Showing that stereographic projection is a homeomorphism

For any $n\geq 0$,the unit $n$-sphere is the space $S^{n}\subset \mathbb{R^{n+1}}$ defined by

$$S^{n}=S^{n}(1) :=\left\{ (x_{1},...,x_{n+1}) \left\vert\,\sum_{i=1}^{n+1} x_{i}^{2}=1\right.\right\}$$ with the subspace topology .The point $P=(0,...,0,1)$ in $S^{n}$ called the north pole. Then we have the following conclusion:

$\qquad\qquad\qquad$For any $n\geq 0$, $S^{n}$ with the north pole removed is homeomorphic to $\mathbb{R^{n}}$.

$\qquad\qquad\qquad\qquad\qquad$ From FIGURE 1-1 ,we define a map $f:S^{n}\backslash\{P\}\rightarrow\mathbb{R^{n}}=\left\{ (x_{1},...,x_{n+1})\in \mathbb{R^{n+1}}\left\vert x_{n+1}=0\right. \right\}$.

(This map is called stereographic projection.)

It can be described geometrically as follows :Given a point $x=(x_{1},...,x_{n+1})\in S^{n} \backslash \{P\}$, stereographic projection sets $f(x) = y,$ where $(y,0)$ is the point where the line through $P$ and $x$ meets the subspace $\mathbb{R^{n}}\times \{0\}$ at $u=(y,0)=(y_{1},...,y_{n},y_{n+1}).$

Hence $$\left\{\begin{matrix} u=\lambda x+(1-\lambda )P \\ y_{n+1}=0 \end{matrix}\right.\quad(\lambda \in\mathbb{R})\Longrightarrow \lambda =\frac{1}{1-{x}_{n+1}}.$$ so the analytical expression of the stereographic projection is $$f(x_{1},...,x_{n+1})=\left(\frac{x_{1}}{1-{x}_{n+1}},...,\frac{x_{n}}{1-{x}_{n+1}},0\right)\in \mathbb{R^{n}}.$$

From the geometric intuition,we can get the map $f:S^{n}\backslash\{P\} \rightarrow\mathbb{R^{n}}$ is bijective ,but how can I prove it theoretically,further I need some rigorous proofs that $f:S^{n}\backslash\{P\}\rightarrow\mathbb{R^{n}}$is a homeomorphism.(We must strictly demonstrate $f$ injectivity,surjectivity,continuous.)

Another question: I know the analytical expression of the inverse $f^{-1}: \mathbb{R^{n}}\rightarrow S^{n}\backslash\begin{Bmatrix} P\end{Bmatrix}$ is $$f^{-1}(y_{1},...,y_{n})=\frac{1}{{\|y\|}^{2} +1}(2y_{1},...,2y_{n},{\|y\|}^{2} -1).\quad {\|y\|}^{2}=y_{1}^{2}+...+y_{n}^{2}.$$ but how can I get this expression .

Who can provide some related materials or solutions about these?(I want use some knowledge of Differential Calculus in Several Variables to solve these questions). Any of your help will be appreciated!

• I think using topological language will be useful and give a cleaner proof than epsilon-delta language in showing the continuity. By considering the composition of f and the projection maps, we can show the continuity clearly. After all, when we are talking about homeomorphisms, we are dealing with topological properties. – Oscar LIU Apr 26 '17 at 10:44

Hint: To show that $f$ is bijective, one can often (as can be managed here) compute $f^{-1}$ explicitly. Then, to show that $f$ is a homeomorphism, by definition it remains to show that $f$ and $f^{-1}$ are continuous, but they are both visibly compositions of continuous functions.
• Is $S^n \setminus \{p\}$ a compact space? I think it isn't because $S^n$ is closed and removing a point from it no longer makes it a closed space. – null May 6 '18 at 4:29
• It is noncompact for $n > 0$ (but notice that for $n = 0$ it is compact, so the reasoning you give cannot hold in general). – Travis May 6 '18 at 16:01
• Right, for $n=0$ , it is just $x^2 = 1$ that is two points +1, -1. I don't want to say $S^n \setminus \{p\}$ is not compact as is it is homeomorphic to $R^n$ which is non-compact. I think going back to the basic definition of compactness, around $\{p\} = (0,0,...1)$ open cover generated by open balls on $S^n \setminus \{p\}$ with fixed n-th co-ordinate $x_n = (1- \dfrac{1}{n}) \ \ \forall n$ have no finite subcover, hence it is non-compact. Please tell me if I make sense. – null May 7 '18 at 4:39
Alternately, stereographic projection maps circles to circles, and so for topological reasons $f$ maps open disks in the planes into open discs on $S^n - \{p\}$ (regions with circular boundary). But the set of open disks in the plane is a basis for the standard topology on the plane, and hence (once you know it exists) $f^{-1}$ is continuous. Similarly, so is $f$.