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Let $\pi:P\longrightarrow M$ be a $G$-principal bundle and $I:=[0, 1]$.

Given a curve $\alpha:I\longrightarrow M$ and $p_0\in \pi^{-1}(\alpha(0))$ how can I show there is a curve $\beta:I\longrightarrow P$ such that $\pi\circ \beta=\alpha$ and $\beta(0)=p_0$?

I thought defining $$\beta(t):=\phi_t^{-1}(\alpha(t), e_G),$$ where $\phi_t:\pi^{-1}(U_t)\longrightarrow U_t\times G$ are local trivializations with $\alpha(t)\in U_t$ for each $t\in I$. This satisfies $\pi\circ \beta=\alpha$ but I don't know how to get the condition $\beta(0)=p_0$.

Thanks.

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2 Answers 2

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Let $\pi^*:\alpha^*P\rightarrow I$ be the pullback of $\pi:P\rightarrow M$ you have the pullback square:

$$ \matrix{\alpha^*P &{\buildrel{f}\over{\longrightarrow}} & P\cr \pi^*\downarrow &&\downarrow \pi\cr I &{\buildrel{\alpha}\over{\longrightarrow}} &M} $$ to $I$, since $I$ is contractible, $\pi^*$ is trivial, take $\beta'$ a global section of $\pi^*$, and $\beta =f\circ \beta'$.

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Once you have any $\beta : I \to P$ such that $\pi \circ \beta = \alpha$, you can use the fact that $G$ acts transitively on each fiber of $\pi$ to find $g \in G$ such that $g\beta(0) = p_0$, and then take $\beta' : I \to P$, where $\beta'(t) = g\beta(t)$, to be your desired lift.

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