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Suppose you have a vector bundle $E \to B$ with projection $p$. Consider a local trivialization $h_\alpha: U_\alpha \times \mathbb{R} \to p^{-1}(U_\alpha)$ and take the map $i_\alpha: U_\alpha \to U_\alpha \times \mathbb{R}$ given by $x \rightsquigarrow (x,0)$. The $U_\alpha$ cover $B$.

Is it true that we can always arrange the local trivializations so that for any $\alpha$, $p \circ h_\alpha \circ i_\alpha=Id_{U_\alpha}$?

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  • $\begingroup$ Oh. Of course you can. The composed map is a homeo $j$, so just let $h'_\alpha(x,t)=h_\alpha(j^{-1}(x),t)$ be your trivialization on $U_\alpha$ $\endgroup$ Commented Sep 21, 2015 at 0:52

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A local trivialisation is not just a homeomorphism $h_{\alpha} : U_{\alpha}\times\mathbb{R}^k \to p^{-1}(U_{\alpha})$, but a homeomorphism such that

  • $p\circ h_{\alpha} = \operatorname{pr}_1$, that is $\pi(h_{\alpha}(x, t)) = x$, and
  • the map $v \mapsto h_{\alpha}(x, v)$ is a linear isomorphism $\mathbb{R}^k \to p^{-1}(x)$.

Given the first condition, we have $p\circ h_{\alpha}\circ i_{\alpha} = \operatorname{pr}_1\circ\, i_{\alpha} = \operatorname{id}_{U_{\alpha}}$.

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