# Do local trivializations fix the base-space?

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 $\alpha$, $p \circ h_\alpha \circ i_\alpha=Id_{U_\alpha}$?

• 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 $U_\alpha$ Commented Sep 21, 2015 at 0:52

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 $$p\circ h_{\alpha}\circ i_{\alpha} = \operatorname{pr}_1\circ\, i_{\alpha} = \operatorname{id}_{U_{\alpha}}$$.