Complement of an hypersurface Let $M$ be a smooth manifold $\mathcal{C}^{\infty}$ and $ H \hookrightarrow M$ an embedded closed submanifold of codimension $1$. 
I should demonstrate that $M-H$ has at most two connected components. It's pretty intuitive ,but I do not really know from where to start.
 A: Assume that $H$ is a connected closed manifold.
(1) Consider a function $h : M\rightarrow \mathbb{R},\ h(x)=
 d(x,H)$ where $d$ is a distance function. Since $H$ is closed, so there is at least one
 point $x_0\in H$ s.t. $d(x,x_0)=d(x,H)$.
(2) Fix a point $y\in H,\ y_\pm \in M-H$ s.t. $y$ is a mid point in
a segment $[y_+y_-]$ and $d(y_+,y_-)\leq 2\varepsilon$.
If any point in $M-H$ can be joined to one of points $y_\pm$ by a curve not
 intersecting $H$, then $M-H$ has at most two components.
If not, there is $x\in M-H$. By (1) we have $x_0\in H$. Since $H$
 is connected, there is a path $\alpha$ in $H$ from $y$ to $x_0$.
 Hence there are $\alpha_\pm$ s.t. (a) $\varepsilon$-tubular neighborhood of $
 \alpha$ contains $\alpha_\pm$, (b) $\alpha_\pm$ is curves from $y_\pm$ to a point near
 $x_0$ and (c) $ \alpha_\pm$ does not intersect $H$.
Hence $ \alpha_\pm$ are in different sides wrt $H$ around $x_0$. If
 $\alpha_+$ are in same side with $[xx_0]$, then we connected them
 so that $x$ can be joined to $y_+$.
A: Here is another approach to this problem. It is more elementary since it avoids using the tubular neighborhood theorem. This proof is part of lemma 4.4 of Hirsch's Differential Topology.
The idea is the following. We will show that if $p$ is a point on $H$, then every neighborhood of $p$ intersects all of the connected components of $M \setminus H$. But locally around any point, $H$ has only two sides because it sits in $M$ like $\mathbb{R}^n$ sits inside $\mathbb{R}^{n+1}$ (here we use that $H$ is embedded). Hence we conclude there must be at most two connected components in $M \setminus H$.
Let $A$ be a connected component of $M \setminus H$. In the following $\partial A$ denotes the topological boundary of $A$ inside $M$.
Step 1: $A$ is open in $M$ and $\partial A$ is non-empty. 
$A$ is open in $M \setminus H$ since it is a connected component and $M \setminus H$ is a manifold (recall that in general connected components are closed but need not be open. However in a locally connected space they are also open). It is also open in $M$ since $M \setminus H$ is open in $M$. 
$\partial A$ is non-empty since otherwise $A$ would be open and closed in $M$, hence we would have $A=M$, a contradiction.
Step 2: $\partial A \subset H$.
Clearly it suffices to show that $\partial A$ cannot intersect another connected component of $M \setminus H$. But if this was the case, then the closure of $A$ inside $M \setminus H$ would also intersect another connected component of $M \setminus H$. But $A$ is closed in 
$M \setminus H$ and connected components are disjoint, so this cannot happen.
Step 3: $\partial A = H$. 
We will show that $\partial A$ is both open and closed in $H$ and use the fact that $H$ is connected to conclude.
$\partial A$ is clearly closed in $M$ since it is a boundary. Hence it is also closed in $H$.
Now we show that $\partial A$ is open in $H$. Let $p$ be a point in $\partial A$. Since 
$p \in H$ and $H$ is embedded, there is a "slice chart" for $H$ around $p$. This is a neighborhood $U$ of $p$ inside $M$ and a diffeomorphism $\phi: U \to \mathbb{R}^{n+1}$ such that 
$\phi^{-1}( \mathbb{R}^n \times \{0 \}) = U \cap H$. Since $p \in \partial A$, there is a point $x \in A$ such that $x \in U$. But then the preimage by $\phi$ of the half-space in 
$\mathbb{R}^{n+1}$ that contains $\phi(x)$ is connected and contains $x$, so it must be contained in $A$ since $A$ is a connected component. Hence $\partial A$ contains the preimage of the boundary of this half-space, which is $U \cap H$. Since $p$ is arbitrary this shows that $\partial A$ is open inside $H$.
Step 4: Conclude that $M \setminus H$ has at most two connected components.
Pick a point $p \in H$ and a slice chart $U$ around $p$ as above. Then $U \setminus H$ has exactly two connected components. By step 3, every connected component of $M \setminus H$ must intersect $U \setminus H$. Now observe that if two connected components of $M \setminus H$ intersect the same component of $U \setminus H$, then they must be equal. Hence there can be at most two connected components of $M \setminus H$.
