By the Jordan curve theorem, if $C \subset S^2$ is (the image of) a simple closed curve, then $S^2 \setminus C$ has precisely two connected components. This statement admits the following "converse".
Converse of Jordan curve theorem: Suppose $M$ is a closed, connected surface such that, if $C \subset M$ is a simple closed curve, then $M \setminus C$ has precisely two connected components. Then, $M$ is homeomorphic to $S^2$.
This statement is a corollary of the classification of surfaces.
- Multiple tori have handles, and it's easy to see that deleting a loop which "grasps" a handle leaves a connected complement.
- Any non orientable surface contains a Mobius band, and the Mobius band minus its "equator" is connected.
Thus, any closed, connected surface $M$, besides $S^2$, admits a simple closed curve $C \subset M$ such that $M \setminus C$ is connected.
My question is whether this also works in higher dimensions. Does the following "converse" to the Jordan-Brouwer separation theorem hold?
Converse of Jordan-Brouwer separation theorem: Suppose $M$ is a compact, connected $n$-dimensional manifold without boundary. Suppose that, for every embedded copy $C \subset M$ of $S^{n-1}$, it occurs that $M \setminus C$ has precisely two connected components. Does it then follow that $M$ is homeomorphic to $S^n$?
This is already false in dimension 3: Take M to be any "homology 3-sphere", not homeomorphic to the 3-sphere. For instance, Poincare homology sphere will do the job.
That is true is that the topological Poincaré conjecture holds in all dimensions.