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

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).

• If you are interested in embeddings in euclidean space then it is easy to see that the complement has one compact connected component and one non-compact connected component (provided that your manifold is connected otherwise it is to generalize this). The compact connected component is your nullbordism. – ThorbenK Feb 13 at 13:22
• @ThorbenK Thanks a lot for your comment. I am an undergraduate student. I am not sure if I get your point. Assumed that $f$ is the embedding from $M$ to $E^{m+1}$, do you mean that $f(M)$ bounds the compact connected component of $E^{m+1}\backslash f(M)$? But how can we find $V$? It has many possibilities, and the complement of $M$ may be connected. And how can we find an embedding $F\colon V\to E^{m+1}$? Thanks. – Yuehuan Yu Feb 13 at 14:23
• I'm still assuming that $E^{m+1}$ is euclidean space. Then the compact connected component is one choice of $V$ and as a subset of euclidean space it is already embedded. Of course there are a lots of manifolds that bound $M$ and not all of them can be embedded in $E^{m+1}$. The fact that the complement is disconnected can be proven in multiple ways and it is a good exercise the easiest prove I can think of uses intersection theory. – ThorbenK Feb 13 at 15:23
• @ThorbenK Yeah, $E^{m+1}$ denotes (m+1)-dimensional Euclidean space. You find a manifold $V\subset E^{m+1}$ bounds $f(M)$, do you mean that $V$ bounds $M$? But $M$ is not the subset of $E^{m+1}$, or we can just regard $M$ as a subset of $E^{m+1}$ since $M$ can be embedded in $E^{m+1}$? – Yuehuan Yu Feb 13 at 15:52

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.

• Thanks for your answer. I have understood what you said, but I have a small question. Assume that $f$ is the embedding from $M$ to $E^{m+1}$. By Jordan-Brouwer Separation Theorem, we can find the "inside" $D_1$ of $f(M)$ and we have $\partial\bar{D_1}=f(M)$. But $M$ is not a subset of Euclidean space, so how can we find the abstract manifold $V$ that bounds $M$? Can we extend the diffeomorphism $f^{-1}\colon f(M)\to M$ to a map $g$ defined on $\bar{D_1}$ and set $V=g(\bar{D_1})$? – Yuehuan Yu Feb 14 at 9:35
• The answer to the question "how can we find the abstract manifold $V$ that bounds $M$?" depends on how you like to write down your manifolds. Some people would be perfectly happy with "It is the compact component of $\mathbb{R}^{m+1}\setminus f(M)$". If you want to write $V$ as a collection of charts, you'll have a bunch of trivial charts that paste together easily. Just take open balls. The only weird charts are on the boundary which you recover from $f$. That manifolds can be abstract is not necessary, it is just considered generic. But, if you're handed an embedding, go ahead and take it! – Prototank Feb 14 at 21:35