Stereographic Projection of $$S^1$$ into $$\mathbb{R}$$

Let $$(x, y)$$ denote coordinates in $$\mathbb{R}^2$$ and suppose that $$x^2 + y^2 = 1$$. We can parametrize the unit circle centered at the origin $$S^1$$ by the equations

x = \sin t,\quad y = \cos t.

This parametrization goes around the circle clockwise starting at the north pole $$N = (0, 1)$$, where $$t = 0$$. Geometrically, stereographic projection of $$S^1 \setminus\{N\}$$ onto $$\mathbb{R}$$ is the map defined by extending a line through the points $$N$$ and $$(x, y)$$ to the $$x$$-axis. The point $$(\xi, 0)$$ where this line and the $$x$$-axis meet is the stereographic projection of $$(x, y)$$ onto $$\mathbb{R}$$ (after dropping the $$y$$-coordinate). This map is given by
\label{eq:Projection}
\xi = \frac{x}{1 – y} = \frac{\sin t}{1 – \cos t}.

To obtain (\ref{eq:Projection}), we proceed as follows. Fix a value of $$t \in (0, 2\pi)$$ and let $$\gamma: \mathbb{R} \rightarrow \mathbb{R}^2$$ be the parametrization of the line through $$N$$ and the point $$(\sin t, \cos t)$$ given by
\label{eq:ParametricLine}
\gamma(u) = (1-u)\,(0,1) + u\,(\sin t, \cos t).

If we write $$\gamma(u) = ( \xi, \eta )$$ for some value of $$u$$, we see that (\ref{eq:ParametricLine}) is equivalent to the pair of equations
\label{eq:XiEta}
\xi = u\,\sin t,\quad \eta = 1 – u + u\,\cos t.

The point $$( \xi, \eta )$$ lies on the $$x$$-axis if $$\eta = 0$$. Solving for $$u$$, we obtain

u = \frac{1}{1 – \cos t}.

Substituting this into the expression for $$\xi$$ in (\ref{eq:XiEta}) gives (\ref{eq:Projection}).

We now turn this calculation around to obtain inverse stereographic projection, which provides us with a parametrization of $$S^1$$ by $$\xi$$. To that end, we define
\label{eq:INverseParametricLine}
\rho:\mathbb{R}\rightarrow S^1\setminus\{N\},\quad \rho(u) = (1-u)\,(0,1) + u\,(\xi, 0).

Writing out the components (as before) we obtain
\label{eq:XY}
x = u\,\xi, \quad y = 1 – u.

In order for $$(x, y)$$ to be a point on $$S^1$$, we must have

x^2 + y^2 = u^2\xi^2 + (1 – u)^2 = 1.

This simplifies to
\label{eq:U}
u\,[u\,(1 + \xi^2) – 2] = 0.

It then follows that
\label{eq:USolution}
Now, a line from $$N$$ to the point $$(\xi, 0)$$ on the $$x$$-axis intersects $$S^1$$ in one other point, which is the point we seek. This point corresponds to the nonzero value of $$u$$ in (\ref{eq:USolution}). Substituting this value for $$u$$ into (\ref{eq:XY}), we find
Remark. The map defined by (\ref{eq:ISP}) is a rational map, that is, the coordinates $$x$$ and $$y$$ of a point on $$S^1$$ are rational functions of the coordinate $$\xi$$ on $$\mathbb{R}$$. Since the rational numbers $$\mathbb{Q}$$ are dense in $$\mathbb{R}$$, an immediate consequence of this is that the points in $$S^1$$ with rational coordinates are dense in $$S^1$$. This follows from the continuity of the map