Construction of explicit atlas on $S^n$ Let's consider $S^n=\{(x_0,\cdots,x_n)\in \mathbb{R}^{n+1}:x_0^2+\cdots+x_n^2=1\}$. One can check that this is an $n$-dimensional manifold using inverse function theorem, but I'd like to construct an explicit atlas on it.
Take $N=(1,0,\cdots,0)$ and $S=(-1,0,\cdots,0)$, and let $U_1=S^n\setminus \{N\}$ and $U_2=S^n\setminus \{S\}$ and define $\phi_1(x_0,\cdots,x_n)=\frac{1}{1-x_0}(x_1,\cdots,x_n)$ and $\phi_2(x_0,\cdots,x_n)=\frac{1}{1+x_0}(x_1,\cdots,x_n)$.
Then the claim is that $(U_i,\phi_i)$ form compatible charts.
But I have several problems seeing this.
I can see that $\phi_i$ are bijective onto $\mathbb{R}^n$. In fact for fixed $x_0$ they map a "circle" on $S^n$ to a "circle" (or sphere?) in $\mathbb{R}^n$
It is clear that $\phi_i$ are continuous.
But here I'm stuck. My questions are

What exactly inverses look like?
If one can't find inverses explicitly (or if it is not really needed), how does one show that inverses are continuous, and the transition map is diffeomorhpic?

 A: The inverse of $\phi_1$ sends a point $P$ in the plane, to the point 
$$
Q = PN \cap S^n
$$
where $PN$ denotes the line segment from $P$ to the north pole $N$, and $Q$ is the point of the intersection that's not $N$.  
Such a point has the form 
$$
sP + (1-s) N
$$
and has length $1$ (or squared length 1), which (since the vectors $P$ and $N$ are orthogonal) that 
$$
s^2 \|P\|^2 + (1-s)^2 = 1 \\
s^2 \|P\|^2 + (1-s)^2 = 1 \\
s^2 \|P\|^2 + s^2 - 2s  = 0 \\
s(s \|P\|^2 + s - 2)  = 0
$$
hence either $s = 0$ (which gives $Q$ = north pole) or 
$$
s(\|P|^2 + 1) = 2 \\
s = \frac{2}{\|P\|^2 + 1}.
$$
So this gives you an explicit formula:
$$
\phi^{-1}(x_1, \ldots, x_n) = \frac{2}{\|P\|^2 + 1} (0, x_1, \ldots, x_n) + (1 - \frac{2}{\|P\|^2 + 1)} (0, \ldots, 0, 1). 
$$
There's a decent chance I've screwed up the algebra here, but I hope you get the main idea and check the details for yourself. 
A: The inverse can we found down explicitly (here for "N"):
We want to map
$$\tag1(x_1,\ldots, x_n)\mapsto (x_0,(1-x_0)x_1,\ldots, (1-x_0)x_n),$$
but need to find suitable $x_0$ that guarantees norm $1$.
That is, we want to solve
$$ x_0^2+(1-x_0)^2(x_1^2+\ldots+x_n^2)-1=0$$
where $x_0=1$ is a trivial, but undesired solution, which we divide out:
$$ x_0+1+(x_0-1)(x_1^2+\ldots+x_n^2)=0,$$
$$\tag2x_0=\frac{x_1^2+\ldots+x_n^2\;-1}{x_1^2+\ldots+x_n^2\;+1}. $$
Note that $x_0$ in $(2)$ and hence the right hand side of $(1)$ is a continuos function of the input.

If you play around with both maps, you should find that the transition map is (both ways) something like
$$ (x_1,\ldots,x_n)\mapsto \frac1{x_1^2+\ldots+x_n^2} (x_1,\ldots,x_n)$$
