What would be counterexamples to the following statement:
It is not true that any $n$-manifold with boundary is a $n$-manifold with finitely many embedded disjoint open disks removed, since that would mean that its boundary is a disjoint union of spheres. E.g. the solid torus $B^2\times S^1$ has boundary the torus $S^1\times S^1$, which is not a disjoint union pf spheres.
Fuzzy question: If one had a classification of manifolds (up to diffeomorphism/homeomorphism/h-equivalence), could one automatically obtain a clasification of manifolds with boundary?
I know that to each manifold $M$ with boundary, there is associated a double manifold $DM$, which is the disjoint union of two copies of $M$, glued along their boundary. But if knew precisely what $DM$ was (in the classification scheme), could we classify $M$ itself?
Is any manifold a double of some other manifold, i.e. can every manifold be split into equal halves?
Simpler, $\mathbb RP^2$, the real projective plane, is not the double of a manifold with boundary.
If a manifold $M$ has boundary, notice that $\partial (M \times [0,1])$ is the double of $M$. So doubles have the property that they are boundaries of higher-dimensional manifolds. In dimension 2, a manifold is a double if and only if it is null-cobordant, which is if and only if it has even Euler characteristic.
You could construct some much sharper obstructions to a manifold being a double. Doubles have an effective action of $\mathbb Z_2$ with co-dimension $1$ fixed point set, the action switching the orientation of the normal bundle. With a little more work you could refine this to an if and only if statement for when a manifold is a double.