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This is a question in obstruction theory. It should be basic but I can't find a reference.

Let's stick to the $C^\infty$ category, so all objects mentioned are smooth. Let $\pi: E \to M$ be a vector bundle over a manifold (everything paracompact, Hausdorff), and let $s(t): [0, 1] \to E$ be a section of $E$ over a curve $\gamma(t): [0, 1] \to M$.

Let $\gamma(t, \lambda): [0, 1]^2 \to M$ be a family of curves such that $\gamma(t, 0) = \gamma(t)$.

Question: is there a lifting of the family of curves $\gamma(t, \lambda)$ to a family of sections $s(t, \lambda): [0, 1]^2 \to E$ such that $s(t, 0) = s(t)$?

My gut feeling is that since the fibers of $\pi$ are contractible and the total space of $E$ deformation retracts onto the image of the zero-section $s_0(M) \cong M$, there should be no obstruction to lifting in the topological category, and then by the usual type of perturbation argument, in the smooth category. But I'd like a more precise argument.

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    $\begingroup$ A simple way to obtain a lift is by using a connection, and then taking the parallel transport along the deformation, with initial data given by $s(t)$ $\endgroup$
    – Sebastian
    Commented May 7 at 1:04
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    $\begingroup$ The geometric argument using connections is probably preferable, but topologically this is immediate from the fact that any fiber bundle (it does not matter that the fibers are contractible) over a paracompact Hausdorff base is a fibration (plus the suggested perturbation argument to make the resulting section smooth). $\endgroup$
    – Thorgott
    Commented May 8 at 13:08

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