$S^{n}$ is an n-dimensional topological manifold I was studying Massey's Algebraic Topology: An introduction, where he explained why $S^n$ is an n-manifold. I have previously proved this using charts and stereographic projection, however, I would like to understand this demonstration.


We can easily prove that the unit $n$-dimensional sphere
$$
S^{n}=\left\{x \in \mathbf{R}^{n+1}:|x|=1\right\}
$$
is an $n$-manifold. For the point $x=(1,0, \ldots, 0)$, the set $\left\{\left(x_{1}, \ldots, x_{n+1}\right) \in\right.$ $\left.S^{n}: x_{1}>0\right\}$ is a neighborhood with the required properties, as we see by orthogonal projection on the hyperplane in $\mathbf{R}^{n+1}$ defined by $x_{1}=0$. For any other point $x \in S^{n}$, there is a rotation carrying $x$ into the point $(1,0, \ldots, 0) .$ Such a rotation is a homeomorphism of $S^{n}$ onto itself; hence, $x$ also has the required kind of neighborhood.


I don't see what the orthogonal projection on the hyperplane in $\mathbf{R}^{n+1}$ defined by $x_{1}=0$ looks like and which would be the rotation carrying $x$ into the point $(1,0,...,0)$. Also, how are these homeomorphisms to the open $n-$dimensional disc $U^n=\{x \in \mathbf{R}^n : |x|<1\}$.
 A: In Massey's definition of an $n$-manifold it is required that each point has a neighborhood homeomorphic to the open disk $U^n \subset \mathbb R^n$ with center $0$ and radius $1$. Usually one requires that each point has a neighborhood homeomorphic to some open subset $U \subset \mathbb R^n$, but it is easy to see that both conditions are equivalent.
For $i = 1,\ldots,n+1$ let us define
$$(S^n_i)^+ = \{(x_1,\ldots,x_{n+1}) \in S^n \mid x_i > 0\},\\ (S^n_i)^- = \{(x_1,\ldots,x_{n+1}) \in S^n \mid x_i < 0\} .$$
These are $2n+2$ open subsets of $S^n$ which cover $S^n$. In fact, each $x = (x_1,\ldots,x_{n+1}) \in S^n$ must have at least one coordinate $x_i \ne 0$, and if $x_i > 0$, then $x \in  (S^n_i)^+$ whereas if $x_i < 0$, then $x \in  (S^n_i)^-$.
Let $p_i : \mathbb R^{n+1} \to \mathbb R^n, p_i((x_1,\ldots,x_{n+1})) = (x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_{n+1})$, denote the projection omitting the $i$-th coordinate. Clearly $p_i((S^n_i)^\pm) \subset U^n$ because for $x \in (S^n_i)^\pm$ we have $\lvert p_i(x) \rvert = \sqrt{\sum_{k \ne i}x_k^2}  < \sqrt{\sum_{k}x_k^2} = 1$ (recall $x_i \ne 0$, i.e. $x_i^2 > 0$).
Now define
$$\phi_i^\pm : U^n \to \mathbb R^{n+1}, \phi_i(y_1,\ldots,y_n) = \left(y_1,\ldots,y_{i-1},\pm\sqrt{1- \lvert y \rvert^2},y_i,\ldots,y_n \right) .$$
This well-defined becaue $\lvert y \rvert < 1$ for $y \in U^n$. Clearly $\lvert \phi_i^\pm(y) \rvert = 1$ and thus $\phi_i^\pm(U^n) \subset (S^n_i)^\pm$.
Consider the restrictions $p_i^\pm : (S^n_i)^\pm \to U^n$ of $p_i$ and $\psi_i^\pm : U^n \to (S^n_i)^\pm$ of $\phi_i^\pm$. We have $\phi_i^\pm \circ p_i^\pm = id_{ (S^n_i)^\pm}$ and $p_i^\pm \circ \phi_i^\pm = id_{ U^n}$. This shows that $p_i^\pm$ is a homeomorphism with inverse $\phi_i^\pm$.
The homeomorphism $p^+_i : (S^n_1)^+ \to U^n$ is that what Massey denotes as orthogonal projection on the hyperplane in $\mathbb R^{n+1}$ defined by $x_1=0$.
Our above considerations show that each point of $S^n$ has a neighborhood homeomorphic to $U^n$ - recall that the sets $(S^n_i)^\pm$ form an open cover of $S^n$. Therefore we do not need rotations carrying arbitrary points $x \in S^n$ to $(1,0\ldots,0)$.
However, if you absolutely want to rotate $S^n$, you must observe that a rotation in $\mathbb R^{n+1}$ is an orthogonal linear map $L$ with positive determinant. Each such map has the property $\lvert L(x) \rvert = \lvert x \rvert$, i.e. $L$ restricts to a homeomorphism  $L' : S^n \to S^n$.  Note that positive determinant is irrelevant to get a homeomorphism  $L' : S^n \to S^n$, we can work with an arbitrary orthogonal $L$.
Now consider $x \in S^n$. We can find an orthonormal basis $\xi_1,\ldots,\xi_{n+1}$ of $\mathbb R^{n+1}$ such that $\xi_1 = x$. Mapping $\xi_k$ to the standard basis vector $e_k$ gives us an orthogonal linear map $L$ such that $L(x) = e_1 = (1,0\ldots,0)$. It may have still have negative determinant, but we can remedy this by composing $L$ with the reflection $R : \mathbb R^{n+1} \to \mathbb R^{n+1}, R(x_1,\ldots, x_{n+1}) = (x_1,\ldots, x_n, -x_{n+1})$. In fact $R\circ L$ is the desired rotation. However, it seems to me that using rotations (or more generally arbitrary orthogonal maps) is overhead in comparison with our above approach.
