I would like to know whether the following tentative generalization of the Jordan curve theorem in higher dimensions and for smooth manifolds are true, and in case I ask for references proving them.
Let $M$ be an orientable connected smooth n-manifold ($n>1$) without boundary. Suppose $\iota: X_{n-1}\rightarrow M$ is a compact connected orientable $n-1$ dimensional submanifold. I expect the following to be true:
- If the map induced in homology $\iota _* : H_{n-1}(X_{n-1})\rightarrow H_{n-1}(M)$ is trivial, then $M-X_{n-1}$ is the disjoint union of two connected n-manifolds $A$, $B$ such that $\partial A=\partial B= X_{n-1}$.
For this statement I essentially have a proof: since $X_{n-1}$ is trivial in homology $X_{n-1}=\partial A$, and then I can define $B=M-\overline{\partial A}$. Is this obvious that $B$ is connected or there might be some subtlety?
Furthermore let us assume $M$ to be also compact. Then I expect the converse to be also true:
- If $M$ is compact, then $M-X_{n-1}$ is the disjoint union of n-manifolds $A$, $B$ such that $\partial A=\partial B= X_{n-1}$ if and only if $\iota _* : H_{n-1}(X_{n-1})\rightarrow H_{n-1}(M)$ is trivial.
Is this statement true? An obvious counterexample to the second statement if we drop compactness is $\mathbb{R}^2-\left\{0\right\}$.
I do not think you have a proof for your statement. It certainly needs a lot more argument to go from the fact that the fundamental class of $X$ is the (algebraic) boundary of a chain, which is all the hypothesis implies, to $X$ as a subset of $M$ being the (topological) boundary of some other subset $A$.
Let $M$ be a compact, connected $n$-dimensional manifold and $X$ a closed (hence itself compact), connected codimension $1$ submanifold. Note that $M$ and $X$ are $\mathbb{F}_2$-orientable. Then, $H^0(M-X;\mathbb{F}_2)\cong\mathbb{F}_2^{\pi_0(M-X)}$ on one hand and $H^0(M-X;\mathbb{F}_2)\cong H_n(M,X;\mathbb{F}_2)$ on the other by a sufficiently general version of Poincaré duality. This fits into the pair sequence $$0\rightarrow H_n(M;\mathbb{F}_2)\rightarrow H_n(M,X;\mathbb{F}_2)\rightarrow H_{n-1}(X;\mathbb{F}_2)\rightarrow H_{n-1}(M;\mathbb{F}_2).$$ Now, $H_n(M;\mathbb{F}_2)\cong H^0(M;\mathbb{F}_2)\cong \mathbb{F}_2$ and $H_{n-1}(X;\mathbb{F}_2)\cong H^0(X;\mathbb{F}_2)\cong\mathbb{F}_2$ by two more applications of Poincaré duality. Thus, the map $H_{n-1}(X;\mathbb{F}_2)\rightarrow H_{n-1}(M;\mathbb{F}_2)$ is injective if and only if it is non-trivial if and only if $M-X$ is (path-)connected. In the other case, $M-X$ has two (path-)components.
If $M$ and $X$ are additionally orientable (this is famously the case if $M$ is simply connected), the above argument can be repeated verbatim with $\mathbb{Z}$-coefficients instead (the example of $\mathbb{RP}^2$ in $\mathbb{RP}^3$ shows that some orientability hypothesis is necessary). There also is a cohomological version of the conclusion if one starts with $H_0$ instead of $H^0$ (the fact that triviality of the induced map on $(n-1)$-th homology and cohomology are equivalent is somewhat surprising). It should be possible to generalize this result to the non-compact case, though I haven't considered the details.
Lastly, let me argue that, in the case $M-X$ is disconnected, the boundary of each of its two components is $X$. Clearly, the boundary is contained in $X$. I will now assume smoothness (I suspect that's unnecessary, but the following argument is the most natural to me). Take a tubular neighborhood $U$ of $X$ in $M$, i.e. $U$ is (diffeomorphic to) the total space of the normal bundle of $X$ in $M$. It is clear that every point of $M-X$ can be connected to a point of $U-X$ by a path in $M-X$ (take a path in $M$ to some point in $U$ and observe it has to meet $U$ before meeting $X$). Thus, $M-X$ having two path-components means that $U-X$ has at least two path-components. Furthermore, $U\setminus X$ is the total space of an $\mathbb{R}\setminus\{0\}$-bundle over $X$, whose long exact sequence is $$\pi_1(X)\rightarrow\pi_0(\mathbb{R}^{\times})\rightarrow\pi_0(U-X)\rightarrow\pi_0(X)=\{\ast\}.$$ It follows that $U-X$ has at most, hence exactly, two path-components. The sequence then implies trivial monodromy, so that the normal bundle (a line bundle) is trivial. Thus, $U\cong N\times\mathbb{R}$ as bundles over $N$. It follows that the closure of each component of $U\setminus N$ in $U$ contains $X$. The previous argument implies that each of these components is the intersection of a component of $M-X$ with $U$, hence their closures in $M$ respectively contain $X$, too.