Topology on the n-dimensional sphere using stereographic projection?

Question

Let the n-dimensional sphere be defined as the set: $$ \mathbb{S}^{n-1} := \{(x_1, x_2,..., x_n)\in \mathbb{R}^{n-1}: x_1^2 + x_2^2+...+x_n^2 = 1 \}$$

If we cover $\mathbb{S}^{n-1}$ with this Atlas of 2 charts $$ \mathscr{A}= \{ (U_N, \phi_N), (U_S, \phi_S)\}$$ using the stereographic projections:

$$\phi_N: U_N:=\mathbb{S}^{n-1}\setminus \{(0,0,...,0,1)\}\rightarrow \mathbb{R}^{n-1}, \\ \phi_N(x_1, x_2,..., x_n) = (\frac{x_1}{1-x_n}, \frac{x_2}{1-x_n},..., \frac{x_{n-1}}{1-x_n}) $$ and $$\phi_S: U_S:=\mathbb{S}^{n-1}\setminus \{(0,0,...,0,-1)\}\rightarrow \mathbb{R}^{n-1}, \\ \phi_S(x_1, x_2,..., x_n) = (\frac{x_1}{1+x_n}, \frac{x_2}{1+x_n},..., \frac{x_{n-1}}{1+x_n}) $$ How can you proof that the topology derived from the differential structure provided by $\mathscr{A}$ is, in fact, the usual topology of $\mathbb{R}^{n-1}$ restricted to $\mathbb{S}^{n-1}$? Moreover, how can you proof that this topology is Hausdorff and also second countable?

Answer 1

To see that you get the usual topology, observe that chart mappings are continuous for the subspace topology on $U_N$ and $U_S$. Since you can easly write down continuous inverses, you conclude that the charts are homeomorphisms from $U_N$ and $U_S$ (endowed with the topology inherited from the subspace topology on $S^{n-1}$). This readily implies that the topologies coincide.

The Hausdorff and second countability properties are also easly verified directly. Given two points such that at least one of them is different from $N$ and $S$, there is one chart containing both points, and you can separate them in that chart. For $N$ and $S$ you can take the unit balls in the two charts as disjoint open neighborhoods. For second countability, you simply proof that the preimages of balls with rational centers and radii in the two charts form a basis for the the induced topology.