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I'm reading about Riemannian Geometry and my question is regarding Manifolds with Boundary.

I want to show a point of a manifold with boundary is either an interior point or a boundary point, so no overlap.

Let $\mathbb{H}_n$ be the closed half space of dimension n. In order to prove the above my aim is to show an open neighborhood U of a point p, in $\mathbb{H}_n$\ $\partial \mathbb{H}_n$ can't be homeomorphic to an open subset U of $\mathbb{H}_n$ when we identify p with a point in $\partial \mathbb{H}_n$.

How should I proceed?

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  • $\begingroup$ You should proceed to find some property that one of the has and the other does not. $\endgroup$ – Mariano Suárez-Álvarez Dec 23 '14 at 22:00
  • $\begingroup$ Right, but I'm having difficulty finding a mutually exclusive event. $\endgroup$ – Mary Dec 23 '14 at 22:03
  • $\begingroup$ Think about the topology on $\mathbb{H}^n$, and how it relates to the topology on $\mathbb{R}^n$. $\endgroup$ – Ashwin Iyengar Dec 23 '14 at 22:04
  • $\begingroup$ Also, you should prove that "being on the boundary" is a property independent of choice of local coordinates --- the inverse function theorem is useful. $\endgroup$ – Ashwin Iyengar Dec 23 '14 at 22:06
  • $\begingroup$ That last part I think I can handle. Can you show me how to answer my original question? I would appreciate it. $\endgroup$ – Mary Dec 23 '14 at 22:13
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As this thread mentions in the comments,

Interior and boundary points of $n$-manifold with boundary

you need the "invariance of domain" theorem:

http://en.wikipedia.org/wiki/Invariance_of_domain

to prove what you want, which is nontrivial, and requires the use of algebraic topology. As far as I can tell, there doesn't seem to be a much simpler proof that's easily accessible.

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  • $\begingroup$ No problem. Good luck! $\endgroup$ – Ashwin Iyengar Dec 23 '14 at 23:33

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