Show that $M^n-N^{n-1}$ has exactly two components with $N^{n-1}$ as the topological boundary of each. 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.
 A: 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.
