There is a coordinate ball in $\mathbb{S}^n$ whose closure is equall to all of $\mathbb{S}^n$.

Question

This is an exercise from John Lee's Introduction to Topological Manifolds.

Using the map of Example 4.55, show that there is a coordinate ball in $\mathbb{S}^n$ whose closure is equal to all of $\mathbb{S}^n$. I cannot figure out how we can use this map $q$ to show that each point on the sphere $\mathbb{S}^n$ has a local homeomorphism from the point's neighborhood into an open ball in $\mathbb{R}^n$. I would greatly appreciate any help.

Answer 1

The map $q : \overline{\mathbb B^n}\to \mathbb S^n$ collapses $\mathbb S^{n-1}$ to the north pole $n = (0,\ldots,0,1) \in \mathbb S^n$. It establishes a homeomorphism $q' : \mathbb B^n \to U = \mathbb S^n \setminus \{ n \}$. Hence $U$ is a coordinate ball in $ \mathbb S^n$ whose closure is $\mathbb S^n$.