Let $(M, g)$ be a complete Riemannian manifold and let $(\tilde{M}, \tilde{g})$ its universal cover. Let $\pi : \tilde{M} \to M$ be the covering map.

Let $\gamma : I \to (M, g)$ be a smooth curve with $I \subset \mathbb{R}$ be an interval. Let $t_0 \in I$ and let $\tilde{x} \in \tilde{M}$ be a point such that $\pi(\tilde x) = \gamma(t_0)$.

Then I want to prove that there exists a unique curve $\tilde{\gamma} : I \to (\tilde M, \tilde g)$ such that $\pi \circ \tilde \gamma = \gamma$ and $\tilde\gamma(t_0) = \tilde x$.

EDIT: I am interested in this question because I'm trying to understand the proof of Theorem69[Structure Theorem for Nonnegative Ricci Curvature] of Section 3.5 of Chapter 9 of Riemannian Geometry book by Petersen. I also can't understand why the geodesics $\gamma_i$ defined in the proof converge to a geodesic $\hat \gamma$. But maybe I will ask this in an other question.

$2^{nd}$ EDIT: I think I have understood why the geodesic $\gamma_i$ converge to $\hat \gamma$. :)

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    $\begingroup$ This is the path lifting property of covering maps and is true in a much more general setting. See e.g. Prop 11.10 of Lee's Topological Manifolds. $\endgroup$ – Anthony Carapetis Jul 4 '16 at 13:54

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