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Problem setting: We define a cylinder with a hole \begin{equation} A_{\delta} = \{(x',x_N) \in \mathbb{R}^{N-1} \times \mathbb{R} \mid \delta \leq |x'| \leq 1 \} \end{equation} for some $\delta \in (0,1)$ and we set $N_{i} = \{(x',x_N) \in \mathbb{R}^{N-1} \times \mathbb{R} \mid |x'|=1, x_N= i\}$ for $i \in \{\pm 1\}$. Then we consider an orientable compact $(N-1)$-dimensional (smooth) manifold $M \subset \mathbb{R}^N$ with $\partial M = N_{+1} \cup N_{-1}$.

Question: Is it true that, if $M \subset A_{\delta}$, then $N_{+1}$ and $N_{-1}$ are in the same connected component of $M$? (If it's true, how can we prove it?)

Thank you for your time!

IgotYourPoint
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  • Hi! It appears you're new to MSE so you may not know, but it is part of the site's rules that you submit your own working/attempts as part of your questions, or an explanation of the parts you have trouble understanding with regards to the topic. Neglecting to do so may attract downvotes or votes to close the question. – H. sapiens rex Sep 24 '23 at 13:50
  • There are trivial counterexamples: for instance, you could take the union of $S^{N-2}\times[-1,1]$ with some small closed $(N-1)$-manifold located elsewhere in $A_\delta$. I guess what you really want to ask is whether $N_{+1}$ and $N_{-1}$ must be in the same connected component of $M$? – Eric Wofsey Sep 24 '23 at 14:19
  • Also presumably you want to require $N>2$. – Eric Wofsey Sep 24 '23 at 14:35
  • Ah, you're right. I forgot to mention that the dimension is larger than 2. And yes, I meant to say that the boundaries are in the same connected component of M. Thanks. – IgotYourPoint Sep 24 '23 at 14:45

1 Answers1

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If $N_{+1}$ and $N_{-1}$ are not in the same connected component, then the connected component of $N_{+1}$ is a compact manifold $M_0\subset A_\delta$ with boundary $N_{+1}$, so it suffices to show that no such manifold exists. Consider the fundamental class $[N_{+1}]\in H_{N-1}(N_{+1})$. Note that the image of $[N_{+1}]$ in $H_{N-1}(A_\delta)$ is nontrivial, since $A_\delta$ deformation-retracts to $N_{+1}$. Thus the image of $[N_{+1}]$ in $H_{N-1}(M_0)$ must be nontrivial as well. But this is false, since $[N_{+1}]$ is the boundary of $M_0$. (Explicitly, for instance, if you pick a triangulation of $M_0$, then the sum of all the $(N-1)$-simplices with appropriate orientations will be an $(N-1)$-chain whose boundary is a representative for $[N_{+1}]$. Or you can work with mod 2 coefficients and not have to worry about orientations, and do not need to assume that $M$ is orientable.)

Eric Wofsey
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  • What do you mean by the image of the fundamental class [$N_{+1}$]? – IgotYourPoint Sep 24 '23 at 15:24
  • I mean the image under the induced map of homology from the inclusion map $N_{+1}\to A_\delta$ (or $N_{+1}\to M_0$). – Eric Wofsey Sep 24 '23 at 15:34
  • Thanks. Another question: why does the nontriviality of the image of [$N_{+1}$] in $H_{N-1}(M_0)$ contradict the fact that $N_{+1}$ is the boundary of $M_0$? I just want to know the detail. – IgotYourPoint Sep 24 '23 at 15:49
  • This is a basic theorem in the theory of homology of manifolds. As I mentioned in the answer, one way to prove it (for smooth manifolds) is to use a triangulation. For another way that works for arbitrary topological manifolds see https://math.stackexchange.com/questions/634810/a-somewhat-conceptual-proof-that-the-boundary-of-a-fundamental-class-of-a-mani. – Eric Wofsey Sep 24 '23 at 17:49
  • I'm sorry, though I've seen the link, I don't still get your point. Could you show me the details or give me a little more references? – IgotYourPoint Sep 24 '23 at 20:37
  • The link shows that if $M$ is a compact $n$-manifold, then the fundamental class $[\partial M]$ is in the image of the connecting homomorphism $\partial$ in the exact sequence on homology $$H_n(M,\partial M)\stackrel{\partial}\to H_{n-1}(\partial M)\to H_{n-1}(M).$$ Thus by exactness, $[\partial M]$ is also in the kernel of the map $H_{n-1}(\partial M)\to H_{n-1}(M)$. – Eric Wofsey Sep 24 '23 at 23:44