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Let $n$ be any given natural number, and let $$ S^n \colon= \left\{ \, \left( x_1, \ldots, x_{n+1} \right) \in \mathbb{R}^{n+1} \, \colon \, \sum_{i=1}^{n+1} x_i^2 = 1 \, \right\}. $$ Let point $\mathbf{p} \in \mathbb{R}^{n+1}$ be given by $$ \mathbf{p} \colon= \left( 0, \ldots, 0, 1 \right). $$ Then of course $\mathbf{p} \in S^n$.

Now let the map $f \colon S^n \setminus p \rightarrow \mathbb{R}^n$ be given by $$ f \left( x_1, \ldots, x_n, x_{n+1} \right) \colon= \frac{1}{1-x_{n+1} } \left( x_1, \ldots, x_n \right). $$

Is this map $f$ a homeomorphism?

My Attempt:

Let $\left( u_1, \ldots, u_n, u_{n+1} \right)$ and $\left( v_1, \ldots, v_n, v_{n+1} \right)$ be any points in $S^n \setminus \mathbf{p}$ for which $$ f\left( u_1, \ldots, u_n, u_{n+1} \right) = f \left( v_1, \ldots, v_n, v_{n+1} \right). $$ Then we have $$ \frac{1}{1-u_{n+1}} \left( u_1, \ldots, u_n \right) = \frac{1}{1-v_{n+1}} \left( v_1, \ldots, v_n \right). $$ So for each $i = 1, \ldots, n$, we have $$ \frac{u_i}{1 - u_{n+1} } = \frac{v_i}{1-v_{n+1} }, $$ which is the same as $$ \frac{u_i}{1 - \sqrt{ 1 - \sum_{j=1}^n u_j^2 } } = \frac{ v_i }{ 1 - \sqrt{ 1 - \sum_{j=1}^n v_j^2 } }, \tag{1} $$ because we have the equalities $$ \sum_{j=1}^{n+1} u_j^2 = 1 = \sum_{j=1}^{n+1} v_j^2. $$

What next? How to show from here that $$ \left( u_1, \ldots, u_n, u_{n+1} \right) = \left( v_1, \ldots, v_n, v_{n+1} \right)? $$

Now let $\left( y_1, \ldots, y_n \right)$ be any point in $\mathbb{R}^n$. We need to find a point $\left( x_1, \ldots, x_n, x_{n+1} \right) \in S^n \setminus \mathbf{p}$ such that $$ f\left( x_1, \ldots, x_n, x_{n+1} \right) = \left( y_1, \ldots, y_n \right). $$

How to find such a point $\left( x_1, \ldots, x_n, x_{n+1} \right) \in S^n \setminus \mathbf{p}$?

We find that if the map $g \colon \mathbb{R}^{n+1} \setminus \mathbf{p} \rightarrow \mathbb{R}^n$ given by $$ g \left( x_1, \ldots, x_n, x_{n+1} \right) \colon= \frac{1}{1-x_{n+1} } \left( x_1, \ldots, x_n \right). $$ is continuous, then the restriction of $g$ to the subset $S^n \setminus \mathbf{p}$ of $\mathbb{R}^n \setminus \mathbf{p}$ is also continuous, and this restriction is of course our map $f$.

How to rigorously show that the map $g$ is indeed continuous?

Finally, how to show that $f^{-1}$ is also continuous? Equivalently, how to show that $f$ is an open (or closed) map?

Saaqib Mahmood
  • 27,542

2 Answers2

2

Goal of this Answer

This isn't a complete solution, rather it serves as some notes to help you get over some of the humps in this analysis. I will cover

  1. Injection of $f$
  2. Surjection of $f$
  3. Obtaining $f^{-1}$
  4. Small Conclusion

hope you find this helpful.

Injection

Use the fact that

$$\sum_{i=1}^{n+1}u_{i}^2 =1 $$

to prove this.

We want to prove that

$$\frac{u_i}{1-u_{n+1}}=\frac{v_i}{1-v_{n+1}} \to u_i=v_i$$

So to do this square both sides of the equation:

$$\frac{u_i^2}{(1-u_{n+1})^2}=\frac{v_i^2}{(1-v_{n+1})^2}$$

and then sum both sides

$$\frac{\sum_{i=1}^{n}u_i^2}{(1-u_{n+1})^2}=\frac{\sum_{i=1}^{n}v_i^2}{(1-v_{n+1})^2}$$

to get

$$\frac{1-u_{n+1}^2}{(1-u_{n+1})^2}=\frac{1-v_{n+1}^2}{(1-v_{n+1})^2}$$

which using some difference of squares gives us:

$$\frac{1-u_{n+1}}{1+u_{n+1}}=\frac{1-v_{n+1}}{1+v_{n+1}}$$

from here this is similar to proving that $h(x)=\frac{1-x}{1+x}$ is injective. After you prove that $u_{n+1}=v_{n+1}$ everything else follows from the identies given.

Surjection

We want to prove that for a fixed $a\in\mathbb{R}$ we can find a $(u_1,...,u_{n+1})\in \mathbb{R}^{n+1}$ such that:

$$\frac{u_i}{1-u_{n+1}}=a$$

This too is also trivial.

Inverse Function

To find the inverse function, we start with the identity:

$$y_i = \frac{u_i}{1-u_{n+1}}$$

The goal here is to write $$u_i = g_i(y_1,...,y_n)$$.

The problem in our way is that $u_{n+1}$ is residual information from a larger space. So we need to find out what it is in $\mathbb{R}^n$ to move ahead. To be specific we need to find $g$ where $$u_{n+1} = g_n(y_1,...,y_n)$$

to do this we use a similar trick to what we did with the injection to obtain:

$$\sum_{i=1}^n y_i^2= \frac{1+u_{n+1}}{1-u_{n+1}}$$

using a similar trick to proving the subjectivity of $h(x)=\frac{1+x}{1-x}$ we get

$$u_{n+1}= \frac{\sum_{i=1}^n y_i^2-1}{\sum_{i=1}^n y_i^2+1}$$

using this and

$$1-u_{n+1}= \frac{2}{\sum_{i=1}^n y_i^2+1}$$

you can get your inverse function.

What is left?

After that all you need to do is prove:

  1. Continuity of $f$
  2. Continuity of $f^{-1}$
  3. Surjection of $f^{-1}$

and you are done.

user160110
  • 2,740
0

It is a classical problem, you must take a look at your map geometricaly, it corresponds to the stereographic projection relative to the south pole of $S^{n}$, you can find a detailed topic on your question in this text: Showing that stereographic projection is a homeomorphism

Joseph L.
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