Prob. 6, Sec. 29, in Munkres' TOPOLOGY, 2nd ed: Is this map a homeomorphism? 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?

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


*

*Injection  of $f$

*Surjection of $f$

*Obtaining $f^{-1}$

*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:


*

*Continuity of $f$

*Continuity of $f^{-1}$

*Surjection of $f^{-1}$
and you are done. 
