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Consider the set $\mathcal{M} = \{\ \mathbf{x} \in \mathbb{R}^{3}\ | \ 1 \leq ||\mathbf{x}|| \leq 2 \ \}$. This is a $3$-submanifold with boundary.

Obviously, we have $\partial \mathcal{M} = \{\ \mathbf{x} \in \mathbb{R}^{3}\ | \ ||\mathbf{x}|| \in \{1,2\} \ \}$.

In my definition for a submanifold with boundary, for the coordinate patches covering the boundary of $\mathcal{M}$, I need a coordinate patch $\alpha:\mathbb{H}^{3} \to V$, where $V \subset \mathcal{M}$.

Here $\mathbb{H}^{3} = \{\ (x,y,z)\in\mathbb{R}^{3}\ |\ z \geq 0\ \}$ is the upper half plane of $\mathbb{R}^{3}$. The idea is that the boundary coincides with the the surface $z=0$ in $\mathbb{H}^{3}$.

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To me it's clear that $f: \mathrm{int}(\mathcal{M}) \to \mathrm{int}(\mathcal{M})$ defined by the identity function $f(x,y,z)=(x,y,z)$ is a coordinate patch covering all of $\mathrm{int}(\mathcal{M})$.

How on earth do I define the coordinate patches over the boundary?

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  • $\begingroup$ Each coordinate patch can only include one of the two walls. $\endgroup$ – T.J. Gaffney Apr 6 '16 at 20:47
  • $\begingroup$ That's what I was thinking, that have a seperate patch for each piece of the boundary (I think I might even need two per piece). However, which function do I need to use? I was thinking something along the line of the stereographic projection... $\endgroup$ – Greg.Paul Apr 6 '16 at 20:58
  • $\begingroup$ I know there should exist some continuous function $[0,\infty)\to[1,2)$ and some continuous function $[0,\infty)\to(1,2]$ $\endgroup$ – T.J. Gaffney Apr 6 '16 at 21:02

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