Manifolds with boundary: definitions and basic properties

Let $H^{n} = \left\{ (x_{1}, \ldots , x_{n}) \in \mathbb {R}^{n} : x_{n} \geq 0\right\}$ and $\partial H^{n} = \left\{ (x_{1}, \ldots , x_{n}) \in \mathbb {R}^{n} : x_{n} = 0\right\}$. An $n$-manifold with boundary is a topological space $M$ such that:

1. there is an embedding of $M$ into $\mathbb {R}^{N}$ for some $N$;

2. given $x \in M$, there is a neighborhood $U$ of $x$ and a homeomorphism $h$ of $U$ onto an open set in $H^{n}$.

Show that if $h: U \to H^{n}$ is a homeomorphism onto an open set in $H^{n}$ and $h(x) \in \partial H^{n}$, then there is a smaller open set $V$ about $x$ and a homeomorphism $h': V \to H^{n}$.

In a manifold with boundary, a point $x$ with $h(x) \in \mathbb {R}^{n} \setminus \partial H^{n}$ (i.e.$h(x)$ in the interior of $H^{n}$) is called an interior point, and a point with $h(x) \in \partial H^{n}$ is called a boundary point. Assume the following two results (to be proved in Chapter 6 using homology theory): first, an open set in $\mathbb {R}^{n}$ is not homeomorphic to an open set in $\mathbb {R}^{m}$ for $m \neq n$; second, if $x \in \partial H^{n}$ and $U$ is an open set in $H^{n}$ about $x$, then $U$ is not homeomorphic to an open set in $\mathbb {R}^{n}$ (or even any $\mathbb {R}^{m}$).

Using these two results, show that the dimension of a manifold is well defined and that an interior point cannot be a boundary point as well, so the concepts of interior point and boundary point are well defined.

Show that the boundary of an $n$-manifold is either empty or an $(n-1)$-manifold.

Suppose $M, N$ are $n$-manifolds with boundary. Show that if $h: M \to N$ is a homeomorphism, then $h|_{\partial M}$ is a homeomorphism from $\partial M$ to $\partial N$.

Show directly that $H^{2}_{+} = \left\{ (x_{1}, x_{2}) : x_{1} \geq 0, x_{2} \geq 0\right\}$ is homeomorphic to $H^{2}$.

Show that the product of an $m$-manifold $M$ and $n$-manifold $N$ is an $(m+n)$-manifold $M \times N$. Conclude that $T = S^{1} \times S^{1}$ (the torus) is a compact, connected surface.

Show that a connected $n$-manifold is path connected.

Show that a compact $n$-manifold has only a finite number of path components, each of which is a compact, connected $n$-manifold. Conclude that a compact $1$-manifold is the disjoint union of a finite number of circles.