Vector bundle example in Hatcher In Hatchers notes on K-Theory he gives the following example of a vector bundle that I don't understand. Define $E:=\{(x,v) \in S^n \times \mathbb{R}^{n+1} \ | \ x \perp v\}$, that is, we regard the inner product $\langle \cdot , \cdot \rangle: \mathbb{R}^{n+1} \times \mathbb{R}^{n+1} \to \mathbb{R}$ restricted to $S^n \times \mathbb{R}^{n+1}$. He then says that one should think of $v$ as a tangent vector $S^n$ but I don't know if that is truly necessary.
He defines $p:E \to S^n, \ (x,v) \mapsto x$ and constructs local trivializations the following way. Choose an $x \in S^n$ and let $U_x \subseteq S^n$ be the open hemisphere containing $x$ and bounded by the hyperplane through the origin orthogonal to $x$. Define $h_x:p^{-1}(U_x) \to U_x \times p^{-1}(x)$ by $h_x(y,v)=(y, \pi_x(v))$ where $\pi_x$ is the orthogonal projection onto the hyperplane $p^{-1}(x)$. Then $h_x$ is a local trivialization since $\pi_x$ restricts to an isomorphism of $p^{-1}(y)$ onto $p^{-1}(x)$ for each $y \in U_x$.
$(1)$ By open hemisphere he means the subset of $S^n$ where the last coordinate is greater than $0$ right? Why is this necessary and why do we not regard the entire hemisphere?
$(2)$ If $n=2$ and $x=(0,0,1)$, then the hyperplane is the $xy$-plane, so nothing changes when restricting the open hemisphere and $U_x$ is simply the open hemisphere, right? So when moving $x$ on $S^2$ it just tilts the hyperplane by the same angle and thus "cuts off" a bit of the open hemisphere, doesn't it?
$(3)$ When defining $h_x$ we have $p^{-1}(U_x)=U_x \times \mathbb{R}^{n+1}$ and one regards $\pi_x: p^{-1}(x) \bigoplus p^{-1}(x)^{\perp} \to p^{-1}(x),$ where $p^{-1}(x) \bigoplus p^{-1}(x)^{\perp}=S^n \times \mathbb{R}^{n+1}.$ Why is $p^{-1}(x)$ a hyperplane?
$(4)$ Why does $\pi_x$ restrict to an isomorphism?
 A: (1) No. Given $x \in S^n$, he considers the hyperplane through the origin orthogonal to $x$. This is the $n$-dimensional linear subspace
$$E_x = \{ v \in \mathbb R^{n+1} \mid \langle v, x \rangle = 0 \} \subset  \mathbb R^{n+1} . \tag{1}$$
Note that
$$E= \bigcup_{x \in S^n} \{x\} \times E_x . \tag{2}$$
$E_x$ separates $S^n$ into the two disjoint open hemispheres
$$H_x^\pm = \{ v \in S^n \mid \pm \langle v, x \rangle > 0 \} .\tag{3}$$
We have $x \in H_x^+$. Thus the open hemisphere containing $x$ and bounded by the hyperplane $E_x$ is the set $U_x = H_x^+$.
(2) First question: $U_x$ is the upper open hemisphere $\{(x_1,x_2,x_3) \in S^2 \mid x_3 > 0\}$. Second question: No. If we move $x$, then $U_x$ is dragged along with $x$. For example, if $x = (1,0,0)$, then $U_x = \{(x_1,x_2,x_3) \in S^2 \mid x_1 > 0\}$.
(3) By $(2)$ and the definiton of $p : E \to S^n$ we have for $x \in S^n$
$$p^{-1}(x) = \{x\} \times E_x . \tag{4}$$
Hatcher identifies $\{x\} \times E_x$ with $E_x$. When he speaks about the orthogonal projection onto the hyperplane $p^{-1}(x)$ he means the linear map
$$\pi_x : \mathbb R^{n+1} \to E_x, \pi_x(v) =  v - \langle v, x \rangle x.$$
Note that

*

*$\pi_x(v) \in E_x$ because $\langle \pi_x(v),x \rangle = \langle v, x \rangle - \langle v, x \rangle \langle x, x \rangle = 0$ (since $\langle x, x \rangle = \lVert x \rVert^2 = 1$ for $x \in S^n$).


