Understanding the definition of manifolds with boundary

Question

One example of a manifold with boundary is said to be the $n$-dimensional disk, so $M:=\{x\in\mathbb{R}^n:||x||\le 1\}$. In order to show this, one has to show there are charts $h_i:U_i\to U_i^\prime\subseteq\mathbb{R}^n_-=\{(x_1,...,x_n)\in\mathbb{R}^n:x_1\le 0\}$ with the properties of an atlas for normal manifolds.

So basically the charts have to homeomorphically map the manifold to open sets in the "half-plane".

To verify the example of the disk, can't I just take the translation $h:M\to M-(1,0,...,0)\subseteq \mathbb{R}^n_-$ as an atlas consisting of just a single chart?

Moreover for any bounded set in $\mathbb{R}^n$, can't I just translate it into $\mathbb{R}^n_-$ and get a manifold with boundary?

Your suggestion for bounded sets won't work since you need charts mapping homeomorphically to open subsets of the half plane. — paul blart math cop, Jun 18 '20 at 20:04
@Kajelad I think it is but to avoid any complications with the boundary I could just shift it by $2$ instead of just $1$. — mathemagician99, Jun 18 '20 at 20:28
No. The boundary has to map to the boundary of the half-space. — Ted Shifrin, Jun 19 '20 at 01:44
No matter what you do, you cannot find single chart for closed ball since it is compact. You should be able to find at least two chart for it, by using stereographic map etc. Maybe this. will help — Kelvin Lois, Jun 19 '20 at 05:58
@Eumenes Yes, that helps a lot. But why can't there be a single chart if the manifold is compact? Is this a general fact or does this only apply to the ball? — mathemagician99, Jun 19 '20 at 09:10
If X is compact then image X under continous map $f(X)$ is also compact. Apply this to chart map $h$ above. — Kelvin Lois, Jun 19 '20 at 10:44
@Eumenes When you write that $\sigma^{-1}(U^\pm_i)$ is homeomorphic to $B^{n-1}$ via the projection, don't you mean $B^n$ because we project from $\mathbb{R}^{n+1}$ to $\mathbb{R}^n$? — mathemagician99, Jun 19 '20 at 19:22
If you know some smooth manifold theory, there is a general technique to show these kind of things. Look here for example, showing that $\mathbb{B}^n$ is a smooth manifold with boundary which also implies it is a topological manifold with boundary. — Kelvin Lois, Jun 20 '20 at 05:30
I vote to close your question. Tell me if you're not agree and have more to ask. Showing that $\bar{\mathbb{B}}^n$ is a manifold with boundary (Lee ITM Probelm 3-4) — Kelvin Lois, Jun 20 '20 at 05:32

Understanding the definition of manifolds with boundary

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