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?

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
