TITLE: The boundary of an $n$-manifold is an $n-1$-manifold
QUESTION [29 upvotes]: 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 $\mathbb{R}^{n-1}.$

REPLY [2 votes]: 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 $u$.