M is an n-dim manifold. $\pi :\tilde M \to M$ the universal cover of M. $\tilde p \in \tilde M$ a lift of p. We choose a measurable section $j:{B_1}\left( p \right) \to {B_1}\left( {\tilde p} \right)$, i.e. $\pi \left( {j\left( x \right)} \right) = x$ for any $x \in {B_1}\left( p \right)$. Let $T = j\left( {{B_1}\left( p \right)} \right)$. Let S be the union of g(T) over all $g \in {\pi _1}\left( M \right)$ such that $g\left( T \right) \cap {B_1}\left( {\tilde p} \right) \ne \emptyset $. Is then ${B_1}\left( {\tilde p} \right) \subset S$?
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$\begingroup$ Yes, assuming that your manifolds are Riemannian and $\pi$ is an isometric covering map; you do not even need "measurable" assumption, it is just a set-theoretic statement. $\endgroup$– MishaCommented Jun 5, 2013 at 4:55
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I assume you are taking $M$ to be Riemannian and $\tilde{M}$ to be endowed with the induced metric?
Let $\tilde{q}\in B_1(\tilde{p})$. We want to show that $\tilde{q}\in S$. Let $q=\pi(\tilde{q})$. There exists a covering transformation $g\in \pi_1M$ such that $\tilde{q}=gj(q)$, so we just need to prove that $q\in B_1(p)$.
Let $\tilde{\gamma}$ be a smooth path from $\tilde{p}$ to $\tilde{q}$ of length less than one. Then $\gamma=\pi\circ\tilde{\gamma}$ is a smooth path from $p$ to $q$ of the same length, because $\pi$ is a local isometry. This finishes the proof.
