Stereographic projection:

Consider the sphere

$$S^n = \Big\{(x_1, x_2, \dots, x_{n+1})\in \mathbb{R}^{n+1} \;\vert \; x_1^2 + x_2^2 + \dots + x_{n+1}^2 = 1\Big\}$$

and the plane $x_{n+1} = 0$ of the equator. Let ${\bf n} = (0, 0, \dots,0, 1)$ and ${\bf x}$ be a general point on the equatorial plane. The line through ${\bf n}$ and ${\bf x}$ is described parametrically by $(1-t){\bf n} + t{\bf x}$ and meets the sphere at points corresponding to the roots of the quadratic equation

$$\langle (1-t){\bf n} + t{\bf x}, (1-t){\bf n} + t{\bf x} \rangle \;= \; 1.$$

The root $t = 0$ corresponds to the point ${\bf n}$ and the second root

$$t = \frac{2(1 - {\bf n}\cdot{\bf x})}{1 + \Vert{\bf x}\Vert^2 - 2{\bf n}\cdot{\bf x}}$$

is continuous with respect to ${\bf x}$ and provides a point $F({\bf x})\in S^n - \{{\bf n}\}$. The map $F$ is a bijective continuous map between the plane $x_{n+1} = 0$ and $S^n - \{{\bf n}\}$. Note that the origin is mapped to the south pole by $F$. The inverse map $G$ is called the stereographic projection. Let us now show that $G$ is also continuous whereby it follows that $F$ is a homeomorphism.

Well, let ${\bf y}$ be a point on the sphere minus the north pole ${\bf n}$. The ray emanating from ${\bf n}$ and passing through ${\bf y}$ meets the plane at the point

$$G({\bf y}) = \Big(\frac{y_1}{1-y_{n+1}}, \frac{y_2}{1-y_{n+1}},\dots, \frac{y_n}{1-y_{n+1}} \Big)$$

We see that $G$ is also continuous and so the sphere minus its north pole is homeomorphic to $\mathbb{R}^n$.

It is useful to note that the stereographic projection takes points ${\bf y}$ close to the north pole to points $G({\bf y})$ of $\mathbb{R}^n$ such that $\Vert G({\bf y})\Vert \rightarrow +\infty$. We summarize the discussion as a theorem.