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Aug 22 at 23:44 vote accept Chris
Apr 5, 2023 at 0:49 comment added Alex Nolte For the first part, no: $P$ admits a local section on $U$. For the second part: the argument given uses the implicit function theorem, which requires restriction, and I don't know off the top of my head. You could perhaps get rid of the restriction by using that Lie group homomorphisms are real-analytic and then using the real-analytic version of the monodromy theorem to patch together lifts.
Apr 4, 2023 at 21:56 comment added Chris If $U$ is contractible, do we need to make $U$ smaller?
Apr 4, 2023 at 4:25 comment added Chris thank you. I think I will need to write it out tmrw for myself to be fully convinced, but I believe this argument works.
Apr 4, 2023 at 4:12 history edited Alex Nolte CC BY-SA 4.0
Edited to address follow-up from poster, then fixed a typo.
Apr 4, 2023 at 4:11 comment added Alex Nolte Ah. I edited the response to explain how to make this compatible with a specific chart on $P'$. In practice, one expects the initial chart for $P'$ to not matter much, since if there are already charts around, something local is being done. The original answer constructs compatible charts in enough neighborhoods to contain every point in $M$.
Apr 4, 2023 at 4:07 history edited Alex Nolte CC BY-SA 4.0
Edited to address follow-up from poster.
Apr 4, 2023 at 3:24 comment added Chris My apologies for the confusion, if I knew how to use tikz on a here I wouldve included a diagram.
Apr 4, 2023 at 3:20 comment added Chris Does this not go the other way? I already have a chart on $P'$, I am trying to essentially lift to a chart on $P$ that makes some diagram commute. In other words, given a section $s$ of $P'$, I would like to find a section $\sigma$ of $P$ such that $F\circ \sigma=s$. This is originally what I was trying to prove, but the existence of charts that satisfy my required property would imply the existence of such sections.
Apr 4, 2023 at 3:18 history answered Alex Nolte CC BY-SA 4.0