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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 $\mathbb{R}^{n-1}.$

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I think you should edit your previous question instead of posting a new one: math.stackexchange.com/questions/552965/… –  Neal Nov 6 '13 at 10:26
    
Since I took the time to answer this one, please delete the old question. –  Christoph Nov 6 '13 at 10:35
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2 Answers

up vote 1 down vote accepted

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.

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Why is $\phi(x)\in V\cap\partial\mathbb H^n$? –  Saal Hardali Nov 6 '13 at 10:39
    
How do you define $\partial M$? –  Christoph Nov 6 '13 at 10:40
    
It's the topological boundary. Or am i missing something? –  Saal Hardali Nov 6 '13 at 10:41
    
For a topological space $X$, the topological boundary $\partial X$ is empty, since $X$ is open and closed. By that definition $\partial M$ would be empty. Hint: One possible definition involves the boundary of $\mathbb H^n$ which is defined as $\{x\in\mathbb H^n : x_n=0\}$ –  Christoph Nov 6 '13 at 10:44
    
I added a paragraph explaining two possible definitions of $\partial M$. Feel free to check their equivalence. –  Christoph Nov 6 '13 at 11:02
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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 $u$.

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