Why does $S^n$ satisfy the local $n$-slice condition? (From Lee's Smooth Manifolds)

Question

Example $5.9$ on page $103$ of John Lee's Smooth Manifolds says the following:

The intersection of $S^n$ with the open subset $\{x:x^i>0\}$ is the graph of the smooth function $$ x^i=f(x^1,\dots,x^{i-1},x^{i+1},\dots,x^{n+1}) $$ where $f\colon B^n\to\mathbb{R}$ is $f(u)=\sqrt{1-|u|^2}$. The intersection of $S^n$ with $\{x:x^i<0\}$ is the graph of $-f$. Since every point in $S^n$ is in one of these sets, $S^n$ satisfies the local $n$-slice condition, this is an embedded submanifold.

The terminology is that if $M$ is a smooth manifold, and $S\subset M$ a subset, then $S$ satisfies the local $k$-slice condition if each point of $S$ is contained in the domain of a smooth chart $(U,\varphi)$ for $M$ such that $S\cap U$ is a single $k$-slice in $U$.

I don't see how this makes $S^n$ satisfy the local $n$-slice condition. Presumably the chart on $\mathbb{R}^{m+1}$ is $(U=\{x:x^i>0\},\mathrm{id})$, so that $S^n\cap\{x:x^i>0\}$ is an $n$-slice of $U$? But this doesn't seem right since $\mathrm{id}(S^n\cap U)$ is a hemisphere of $S^n$, but that's not a $n$-slice in the corresponding half-place $\mathrm{id}(U)$?

Answer 1

I'm also confused by that. But after read Prof. Lee and koch comments i realized that maybe what it means is this :

$\mathbb{S}^n$ covered by the graphs of smooth function. The graph of smooth function is embedded submanifold. But an embedded submanifold satisfy local k-slice condition. So each graph that cover $\mathbb{S}^n$ satisfy local slice condition. Because each point in $\mathbb{S}^n$ is in these graphs, $\mathbb{S}^n$ satisfy local slice condition. Therefore $\mathbb{S}^n$ is embedded submanifold.

Answer 2

The chart is

$$ \{(x_1, \dots , x_{n+1}) \mid (x_1, \dots x_{i-1}, x_{i+1}, \dots x_{n+1}) \in B^n \text{ and } x_i > 0\} \to \Bbb{R}^{n+1}, (x_1, \dots,x_{n+1}) \mapsto (x_1, \dots, x_{i-1},x_i - f(x_1, \dots x_{i-1}, x_{i+1}, \dots x_{n+1}), x_{i+1}, \dots x_{n+1}) . $$