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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?

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    These charts are actually simpler than the stereographic projection. You need to verify that the collection of $U_i^\epsilon$ covers the entire sphere, and that all transition maps are smooth – Amitai Yuval Oct 20 '18 at 15:02
  • Yes, but how do I show exactly this? How do I show if it covers the sphere? If I want to show that the charts cover the sphere, I have to compute the inverse and show, that the inverse is defined on the set that defines the sphere. And if I can't give an explicit formula to the transition maps, how can I show that they are smooth? –  Oct 20 '18 at 15:13

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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 $\epsilon:=\mathrm{sgn}(x_i) =x_i/|x_i|$.

Berci
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  • Thanks a lot Berci. So I choose the inverse such that my condition to be on the unit circle is fullfilled, and the map can either be $\pm$. Now it is clear! Thanks. –  Oct 21 '18 at 10:14
  • Remark: It should be a linear map, right? Not smooth but continuous. –  Oct 21 '18 at 10:45
  • No, it's not linear (at least the inverse is certainly not). If the transition maps are all smooth, we talk about smooth manifold (this is the case here), and if they are only continuous, we talk about topological manifold.. – Berci Oct 21 '18 at 11:46
  • Okay I managed to proof it by myself, that it is indeed from $\mathcal{C}^\infty$. Thanks! –  Oct 22 '18 at 07:38