Why can we replace the euclidean space with the upper half space as model space for a manifold

Question

Reading the Tu.'s book "introduction to manifolds" it first defines manifolds as topological spaces which are locally euclidean (which means locally they are homeomorphic to the euclidean space $R^n$). Then, when it introduces manifolds with boundaries, it takes as a model space the upper half space $H^n$ (defined as all the points of $R^n$ having the n-th coordinate $x_n \ge 0$). But the space $H^n$ is not homeomorphic to $R^n$ (as you can see from this post). So I don't understand, how we are allowed to substitute the model space like this (in the sense that the manifolds using $R^n$ as model space woud still be manifolds under this substitution)? Thank you.

Answer 1

This is because $H^n$ contains the open half-space $\mathbb R^n_+ = \{(x_i) \in \mathbb R^n \mid x_n > 0 \}$ which is open in $\mathbb R^n$. It is homeomorphic (even diffeomorphic) to $\mathbb R^n$, but this is irrelevant here.

A chart $(U,\phi)$ with model $H^n$ is a homeomorphism $\phi : U \to V$ from an open subset of $M$ to an open subset of $H^n$. There are two types of charts around points $p \in M$:

1. Charts $(U,\phi)$ such that $\phi(p) \in \partial H^n$

2. Charts $(U,\phi)$ such that $\phi(p) \notin \partial H^n$

A subtype of 2. is

3. Charts $(U,\phi)$ such that $\phi(U) \subset \mathbb R^n_+$

All charts of type 3. are also "ordinary" charts with model $\mathbb R^n$.

In the second case you can always shrink $U$ to an open subset $U'$ containing $p$ such that $V' := \phi(U') \subset \mathbb R^n_+$ which gives you the chart $(U',\phi \mid_{U'} : U' \to V')$ of type 3.

This partitions $M$ into interior points (having an "ordinary" chart around it) and boundary points (having no "ordinary" chart around it).

It is clear that the set of interior points is a manifold without boundary.

The essential point is this: For a manifold $M$ without boundary all charts with model $H^n$ have type 3. This is a non-trivial fact; Tu proves it for charts in a smooth atlas (but it is true for any chart on a topological manifold). Thus the charts on $M$ with model $H^n$ are also charts with model $\mathbb R^n$. There are less charts with model $H^n$ because the range of such charts is contained in $\mathbb R^n_+$. The collection of all these charts is still an atlas, but is not the maximal atlas in the sense of the standard definition with model $\mathbb R^n$.