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Let $\pi\colon P\to X$ be some $G$-principal bundle; $\{U_\alpha\}$ a cover of $X$; and $\{s_\alpha\colon U_\alpha\to\pi^{-1}(U_\alpha)\}$ a collection of local sections.

Claim: The pullback $\pi^*P$ of $P$ over itself is trivial, as witnessed by the global section $\pi^* s_\alpha$.

My question: What does the above statement mean (if it makes sense at all)? That is, we know that the local sections differ by some transition $g_{\alpha\beta}\colon U_\alpha\beta\to G$, so how do we obtain a single global section from these local ones on the pullback?

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The claim is correct, the local sections are irrelevant.
Proof of claim:
The pulled-back bundle $\pi^*(P) $ consists of those ordered pairs $(p,p')\in P\times P$ which satisfy $\pi(p)=\pi(p')\in P$.
The projection $\Pi:\pi^*(P)\to P $ of that bundle to its base is given by the formula $\Pi(p,p')=p.$
This projection has a canonical global section $\sigma:P\to \pi^*(P)$ given by $\sigma (p)=(p,p)$.
And thus the $G$-bundle $\pi^*(P)$ is trivial, like all principal bundles having a global section.

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  • $\begingroup$ Sorry, I didn't make my question very clear: I know that this bundle is trivial, but I want to understand what happens to these local sections $\endgroup$ – Tim May 17 '18 at 18:54
  • $\begingroup$ These local sections $s_\alpha$ give rise to local sections $S_\alpha:\pi^{-1}(U_\alpha)\to \pi^*P: p \mapsto (s_\alpha(\pi p),s_\alpha(\pi p))$ of the pulled-back bundle $\pi^*P$. But these new local sections $S_\alpha$ will not glue unless the old sections $s_\alpha$ already glued, in which case the old $G$-bundle $P$ already was trivial and there was no need for your construction... $\endgroup$ – Georges Elencwajg May 17 '18 at 19:12

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