pi, \pi[$.

In general, a sphere with a point deleted is diffeomorphic to $\Bbb{R}^n$, which means that any sphere can be defined using only two coordinate patches (one for the entire sphere minus the point, and another for a neighborhood of the point... or the entire sphere minus a different point).

Answer 2

Let $N$ denote the north pole, $(0, 1)$, of the unit circle $\Bbb S^1 \subset \Bbb R^2$ centered at the origin. The map $$\phi : \Bbb R \to \Bbb S^1 - \{N\}$$ defined by projecting each point $X \in \Bbb R \leftrightarrow \Bbb R \times \{0\} \subset \Bbb R^2$ to the unique point where the line $\overline{NX}$ intersects $\Bbb S^1$; this map is called stereographic projection and generalizes readily to higher-dimensional spheres. Explicitly, this map is $$\phi: X \mapsto \left(\frac{2 X}{X^2 + 1} , \frac{X^2 - 1}{X^2 + 1}\right) .$$

One can just as well define a parameterization of all of $\Bbb S^1$ less another point, and hence cover $\Bbb S^1$ with just two such maps. One way to do this is to compose $\phi$ with the antipodal map $\alpha: \Bbb S^1 \to \Bbb S^1$ defined by $Z \mapsto -Z$, the result of which is stereographic projection w.r.t. the south pole, $-N$: $$\alpha \circ \phi : \Bbb R \to \Bbb S^1 - \{-N\}.$$