I am having trouble proving the following about vector bundles. I would think it would be rather easy, but I can't think of how. This isn't homework, but something I want to be true so that I can make a proof go easier in a book I am reading.
So, if $(E_1\times I, B \times I, \pi_1)$ and $(E_2\times I, B \times I, \pi_2)$ are isomorphic vector bundles, is it true that $\pi_1|(B\times 0) = \pi_2|(B\times 0)?$ Does anyone know how to show this? Thank you.
Also, it is even a little easier than this. I can assume that $(e, t) \rightarrow (b, t)$ in each of the maps. That is, it is the identity on the second coordinate.
We are starting with two bundles over $B$, $E_1\stackrel{\pi_1}{\to}B$, and $E_2\stackrel{\pi_2}{\to}B$, and further, we are given that when we cross these bundles with $I$, there is an isomorphism $E_1 \times I \stackrel{\varphi}{\to} E_2 \times I$. With this slight change in notation, the question is whether $\pi_1=\pi_2$.
Since our bundles have "different" total spaces, it doesn't quite make sense to ask this question. Instead, we should ask if $E_1$ and $E_2$ are isomorphic vector bundles. We already have an isomorphism lurking around that we might want to try to modify, namely $\varphi$. Becasue $\varphi$ is a map of bundles, we have $\pi_2\times \operatorname{id}(\varphi((x,t))=\pi_1\times \operatorname{id}((x,t))=(\pi_1(x),t)$ and since $(\pi_2\times \operatorname{id})^{-1}B\times\{t\}=E_2\times\{t\}$, we have that $\varphi$ preserves the second coordinate, and hence we have a restricted map $\varphi':E_1\times\{0\}\to E_2\times\{0\}$.
You should prove that this map is a bundle isomorphism. As a hint, note that the above argument can be applied to $\varphi^{-1}$.