# Jordan Curve Theorem for Manifolds

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 $D^3$.

• If you take as $C$ a meridian of a $2$-torus $M\subset \mathbb{R}^3$, it seems to me that $M−C$ is connected – Francesco Polizzi Jun 11 '11 at 13:37
• For (1) I think you need $f$ to act trivially on the nth (co)homology group. Think about embedding a circle in a torus. I need not break it into two components. – George Lowther Jun 11 '11 at 13:40
• The general theorem of the form of (1) is called the Jordan-Brouwer Separation theorem. See en.wikipedia.org/wiki/Jordan_curve_theorem also the Differential Topology text of Guillemin and Pollack. General theorems of the type (2) follow directly from elementary transversality theorems, see also Guillemin and Pollack. – Ryan Budney Jun 11 '11 at 15:16
• James, IMO this question is a great math.stackexchange.com question. – Ryan Budney Jun 11 '11 at 15:35
• Ryan: I don't think transversality and Guillemin-Pollack suffice. The question is about topological embeddings. – Sergey Melikhov Jun 11 '11 at 15:44

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.
• Yes it does seem like huge overkill. $D^3$ is contractible for one thing. – George Lowther Jun 11 '11 at 14:12
• $H_1(M; \mathbb Z_2)$ being trivial suffices rather than the Betti number condition. Weaker still, the image of the fundamental class of $S^n$ under $f$ in $H_n (M;\mathbb Z)$ being trivial would give you the "if and only if" statement. – Ryan Budney Jun 11 '11 at 15:21
• Small note, the Betti number condition answers the question but if you replace the sphere $S^n$ from the question with an arbitrary $n$-dimensional closed connected submanifold, the Betti number does not suffice and it becomes a mod-2 homology issue. If the manifold is non-orientable then you need that the image of its fundamental class in $H_n(M;\mathbb Z_2)$ is trivial. – Ryan Budney Jun 11 '11 at 15:34
Alexander's horned sphere (Wikipedia) shows that even when the first part of your conjecture (1) holds, you cannot expect the second part to. The horned sphere is a continuous embedding $\mathbb S^2 \to \mathbb S^3$ that does separate $\mathbb S^3$ into two pieces, one of which is homeomorphic to the open ball. But the other is not simply connected: Schoenflies' half of the Jordan theorem fails in higher dimensions. See Schoenflies problem (Wikipedia); in particular, if you add a "local flatness" condition that the map $\mathbb S^2 \to \mathbb S^3$ extend to a thickened $\mathbb S^2$, then you do get the desired result for any value of $2$.