Introduction: Given a Riemann surface whose universal cover is $\mathbb H$, we have the covering map $\pi:\mathbb H \rightarrow X$. There is a complete hyperbolic metric on $\mathbb H$ where $\operatorname{Bihol}(\mathbb H)=\operatorname{Isom}(\mathbb H)$. We can pushforward this metric to $X$, making $X$ into a hyperbolic surface $(X,\rho)$. This metric will be constant curvature $-1$ because locally the metric tensor is that of $\mathbb H$.
Question: Is $(X,\rho)$ a complete metric space? If not, what are some examples where $\rho$ is not a complete metric?
Motivation: Hubbard's book on Teichmuller theory states the collaring theorem:
The collaring theorem: Let X be a complete hyperbolic surface, and let $\Gamma=\{\gamma_1,\gamma_2,\dots\}$ be a(finite or infinite) collection of disjoint simple closed geodesics, each $\gamma_i$ of length $l_i$. Then $A_{\eta(l_i)}(\gamma_i)$ are collars around $\gamma_i$ and they are disjoint.
It is natural to wonder whether this holds true for all Riemann surfaces with their hyperbolic metric. If the hyperbolic metric were complete then the collaring theorem would apply!
Attempt at proof: Assume we have a fundamental domain $F$ for the action of deck-transformations of $\pi:\mathbb H \rightarrow X$. Then given a Cauchy sequence $\{a_n\}\subset X$, we could lift it to a Cauchy sequence in $F$ and use completness of $\mathbb H$ to get hold of a limit!
Question: If the lifted sequence goes near/on $\partial F$, it is a bit difficult to say the lifted sequence is Cauchy.
This is a very general fact of Riemannian geometry: If $M\to N$ is a Riemannian covering, then $M$ is complete if and only if $N$ is. The proof is by the path-lifting property. You can find it in Kobayashi-Nomizu.
Briefly, in the direction you are interested in: Let $c: [0,T)\to N$ be a geodesic. Lift it to a geodesic $\tilde{c}: [0,T)\to M$. By the completeness of $M$, the lifted geodesic extends to $T$. Now, project back to $N$. Thus, $c$ extends to $T$.