How to show that a collection of charts is an atlas? If I look at the n-sphere
$$M = S^n = \{x \in \mathbb{R}^{n+1} : x_1^2 + x_2^2 + \ldots + x_{n+1}^2 = 1\}$$
with the subspace topology of $\mathbb{R}^{n+1}$.
Then a chart for $1 \leq i \leq n + 1$ and $\epsilon = \pm 1$ is defined:
$$U^{\epsilon}_i = \{x \in S^n : \epsilon x_i > 0\}$$$$\varphi^{\epsilon}_i(x) = (x_1,\ldots, \hat{x}_i, \ldots, x_{n+1})\in \mathbb{R}^n$$
How can I show that $\mathcal{A} = (U_i^{\epsilon},\varphi_i^{\epsilon})_{i,\epsilon}$ is an Atlas for $M$?
So as I think this should be something similar to the stereographic projection. But the equator, as it is written there, shouldn't be in the charts. And I can't think of an explicit formula, otherwise wouldn't it be enough to show that the chart is homeomorphic to a subspace of $\mathbb{R}^n$ as well as its inverse?
 A: It always worth to consider low-dimension examples.
What are the given charts for $S^1$? 
These are the 4 half circles, cut along the coordinate axes, and each half circle is coordinated by the horizontal or vertical coordinates of its points, respectively.
One of these charts, specifically $(U_2^{+1}, \varphi_2^{+1})$, is basically the graph of the familiar function $y=\sqrt{1-x^2}$, which is just the inverse of the chart map $(x,y)\mapsto x$ whose domain is the upper half circle. 
For $S^2$, we get 8 hemispheres along the coordinate axes, each projected to a disk in the appropriate coordinate plane. 
To the general case, we have $(\varphi_i^\epsilon)^{-1}=(u_1,\dots, u_n)\mapsto (u_1,\dots, u_{i-1},\ \epsilon\sqrt{1-(u_1^2+\dots+u_n^2)}, \ u_i, \dots, u_n) $, which is a smooth map over the open unit ball of $\Bbb R^n$.
To see each point is covered by these charts, assume an arbitrary point $x\in S^n$ is given, then it has at least one nonzero coordinate, say $x_i$, then choose $\epsilon:=\mathrm{sgn}(x_i) =x_i/|x_i|$. 
