In Lee's "Introduction to Smooth Manifolds" he provides the following example showing that $\mathbb{S}^n$ is a topological $n$-manifold.
For each integer $n \geq 0$, the unit $n$-sphere $\mathbb{S}^n$ is Hausdorff and second-countable because it is a topological subspace of $\mathbb{R}^{n+1}$. To show that it is locally Euclidean, for each index $i = 1, \ldots, n+1$ let $U_i^+$ denote the subset of $\mathbb{R}^{n+1}$ where the $i$th coordinate is positive:
$$U_i^+ = \big\{(x^1, \ldots, x^{n+1}) \in \mathbb{R}^{n+1}: x^i > 0\big\}.$$
Similarly, $U_i^-$ is the set where $x^i < 0$.Let $f: \mathbb{B}^n \rightarrow \mathbb{R}$ be the continuous function
$$f(u) = \sqrt{1-|u|^2}.$$
Then for each $i = 1, \ldots, n+1$, it is easy to check that $U_i^+ \cap \mathbb{S}^n$ is the graph of the function
$$x^i = f(x^1, \ldots, \hat{x}^i, \ldots, x^{n+1}),$$
where the hat indicates that $x^i$ is omitted. Similarly, $U_i^- \cap \mathbb{S}^n$ is the graph of the function
$$x^i = -f(x^1, \ldots, \hat{x}^i, \ldots, x^{n+1}).$$
Thus, each subset $U_i^\pm \cap \mathbb{S}^n$ is locally Euclidean of dimension $n$, and the maps $\varphi_i^\pm: U_i^\pm \cap \mathbb{S}^n \rightarrow \mathbb{B}^n$ given by
$$\varphi_i^\pm(x^1, \ldots, x^{n+1}) = (x^1, \ldots, \hat{x}^i, \ldots, x^{n+1})$$
are graph coordinates for $\mathbb{S}^n$. Since each point of $\mathbb{S}^n$ is in the domain of at least one of these $2n+2$ charts, $\mathbb{S}^n$ is a topolgoical $n$-manifold.
My question is perhaps more on notation, but how is $U_i^+ \cap \mathbb{S}^n$ (likewise for $U_i^- \cap \mathbb{S}^n$) a graph using the given definition? As I understand it, we are omitting $x^i$ when passing the point in $\mathbb{R}^{n+1}$ into $f$, is the result of the function the new value for $x^i$? However, this seems incorrect as $\varphi$ doesn't seem to involve $f$ at all, so how does it follow that the maps are indeed graph coordinates (or more generally, coordinate maps)?
Best Answer
As discussed in the comments above, if $(x^1, \dots, x^{n+1}) \in U_i^{\pm}\cap S^n$, then $(x^1)^2 + \dots + (x^{n+1})^2 = 1$ so $(x^i)^2 = 1 - \left[(x^1)^2 + \dots + (x^{i-1})^2 + (x^{i+1})^2 + \dots + (x^{n+1})^2\right]$ and hence $$x^i = \pm\sqrt{1 - \left[(x^1)^2 + \dots + (x^{i-1})^2 + (x^{i+1})^2 + \dots + (x^{n+1})^2\right]} = \pm f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1}).$$
Therefore $(x^1, \dots, x^{n+1}) = (x^1, \dots, x^{i-1}, \pm f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1}), x^{i+1}, \dots, x^{n+1})$. As $(x^1, \dots, \widehat{x^i}, \dots, x^{n+1}) = \varphi_i^{\pm}(x^1,\dots, x^{n+1})$, we see that $U_i^{\pm}\cap S^n$ is the graph of $\pm f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1})$ over $\varphi_i^{\pm}(U_i^{\pm}\cap S^n)$.