The boundary of an $n$-manifold is an $n-1$-manifold

The following problem is from the book "Introduction to topological manifolds".

Suppose $M$ is an $n$-dimensional manifold with boundary.
Show that the boundary of $M$ is an $(n-1)$-dimensional manifold (without boundary) when endowed with the subspace topology.

So far I’ve manged to prove that the boundary is a second countable Hausdorff space (when endowed with the subspace topology) but I’m stuck with proving it's locally homeomorphic to $\mathbb{R}^{n-1}.$

Let $x\in\partial M$ be any point on the boundary of $M$. Since $M$ is a manifold with boundary, there is an open neighborhood $U$ of $x$ that is homeomorphic to an open subset $V$ of $\mathbb H^n=\{ x\in\mathbb R^n : x_n\ge 0\}$ via a homeomorphism $\phi: U\to V$. Since $x\in\partial M$ we know that $\phi(x)\in V\cap\partial\mathbb H^n$, i.e. $(\phi(x))_n=0$. Since $\phi:U\to V$ is a homeomorphism, it restricts to a homeomorphism $$\phi^{-1}(V\cap\partial\mathbb H^n) \to V\cap\partial\mathbb H^n.$$ We know that $V\cap\partial\mathbb H^n$ is open in the subspace topology on $\partial\mathbb H^n$ since $V$ is open in $\mathbb H^n$. Thus $V\cap\partial\mathbb H^n$ is an open neighborhood of $\phi(x)$ in $\partial\mathbb H^n$.

Now the open neighborhood $U\cap \partial M$ of $x$ in $\partial M$ is exactly $\phi^{-1}(V\cap\partial\mathbb H^n)$, which as we showed, is homeomorphic to an open subset of $\partial\mathbb H^n\cong\mathbb R^{n-1}$.

On $\partial M$: The most simple definition of $\partial M$ would be the set of all points of $M$ that don't have open neighborhoods which are isomorphic to open sets in $\mathbb R^n$. This gives you that any coordinate chart $\phi:U\to V$ of a manifold with a boundary has to send points of $\partial M$ to points of $\partial \mathbb H^n=\{x\in\mathbb H^n : x_n=0\}$, since all other points in $\mathbb H^n$ have open neighborhoods in $\mathbb R^n$ which can be pulled back to open neighborhoods in $M$ via $\phi$.

Another definition would be that $\partial M$ consists of all points of $M$ that are mapped to points of $\partial\mathbb H^n$ via all (or equivalently one) coordinate chart.

• I added a paragraph explaining two possible definitions of $\partial M$. Feel free to check their equivalence. – Christoph Nov 6 '13 at 11:02
• It's not clear why $\phi^{-1}(V\cap\partial\Bbb H^n)$ is equal to $U\cap\partial M$. – Camilo Arosemena-Serrato Apr 22 '17 at 15:09
• @CamiloArosemena, this follows from the definition of boundary points of $M$. A point of $M$ is a boundary point if and only if it is mapped to $\partial\mathbb H^n$ by any chart. – Christoph Apr 23 '17 at 11:49
• @CamiloArosemena Consider two charts $\varphi,\psi\colon U\to\mathbb H^n$, by definition, $\psi\circ\varphi^{-1}$ is a homeomorphism $\varphi(U)\to\psi(U)$ sending $\varphi(x)$ to $\psi(x)$. For $x\in U$ you can now verify that $\varphi(x)\in\partial\mathbb H^n \Leftrightarrow \varphi(U)\setminus\{\varphi(x)\}$ is simply connected $\Leftrightarrow \psi(U)\setminus\{\psi(x)\}$ is simply connected $\Leftrightarrow$ $\psi(x)\in\partial\mathbb H^n$. Hence, either none or all charts send $x$ to $\partial\mathbb H^n$. – Christoph Aug 8 '17 at 12:49
• @CamiloArosemena This is a consequence of Brouwer invariance of domain theorem. If one chart sends $x$ to $\partial \Bbb H^n$, then every other chart in the neighborhood of $x$ does the same. – Desura Jan 3 '18 at 18:50

A point in an $n$-manifold $M$ with boundary is by definition either contained in an open set homeomorphic to $\mathbb R^n$ or homeomorphic to the upper half plane $\mathbb R^n_{\ge 0} = \{ (x_1, ... , x_n) | x_n \ge 0 \}$. The boundary points $u$ in $M$ are precisely the points contained in an open set homeomorphic to the upper half plane. From this you can show (by a contradiction argument) that $u$ must have coordinates of the form $u = (u_1, ..., u_{n-1}, 0)$.

To finish the proof it only remains to be shown that $u$ has a neighbourhood $N$ that is homeomorphic to $\mathbb R^{n-1}$. To this end note that if $N$ is any neighbourhood it contains an open set that is homeomorphic to an open set $U$ in the upper half plane. Let $f$ denote a homeomorphism between $U$ and the open set containing $u$.

Note that $B = \{(x_1, ...,x_{n-1}, 0) | x_i \in \mathbb R \}$ is homeomorphic to $\mathbb R^{n-1}$ and that it is a subspace of the upper half plane. Since $U$ is open in the upper half plane it follows that $U \cap B$ is open in $B$. To conclude the proof it is enough to note that $f(U \cap B)$ is open in $\partial M$ and contains $u$.