Universal cover of boundary

Question

Let $M$ be a compact manifold-with-boundary and $B$ a component of $\partial M$. Let $\tilde{M}$ be the univeral cover of $M$ with infinite-sheeted covering map $p:\tilde{M} \to M$. I wonder about the following: Is it true that each component of $p^{-1}(B)$ is a universal covering space of $B$ ?

Answer 1

No, this would imply that the boundary of a simply connected manifold is simply connected. But this isn't true for a number of reasons; there's the disc $D^2$, say. If you demand higher-dimensional things, simply connected 3-manifolds have simply connected boundary automatically, but as soon as you're in dimension 4 there are contractible manifolds whose boundaries are not $S^3$. (If its boundary was $S^3$, by the same circle of ideas we've talked about before, you could prove the manifold was homeomorphic to $D^4$.) These are called Mazur manifolds and are popular objects of study.

The components of $p^{-1}(B)$ are, in general, covering spaces of $B$. You can prove this by hand. But they won't necessarily be the universal cover.