I suspect I will show my ignorance here, but this 'theorem' I would consider to be intuitively sensible, but I cannot find anything similar by looking through a few books or on the web. If would seem true in principal, but it probably needs some modification to how I have formulated it below. I was wondering if anyone know where I might find a proof of such a thing.


Let $M$ be a connected manifold of dimension $>n$ . Let $f:\mathbb{S}^{n} \rightarrow M$ be a map that is a homeomorphism onto its image $C=f(\mathbb{S}^{n})$ . Then 


(1)- if $M$ has dimension $n+1$, then $M-C$ is the disjoint union of two open sets $A,B$ , each of which is path connected. 


(2)- if $M$ has dimension $\geq n+2$ , then $M-C$ is path connected. 


Notice that in (1) the usual Jordan curve theorem would say that $A$ is bounded and $B$ is unbounded, but this wouldn't seem to hold in the generalised case.


My motivation for the above is that it would give a nice way to show that a $S^2$ is not homeomorphic to the disk $D^3$.



Jordan Curve Theorem for Manifolds
As pointed out by Francesco, part (1) is false in general; however, it is true when the first Betti number of $M$ is 0. Part (2) is correct. All this follows easily from Alexander duality, stating that if $d$ is the dimension of $M$, we have $\mathrm H\_{d-1}(S^n) \simeq \mathrm H^{1}(M, M \smallsetminus S^n)$.


Of course, using this to show that $S^2$ is not isomorphic to $D^3$ is a big overkill.


