# 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 Commented Jun 11, 2011 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. Commented Jun 11, 2011 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. Commented Jun 11, 2011 at 15:16
• James, IMO this question is a great math.stackexchange.com question. Commented Jun 11, 2011 at 15:35
• Ryan: I don't think transversality and Guillemin-Pollack suffice. The question is about topological embeddings. Commented Jun 11, 2011 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. Commented Jun 11, 2011 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. Commented Jun 11, 2011 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. Commented Jun 11, 2011 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 $2$.