Triviality of a tautological bundle I am trying to solve the following exercise.

What I know:
$\tau$ is a vector bundle of dimension $n$ over $\mathbb{R}P^n$. The same is true for the trivial bundle $\mathbb{R}P^n \times \mathbb{R}^n$.
Then we find surjective smooth maps
$$
\pi: \tau \to \mathbb{R}P^n \\
\tilde{\pi}:\mathbb{R}P^n \times \mathbb{R}^n \to \mathbb{R}P^n
$$
such that for each $[p] \in \mathbb{R}P^n$ the fibers
$$
E_{[p]}=\pi^{-1}(\{[p]\})         \\
\tilde{E}_{[p]}=\pi^{-1}(\{[p]\})
$$
come with the structure of a k-dimensional real vector space. Moreover, we find open sets $U,\tilde{U}$ of $\mathbb{R}P^n$ and diffeomorphisms
$$
\varphi: \pi^{-1}(U) \to U \times \mathbb{R}^k 
$$
$$
\tilde{\varphi}: \tilde{\pi}^{-1}(\tilde{U}) \to \tilde{U} \times \mathbb{R}^n 
$$
such that
$$
\pi_n \circ \varphi=\pi \\
\tilde{\pi}_n \circ \tilde{\varphi}=\tilde{\pi} 
$$
where
$$
\pi_n: U \times \mathbb{R}^n \to U, ([p],v) \mapsto [p] \\
\tilde{\pi}_n: \tilde{U} \times \mathbb{R}^n \to \tilde{U}, 
([p],v) \mapsto [p]. 
$$
In particular the maps
$$
\pi_{\mathbb{R}^n} \circ \varphi_{E[p]}: E_{[p]} \to \mathbb{R}^n \tag{1}
$$
$$
\pi_{\mathbb{R}^n} \circ \tilde{\varphi_{E[p]}}: \tilde{E}_{[p]} \to \mathbb{R}^n \tag{2}
$$
are vector space isomorphisms.
Now suppose the claim is true. Then there is a diffeomorphism $\phi: \tau \to \mathbb{R}P^n \times \mathbb{R}^n$ such that $\tilde{\pi} \circ \phi=\pi$ and for each $[p] \in \mathbb{R}P^n$ the map $\phi |_{E_{[p]}}: E_{[p]} \to \tilde{E}_{[p]}$ is a vector space isomorphism.
Then there is a section $f: \mathbb{R}P^n \to \tau, [p] \mapsto f_{[p]}$ that vanishes nowhere.
But I am a bit at a loss about how to find a contradiction.
 A: Consider a section $s$ of this line bundle.  Let $\phi\colon [0,1] \to S^n$ be a path from a point $p\in S^n$ to $-p$.
We have a unique continuous function $\lambda\colon [0,1]\to \mathbb{R}$ such that $s([\phi(t)])=\lambda(t)\phi(t)$ for all $t\in[0,1]$.
We have: $$\lambda(0)p=s([\phi(0)])=s([\phi(1)])=-\lambda(1)p$$
Thus $\lambda(0)=-\lambda(1)$ and by the intermediate value theorem $\lambda(t)=0$ for some $t\in [0,1]$.
Thus this bundle has no non-vanishing sections, and is different to the trivial line bundle.
A: Let $f : \mathcal{T} \to \mathbb{R}P^n \times \mathbb{R}$ be a bundle isomorphism. The quotient map $\mathbb{R}^{n + 1} \to \mathbb{R}P^n$ is continuous, so we have a continuous inclusion
$$
\iota : \mathbb{R}^{n + 1} \setminus \{0\} \to \mathbb{R}P^n \times \mathbb{R}^{n + 1}
$$
which does the quotient map on the first factor and is the identity on the second factor. If we restrict the codomain of $\iota$ to the complement of the image of the zero section $\pi^{-1}(0)$ in $\mathcal{T}$, then $\iota$ is a homeomorphism $\mathbb{R}^{n + 1} \setminus \{0\} \to \mathcal{T} \setminus  \pi^{-1}(0)$. Similarly $f$ restricts to a homeomorphism
$$
f : \mathcal{T} \setminus \pi^{-1}(0) \to \mathbb{R}P^n \times (\mathbb{R} \setminus \{0\}).
$$
Composing these two maps gives a homeomorphism $f \circ \iota : \mathbb{R}^{n + 1} \setminus \{0\} \to \mathbb{R}P^n \times (\mathbb{R} \setminus \{0\})$. But of course the former space is connected, while the latter is not: hence the map $f$ cannot exist.
