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

The chart is

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

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.