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!
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.)