Prove that the unit circle,
$$S^1:=\{(x,y)\in \mathbb{R}^2: x^2+y^2=1\}$$,
Is a one-dimensional manifold in $\mathbb{R}^2$ by using the following coordinate chars, known as stereographic projections, and completing the following steps.
a)Show that $U_N:=S^1$\ $\{(0,1)\}$ and $U_S:=S^{-1}$\ $\{(0,1)\}$ are open subsets of S^1. Here $(0,1)\in\mathbb{R}^2$ is the north pole of $S^1$ and $(0,-1)$ is the south pole.
b) Find an explicit formula for the function $\psi_S:=U_S\rightarrow\mathbb{R}$ that sends a point $(x,y)\in U_S$ to the horizontal coordinate obtained by the straight line that passes from the north pole through $(x,y)$ and intersectimg the line $y=-1$.
My attempt
a)
Let $\gamma_1:\{S^1-\{0,1\}\}\rightarrow \mathbb{R} \times \{0\}$
Be defined by: ?
Let $\gamma_2:\{S^1-\{0,-1\}\}\rightarrow \mathbb{R} \times \{0\}$ Be defined by: ?
Let $U\subset \mathbb{R}^2=\{S^1-\{0,1\}\}$
Let $V\subset \mathbb{R}^2=\{S^1-\{0,-1\}\}$
$U\cup V=S^1$ so for $c\in U\cup V=S^1$, $c\in U$ or $c\in V$. If $c\in U$ then $\gamma_1(U)=\mathbb{R}$ and likewise if $c\in V$ then $\gamma_2(V)=\mathbb{R}$, which are both open in $\mathbb{R} \times \{0\}$
If my attempt for part a is right, how should i define the functions and why?
b) If $\gamma_1$ and $\gamma_2$ are diffeomorphisms then we know $S^1$ is a smooth manifold of dimension $1$. Now we have to show that they are invertable.
Define $\phi_1:\mathbb{R}\rightarrow U$ by?
Define $\phi_2:\mathbb{R}\rightarrow V$ by?
As long as $\gamma_1$ and $\phi_1$ are inverses and $\gamma_2$ and $\phi_2$ are inverses, then we proved that the unit circle is a smooth manifold of dimension $1$.
Best Answer
Hint :
For your first question (a), the openness follows from the definition of the induced topology on $S^1$.
Concerning your second question (b), the simplest way is to draw a picture and work out the explicit maps from there. I've sketched the situation below.
Recall that the north pole is $N=(0,1)$, take any point $A$ on the circle and compute the coordinates of the intersection of $(NA)$ and the line $\{y=-1\}$.