Let $M$ be one smooth $n$-dimensional manifold. Let $\Sigma,\Sigma'\subset M$ be two hypersurfaces on $M$.
Intuitively sometimes it may be possible to connect $\Sigma$ and $\Sigma'$ and obtain a submanifold with boundary.
One example of this is: pick $\mathbb{R}^n$ and the hyperplanes $\Sigma = \{x\in \mathbb{R}^n : x^n = 0\}$ and $\Sigma' = \{x\in \mathbb{R}^n : x^n = h\}$ for some fixed $h\in \mathbb{R}$.
With this we can build the boxes
$$B(a_1,b_1,\dots,a_{n-1},b_{n-1}) =[a_1,b_1]\times[a_2,b_2]\times\cdots [0,h]$$
representing pieces of the region between $\Sigma$ and $\Sigma'$. Also, we can send $a_i\to \infty$ and $b_i \to \infty$ to obtain the whole region between $\Sigma$ and $\Sigma'$.
I want to know when it is possible to do things like this in one arbitrary manifold $M$.
I believe the point is that any two points $p\in \Sigma$ and $q\in \Sigma'$ should be connected by a path $\gamma : [0,1]\to M$ with $\gamma(0)=p$ and $\gamma(1)=q$. But this doesn't help so much, in understanding when this can be done.
So my question is: is there some criteria that allows us to know when this construction can be carried out? Furthermore, when it is possible, is there some way to actually construct the resulting submanifold, or we can just know it exists, without a way to describing its construction?
Suppose $\Sigma_1, \Sigma_2 \subset M$ are disjoint hypersurfaces with trivial normal bundles. Take disjoint tubular neighborhoods $$U_i \simeq \Sigma_i \times \mathbb{R}, \Sigma_i \simeq \Sigma_i \times \{0\}$$ and set $$W:= \{x\in M \mid x\notin \Sigma_i \times (0,1)\text{ for }i=1,2\}. $$ Then $W\subset M$ is a codimension $0$ submanifold with boundary $\partial W\supset \Sigma_1 \cup \Sigma_2$.
Notice that if $\Sigma_1$ and $\Sigma_2$ both lie in the same connected component, we can make $W$ connected by taking either $(-1,0)$ or $(0,1)$ in the construction above.
On the other hand, if $\Sigma\subset M$ is a hypersurface such that there is a submanifold $W\subset M$ with $\partial W \supset \Sigma$, then the normal bundle $N\Sigma$ of $\Sigma$ in $M$ is trivial. Indeed, we can always construct a nonvanishing inward pointing vector field $\nu$ along the boundary $\Sigma$ in $W$. Since $W\subset M$ is embedded, this $\nu$ is a nonvanishing section along $N\Sigma$ as well. Therefore $N\Sigma$ is trivial.
CONCLUSION: There is a $W\subset M$ with $\partial W \supset \Sigma_1 \cup \Sigma_2$ if and only if both $N\Sigma_1$ and $N\Sigma_2$ are trivial.
If we want $\partial W=\Sigma_1 \cup \Sigma_2$ there are obstructions like the cobordism class...