$C^\infty$ change of charts I am stuyding a bit about manifolds and, for this, consider the unitary sphere $\mathbb{S}^{n}\subset \mathbb{R}^{n+1}$.
Consider the atlas $\mathcal{A}=\{(\phi_j^{+-},U_j^{+-})|j=1,...,n+1\}$, where $$\phi_j^{+-} (x_1,...,x_{n+1}) = (x_1,...,{x_j^*},...,x_{n+1})$$, where $x_j^*$ means that we "delete" the j-th coordinate and $$U_j^{+} = \{(x_1,...,x_{n+1}) \in \mathbb{S}^n | x_j>0\}$$ and $U_j^{-}$ means the same, but with $x_j<0$.
My attempt is to show that the change of charts is $C^\infty$, but honestly, none of my attempts seem to be nice. Could anyone give me a hand?
 A: Define
$$U_j^\epsilon = \{(x_1,\ldots,x_{n+1}) \in S^n \mid \epsilon x_j > 0 \}\ $$
where $\epsilon$ stands for a $+$ or $-$ sign. The map $\phi_j^\epsilon$ deletes the $j$-th coordinate; its image is the open unit ball $D^n = \{(y_1,\ldots,y_n) \mid \sum_{i=1}^n y_i^2 < 1 \}$. Clearly $\phi_j^\epsilon$ is the map obtained from the projection map $p^j : \mathbb R^{n+1} \to \mathbb R^n$ which deletes the $j$-th coordinate by restricting domain and codomain to $U_j^\epsilon$ and $D^n$, respectively. The map $p^j$ is $C^\infty$.
The inverse $(\phi_j^\epsilon)^{-1} : D^n \to U_j^\epsilon$ is obtained from
$$\psi_j^\epsilon : D^n \to \mathbb R^{n+1}, \psi_j^\epsilon(y_1,\ldots,y_n) = (y_1,\ldots,y_{j-1},\epsilon\sqrt{1- \sum_{i=1}^n y_i^2}, y_j,\ldots, y_n)$$
by restricting codomain to $U_j^\epsilon$. Also $\psi_j^\epsilon$ is $C^\infty$.
We have $U_j^+ \cap U_j^- = \emptyset$ and $U_j^\epsilon \cap U_i^\delta =  \{x = (x_1,\ldots,x_{n+1} \in S^n \mid \epsilon x_j > 0, \delta x_i > 0 \} \ne \emptyset$ for $i \ne j$. We do not need the explicit descriptions of $\phi_j^\epsilon(U_j^\epsilon \cap U_i^\delta)$ and $\phi_i^\delta(U_j^\epsilon \cap U_i^\delta)$; it suffices to know that they are open subsets of $D^n$. The transition map
$$\tau = \phi_i^\delta \circ (\phi_j^\epsilon)^{-1} : \phi_j^\epsilon(U_j^\epsilon \cap U_i^\delta) \to \phi_i^\delta(U_j^\epsilon \cap U_i^\delta)$$
is given by
$$\tau(y_1, \ldots,y_n) =  p^i(\psi_j^\epsilon(y_1, \ldots,y_n)) .$$
By the chain rule $p^i \circ \psi_j^\epsilon : D^n \to \mathbb R^n$ is $C^\infty$. But $\tau$ is obtained from $p^i \circ \psi_j^\epsilon$ by restricting domain and codomain to open subsets of $D^n$ and $\mathbb R^n$, respectively. Thus $\tau$ is $C^\infty$.
Instead of invoking the chain rule one can also explicitly compute
$$\tau(y_1, \ldots,y_n) = (y_1,\ldots,y_{j-1},\epsilon\sqrt{1- \sum_{i=1}^n y_i^2}, y_j,\ldots, y_n)_{*i}$$
where the index $*i$ means that the $i$-th coordinate of the vector is deleted. All coordinate functions of $\tau$ are obviously $C^\infty$, thus $\tau$ is $C^\infty$.
