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Eduardo Longa
Eduardo Longa
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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?

P.S.: This leads me to another question: how do we define completeness for manifolds with boundary?

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?

P.S.: This leads me to another question: how do we define completeness for manifolds with boundary?

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?

Source Link
Eduardo Longa
Eduardo Longa
  • 2.1k
  • 12
  • 11

Local isometry implies covering map: nonempty boundary case

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?

P.S.: This leads me to another question: how do we define completeness for manifolds with boundary?