Let N be a connected properly embedded submanifold of M, where M is simply connected. Suppose N has codimension 1, then I want to prove that the complement of N has exactly two components.
I think I should use something about tubular neighborhood and intersection number, but I don't where to start. I would like some guidance on how this can be solved. Thanks in advance.
Best Answer
Let me give two rough sketches.
The quick and dirty way is algebraically. Namely, Poincaré duality yields $H_0(M\setminus N;\mathbb{Z}/2\mathbb{Z})\cong H_c^n(M\setminus N;\mathbb{Z}/2\mathbb{Z})$ and the latter can be calculated by again applying Poincaré duality to the other terms in the exact localization sequence \begin{equation*} 0=H_c^{n-1}(M;\mathbb{Z}/2\mathbb{Z})\rightarrow H_c^{n-1}(N;\mathbb{Z}/2\mathbb{Z})\rightarrow H_c^n(M\setminus N;\mathbb{Z}/2\mathbb{Z})\rightarrow H_c^n(M;\mathbb{Z}/2\mathbb{Z})\rightarrow H_c^n(N;\mathbb{Z}/2\mathbb{Z})=0. \end{equation*} (Furthermore, $M$ is orientable as it is simply connected and hence $M\setminus N$ is orientable too. An analysis of the localization sequence with $\mathbb{Z}$-coefficients then implies that $N$, too, is orientable.)
In case our manifolds are smooth, I offer up a geometric argument that is less efficient, but somewhat less high-powered. Following your intuition, let $N\subseteq U\subseteq M$ be a tubular neighborhood. Argue using the simple connectedness of $M$ that the inclusion $U\setminus N\rightarrow M\setminus N$ induces a bijection between their components, e.g. by analyziging a fitting Mayer-Vietoris sequence. Now, $U\setminus N$ is the total space of an $\mathbb{R}^{\times}$-bundle over the connected space $N$, so it is connected if and only if that bundle has no monodromy if and only if the normal bundle of $N$ in $M$ is trivial if and only if $N$ is orientable (these equivalences are all general theory of normal/line bundles). The latter is the case for any closed (equivalent to properly embedded) codimension $1$ submanifold. This is "well-known", but proofs aren't as common as they should be. You can adapt this argument by Georges Elencwajg or this argument by Samelson.