The following theorem is well known in the literature:
Let $M$ and $N$ be riemannian manifolds and let $f : M \to N$ be a local isometry. If $M$ is complete and $N$ is connected, then $f$ is a covering map.
My question is: does the same theorem hold when we assume that $M$ and $N$ are now riemannian manifolds with boundary?
One needs to define a notion of local isometry for manifolds with boundary in such a way that it becomes a local diffeomorphism between the manifolds. Then the thing to check is that map $f: M\to N$ satisfies the path-lifting property (which is a necessary and sufficient condition for a local homeomorphism to be a covering map). The proof of a path-lifting property is essentially the same as for manifolds without boundary: Take a rectifiable map $p: [0,1]\to N$ which admits a lift $\tilde{p}: [0,1)\to M$ and extend it continuously to $[0,1]$ using the fact that for each sequence $t_i\to 1-$, the sequence $(p(t_i))$ and, hence $(\tilde{p}(t_i))$ is Cauchy.