*$\pi_x(v) = v$ for $v \in E_x$ because in that case $\langle v, x \rangle = 0$.


*$\ker \pi_x = Span(x)$ = one-dimensional subspace of $\mathbb R^{n+1}$ spanned by $x$. In fact, $v \in \ker \pi_x$ means that $v = \langle v, x \rangle x \in Span(x)$ and $v = \lambda x$ with $\lambda \in \mathbb R$ gives $\pi_x(v) = \lambda x - \langle  \lambda x, x \rangle x = \lambda x - \lambda  \lVert x \rVert^2 x = 0$.
The proper definition of $h_x$ is then
$$h_x : p^{-1}(U_x)  \to U_x \times E_x, h_x(y,v) = (y,\pi_x(v)) .$$
(4) The claim is that for each $y \in U_x$ we get an isomorphism
$$h_{x,y} : p^{-1}(y) = \{y\} \times E_y \stackrel{h_x}{\to} \{y\} \times E_x .$$
This means that the restriction
$$\pi_{x,y} :  E_y \stackrel{\pi_x}{\to} E_x $$
is an isomorphism of vector spaces. Since $E_y, E_x$ are both $n$-dimensional, it suffices to show that it is surjective. Consider the equation
$$w = v - \langle v, x \rangle x .\tag{5}$$
We have to find a solution $v \in E_y$ (which means $\langle v, y \rangle = 0$). Applying the scalar product with $y$ to $(5)$, we get
$$\langle w, y \rangle = -\langle v, x \rangle \langle x, y \rangle$$
and therefore
$$\langle v, x \rangle = - \frac{\langle w, y \rangle}{\langle x, y \rangle} .$$
This gives
$$v = w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x . \tag{6}$$
Defining
$$\rho_{x,y} : E_x \to E_y, \rho(w) = w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x , \tag{7}$$
we see that
$$\pi_{x,y} \circ  \rho_{x,y} = id .\tag{8}$$
This can of course be verified directly without solving $(5)$. If you want, you can also verify that $\rho_{x,y} \circ  \pi_{x,y} = id$, but it is unnecessary.
By the way, we can now easily see that $h_x$ is a homeomorphism. Its inverse is given by
$$h^{-1}_x(y,w) = (y,\rho_{x,y}(w)) = (y,  w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x)$$
which is obviously continuous.
Remark.
To clear doubts about domain and range of $h_x$, let us summarize what we did above.

*

*$p^{-1}(U_x)$ is a subpace of $U_x \times \mathbb R^{n+1}$.


*The map $H_x : U_x \times \mathbb R^{n+1} \to U_x \times E_x, H_x(y,v) = (y, v - \langle v, x \rangle x) = (y, \pi_x(v))$, is well-defined and continuous.


*$h_x = H_x \mid_{p^{-1}(U_x)} : p^{-1}(U_x) \to U_x \times E_x$, is well-defined and continuous.


*The map $\bar \rho_x : U_x \times E_x \to U_x \times \mathbb R^{n+1}, \bar \rho_x(y,w) = (y, w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x)$, is well-defined and continuous. Since $\langle w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x, y \rangle =0$, we have $w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x \in E_y$. Therefore $\bar \rho_x ( U_x \times E_x) \subset p^{-1}(U_x)$.


*$\rho_x :  U_x \times E_x \stackrel{\bar \rho_x}{\to} p^{-1}(U_x)$, is well-defined and continuous.


*$\rho_x \circ h_x = id$ and $h_x \circ \rho_x = id$. Hence $h_x$ is a homeomorphism.
