Proof of local triviality of the vector bundle $\gamma_n^{1}$ for the base space $\mathbb{P}^n$

Question

Milnor defines the total space of $\gamma_n^1$ as follows

Let $E\left(\gamma_n^1\right)$ be the subset of $P^n \times R^{n+1}$ consisting of all pairs $(\{\pm x\}, v)$ such that the vector $v$ is a multiple of $x$. Define $\pi: E\left(\gamma_n^1\right) \rightarrow P^n$ by $\pi(\{\pm x\}, v)=\{\pm x\}$. Thus each fiber $\pi^{-1}(\{\pm x\})$ can be identified with the line through $x$ and $-x$ in $R^{n+1}$. Each such line is to be given its usual vector space structure. The resulting vector bundle $\gamma_{\mathrm{n}}^1$ will be called the canonical line bundle over $\mathrm{P}^{\mathrm{n}}$.

I am studying Milnor's Characteristic classes. While trying to prove that the vector bundle $\gamma_{n}^{1}$ is locally trivial (for the base space $\mathbb{P}^n$), it claims that:

\begin{equation} h : U \times \mathbb{R} \rightarrow \pi^{-1}(U) \\ (\{\pm x\}, t) \mapsto (\{\pm x\}, tx) \end{equation}

is a homeomorphism. I can feel that it is a homeomorphism but I am having a hard time proving it explicitly.

My Attempt:

Clearly, $h$ is continuous where $h (a,b) = (h_1 (a,b), h_2(a,b))$ where bothe $h_1$ and $h_2$ are continuous. Also, easy to show that it is bijective. But how do I show that the inverse function is continuous?

Answer 1

Milnor starts the proof as follows

Proof that $\gamma_{\mathrm{n}}^1$ is locally trivial. Let $\mathrm{U} \subset \mathrm{S}^{\mathrm{n}}$ be any open set that is small enough so as to contain no pair of antipodal points, and let $U_1$ denote the image of $U$ in $P^n$. Then a homeomorphism $$ \mathrm{h}: \mathrm{U}_1 \times \mathrm{R} \rightarrow \pi^{-1}\left(U_1\right) $$ is defined by the requirement that $$ h(\{\pm x\}, t)=(\{\pm x\}, t x) .$$

As in $U_1$ we have a well defined section $s:U_1 \to U$ given by $s(\{\pm x\})=x$. Define the inverse of $h$ to be the map $f:\pi^{-1}\left(U_1\right)\to \mathrm{U}_1 \times \mathrm{R}$ defined by $$(\{\pm x\}, v) \mapsto (\{\pm x\}, \langle s(\{\pm x\}), v\rangle),$$ where $\langle\cdot,\cdot\rangle$ is ordinary inner product on $\mathrm{R}^n$. This map is continuous as it is the composition of continuous maps. Its also the inverse of $h$ as $$f\circ h((\{\pm x\},t))=f\left(\{\pm x\},ts\left(\{\pm x\}\right)\right)=\left(\{\pm x\},\langle s(\{\pm x\}), ts\left(\{\pm x\}\right)\rangle\right)=(\{\pm x\},t).$$ The last equality is derived from the fact that $s(\{\pm x\}) \in S^{n}$, thus $\langle s(\{\pm x\}), s\left(\{\pm x\}\right)\rangle=1.$