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I came across this problem about showing the triviality of a fibre bundle.

The question is as follows:

If $\xi$ is a fibre bundle and it is given by $p:E\rightarrow B$ and $f:X\rightarrow B$ is any map that is continuous. Let ${U_{\alpha}}$ be an open cover of B such that the fibre bundle $\xi$ restricted to each $U_{\alpha}$ is trivial. I need to show that the induced bundle $f^{*}\xi$ restricted to each $f^{-1}(U_{\alpha})$ is trivial. Then from this I also need to show that if $\xi$ has an atlas of countable finite charts, then so does any induced fibre bundle of $\xi$.

How do I start for this problem? Any advice?

LanaDR
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1 Answers1

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Let $F$ be the fiber. The restriction of $f$ on $U_\alpha$ is trivial is equivalent to saying that there exists an isomorphism of fiber bundles $g_\alpha:E_{\mid U_\alpha}\rightarrow U_\alpha\times F$. We denote by $p_F$ the projection on the second factor.

$f^*\xi$ restricted to $f^{-1}(U_\alpha)$ is $\{(x,y): y\in E, x\in f^{-1}(U_\alpha): f(x)=p(y)\}$. You can defined $h_\alpha:(f^*\xi)_{\mid U_\alpha}\rightarrow f^{-1}(U_\alpha)\times F$ by $h_\alpha(x,y)=(x,p_F(g_\alpha(y))$.

Tsemo Aristide
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  • Thanks! May I clarify something... When you say "The restriction of $f$ on $U_{\alpha}$ is trivial is equivalent...", it should be restriction of $\xi$ on $U_{\alpha}$ am I right? And am I correct to say the base space of the $f^{*}\xi$ is $f^{-1}(U_{\alpha})$ and your are trying to show that there exist an isomorphism? – LanaDR Sep 11 '17 at 03:24
  • Also, for the last part, I know that a chart is a pair $(U,\phi)$ such that $U$ is an open set and $\phi$ is a homeomorphism from $U$ onto an open set in $R^{n}_{+}$. How does the triviality help here? – LanaDR Sep 11 '17 at 03:38