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This is the problem 6.8.1. from "Topology and Geometry" by Glen E. Bredon.

The problem is,

If $M^n$ is a connected, orientable, and compact $n$-manifold with $H_1(M^n;\mathbb{Z}) = 0$ and if $N^{n-1} \subset M^n$ is a compact connected $(n-1)$-manifold, then show that $M^n-N^{n-1}$ has exactly two components with $N^{n-1}$ as the topological boundary of each.

By Using the Corollary 8.8.(Generalized Jordan Curve Theorem), I can easily prove that $M^n-N^{n-1}$ has exactly two components. For any $p \notin N^{n-1}$, we can have an open neighborhood of $p$ which does not touch $N^{n-1}$. Therefore a boundary of a component is contained in $N^{n-1}$. But it is difficult to show that a point in $N^{n-1}$ is contained in a boundary of a component.

Can anybody give a hint?

Thank you.

ljh8372
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  • Maybe I'm missing something subtle, but the topological boundary is just the closure minus interior, and $p\in N^{n-1}$ is definitely in the closure minus interior, of the components of $M^n\setminus N^{n-1}$ (in some chart $U_p\to \mathbb{R}^n$, a neighborhood of $p$ in $N^{n-1}$ corresponds to ${0}\times\mathbb{R}^{n-1}\subseteq \mathbb{R}^n$). Or is $N^{n-1}$ some wild submanifold (not embedded)? – Peter Franek Dec 20 '14 at 15:04
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    $N^{n-1}$ is an embedded one. Along the smooth manifold, we have a slice condition for an embedded submanifold such like Peter Franek explains. But I wonder if we have a slice condition even in a topological manifold. – ljh8372 Dec 20 '14 at 15:10

1 Answers1

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If $M^n-N^{n-1}$ has two components, say $U,V$ you know that $U \cup V \subset M$ is open, since $N$ is closed, obviously $U\cap V = \emptyset$ and $U,V\subset M$ are both open subsets. Hence you know that $int( U)= U \subset cl(U) \subset U \cup N = M-V $, since $M-V$ is closed (note that this already gives one inclusion and amounts to saying what you did, just without picking elements, i.e. slightly more elegant.). We want to show: $N = cl(U) - int(U)$, then the problem follows by symmetry. By the set-inequality this is equivalent to show $cl(U) = U\cup N$. To this this pick $x\in N$ and pick a chart $(\psi,X)$ for $M$ around $x$, such that $\psi(X)=\mathbb R^n$ and $\psi (X\cap N)=\mathbb R^{n-1}\times \{0\}$ and $x \stackrel \psi \mapsto 0$. Then $U\cap X,V \cap X$ each map onto one of the two components of $\mathbb R^n - (\mathbb R^{n-1} \times 0)$. Every neighborhood of $0$ intersects those two components, hence every neighborhood of $x$ (contains a neighborhood which lies in $X$ and hence) intersects $U$ (and $V$). So $x \in cl(U)$.

However, for the argument we need $N$ to be properly embedded.

Daniel Valenzuela
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