Why does any connected closed $m$-manifold that can be embedded in $E^{m+1}$ bounds a compact connected $(m+1)$-manifold

algebraic-topologycompact-manifoldsdifferential-topologygeneral-topologymanifolds-with-boundary

I am reading Sheila Carter and S.A. Robertson's paper Relations Between a Manifold and its Focal Set. In this paper, they use the following facts:

Any closed $m$-manifold $M$ that can be embedded in $E^{m+1}$ bounds a compact connected $(m+1)$-manifold $V$, which of course need not be unique. Thus it is natural to consider all such $(m+1)$-manifolds $V$ and their embeddings $F\colon V\to E^{m+1}$.

In that paper, all manifolds and embeddings are assumed to be smooth. By a closed $m$-manifold they mean a compact connected manifold of dimension $m$, without boundary.

How to prove that? I think it may be related to cobordism theory and read this related question, but I didn't find a solution. Perhaps the key point is that $M$ can be embedded in $E^{m+1}$, which is not always true(considering the Klein Bottle).

Best Answer

I think you're looking for the Jordan-Brouwer Separation Theorem. The proof is non-trivial, but not impossible. In any case, I don't think it would be appropriate to give a proof here. Guilleman and Pollack (G&P) give an outline for the proof p.87-89. I believe these notes build up the background necessary (and are based on G&P) and then show how the theorem is proved.

EDIT: Just to confirm @ThorbenK, this specific proof requires some intersection theory. The role intersection theory plays here is giving us a smoothly constructed "characteristic function" which can tell the difference between the two components.

Related Question