Followings are given problems.
Let $f:(M,g)\rightarrow (N,h)$ a covering map that is a local isometry, and let $p\in M$. If $\gamma:[0,1] \rightarrow N$ is a geodesic such that $\gamma(0)=f(p(0))$, then we can lift $\gamma$ to a geodesic $\tilde{\gamma}$:[0,1]→M a geodesic with $\tilde{\gamma}(0)=p(0)$
of course one can see that $\gamma(t)=f(p(t))$, so that $\tilde{\gamma}(t) = p(t)$ for $t \in[0,1]$
I want to prove this statement. How we can prove this?
A possible proof is as follows. The fact that $f$ is a covering map means that $f$ is a topological covering map (and thus a local homeomorphism), and also a local diffeomorphism and a local isometry. The theory of (topological) coverings tells us that $\gamma$ can be lifted to a continuous map $\widetilde\gamma:[0,1]\rightarrow M$, i.e. we have $f\circ \widetilde\gamma = \gamma$. Now, the fact that $f$ is a local diffeomorphism tells us that $\widetilde\gamma$ is smooth, since $\gamma$ is smooth and being smooth is a local property. Finally, the fact that $f$ is a local isometry tells us that $\widetilde\gamma$ is geodesic, since $\gamma$ is geodesic and being geodesic also is a local property.