Lets $ \mathbb{R_{+}^{n}} = \mathbb{R^{n-1}} \times [0;+\infty[ $
Basically in my course I have this statement within the definition of a manifold with boundary:
- $ \forall x \in M, \exists U_x $ an open neighborhood of $x$ in $M$ and a homeomorphism $\phi: U_x \rightarrow A \subseteq \mathbb{R_{+}^{n}}$, where $A$ is an open in $ \mathbb{R_{+}^{n}} $
Later it is written that:
- a point $x$ is said to be a boundary point if for one (and hence for all) homeomorphism $\phi: U_x \rightarrow A \subseteq \mathbb{R_{+}^{n}} $, with $U_x \in V(x)$ and $A$ is an open in $ \mathbb{R_{+}^{n}}$ we have $\phi(x) \in \mathbb{R^{n-1}} \times \{0\} $
The above became problematic to me once I tried to prove:
- let M be $[0;+\infty[ \times [0;+\infty[ $, then its boundary $\partial M = ([0;+\infty[ \times {0}) \cup ( {0} \times [0;+\infty[) $
It seems to me that for any boundary points in $M$, the only open neighborhood available is $M$ itself (as it is the whole space it is open by the very definition of a topology). Indeed, no other open set can contain one of these boundary points, for example $(0,0)$.
Accepting that I have to exhibit a homeomorphism between $\mathbb{R_{+}^2}$ and $M$. Basically between the upper half of the plane and the upper right quarter of the plane. And then I'll have to prove $\partial M $ to be the set of point ending up on the $x$-axis.
- I guess the mapping $z \rightarrow z^2$ in the complex plane will do the job. It will double angles, mapping the boundaries of the right set on the boundaries of the left one.
My question:
- Is the above correct?
- If so, since boundary points will never be contained in an open set other than $\mathbb{R_{+}^{n}}$ itself, the above local homeomorphisms will always have to be global as far as boundary points are concerned ?
This is not true. Recall that by the definition of induced topology, the intersection $U\cap M$ is open in $M$ for every open set $U\subset\mathbb{R}^2$. In particular, the set $$ \{(x,y)\in M: x<1, y<1\} $$ is an open subset of $M$ containing $(0,0)$.
But you correctly observed that the map $z\mapsto z^2$ gives a global homeomorphism between $M$ and the upper half-plane, which suffices to prove that $M$ is a (topological) manifold with boundary. You did not have to find a global homeomorphism, but since one is easily found, may as well use it.
More generally, the product of topological manifolds $M,N$ with boundary is itself a topological manifold with boundary, and $\partial (M\times N) = \partial M\times N\cup M\times \partial N$. (This fails for smooth manifolds.)