Let $\widetilde{M}$ be a covering space of a Riemannian manifold $M$. Show that $\widetilde{M}$ has a Riemannian metric such that the covering map $\pi:\widetilde{M}\to M$ is a local isometry. Then $\widetilde{M}$ is complete if and only if $M$ is complete.
The obvious choice for a metric in $\widetilde{M}$ is: $$\left<{u,v}\right>_p:=\left<{d\pi_p(u),d\pi_p(v)}\right>_{\pi(p)}$$
for all $p\in\widetilde{M}$ and $u,v\in T_p\widetilde{M}$.
I solved the "if" part (I think this was the hard part, though).
On the other hand, suppose $\widetilde{M}$ is complete. My question is: is it true that $d_{\widetilde{M}}(p,q)\le d_M(\pi(p),\pi(q))$? If so, then if we take a Cauchy sequence in $M$ that would give us a Cauchy sequence in $\widetilde{M}$ which is complete.
I tried this. Let $\gamma:[0,1]\to \widetilde{M}$ be a geodesic that joins $p$ and $q$. $\pi\circ\gamma$ is a geodesic in $M$ that joins $\pi(p)$ and $\pi(q)$. These two lenghts are equal by the definition of the metric, hence $d_{\widetilde{M}}(p,q)\le L(\pi\circ\gamma)$ ($L=$ lenght). I don't know if we can conclude $d_{\widetilde{M}}(p,q)\le d_M(\pi(p),\pi(q))$.
Any hint? Thank you.