Let $O(n)$ denote the $n$ - orthorgonal group consisting of all $n \times n$ real matrices $A$ with $A^{t}A = E_{n}$. Let $e_{n}$ is the column vector $(0,...,1)$ then $Ae_{n}$ is a unit vector in $\mathbb{R^{n}}$, hence lie in the sphere $S^{n-1}$. I want show that
$$p : O(n) \to S^{n-1}$$ $$A \mapsto Ae_{n}$$
is a locally trivial bundle with fiber $O(n-1)$. So I consider the open cover $U = S^{n-1} - e_{n}$ and $V = S^{n-1} - (-e_{n})$. I think this cover will work but I can't explicit the trivialization ( homeomorphism ) on each open set $U$ or $V$, i.e
$$\phi_{U} :U \times O(n-1) \to p^{-1}(U)$$
$$p\phi_{U}(u,A) = u \forall (u,A) \in U \times O(n-1)$$
And similiar with $V$. I tried to define a map $\alpha_{U} : U \to p^{-1}(U)$ and then define $\phi_{U}(u,A) = \alpha(u) \begin{pmatrix}A & 0 \\0 & 1 \end{pmatrix}$. In order to do that I need $\alpha(u)e_{n}=u$ and $\alpha(u) \in O(n)$. Can someone help me this point ?
In general, how can I know that there is a trivial fibre bundle of Stiefel manifold without using other concepts in theory of bundle ?