# A naïve attempt at an elementary proof of Jordan-Brouwer separation theorem

I've been trying to construct an elementary proof of the Jordan-Brouwer separation theorem, utilizing the concept of a topological cone. I understand that it is convenient to prove this theorem using the tools of algebraic topology, but I've yet to have a course in it; it is also obligatory to ask questions of this kind in order to test your own knowledge of the subject.

The idea behind my attempt is extremely simple, even naïve (that's why I think that I must be in the wrong). We know that there are exactly two components of $\mathbb R^n \setminus \mathbb S^{n-1}$, namely $\mathbb R^n\setminus \mathbb B^n$ and $B(\mathbf 0,1)$ (open unit ball in $\mathbb R^n$). If we are able to construct (or otherwise prove the existence of) a homeomorphism $F\colon \mathbb R^n\rightarrow \mathbb R^n$ such that $F|_{\mathbb S^{n-1}}\equiv f$ for a given embedding $f\colon\mathbb S^{n-1}\rightarrow \mathbb R^n$, then the conclusion of Jordan-Brouwer theorem would immediately follow, as homeomorphisms preserve the cardinality of components (which is easy to see using the fact that the connectedness is preserved under continuous maps and homeomorphisms are injective - this might help, but injectivity is never used, and should be to conclude that the cardinality is preserved).

I went on to imagine the problem for $n=2$, which gave me the idea of using topological cones - actually, a slight modification of them. Define $X=\mathbb S^{n-1}\mathbb\times (-\infty, 1]$ and $Y=f(\mathbb S^{n-1})\mathbb\times (-\infty, 1]$. It can straightforwardly be checked that the homeomorphism $\overline f\colon X\rightarrow Y$ given by $\overline f =(f,\mathrm{id}_{(-\infty,1]})$ induces a homeomorphism $$X/\left(\mathbb S^{n-1}\times \{1\}\right)\cong Y/\left(\mathbb S^{n-1}\times \{1\}\right).$$

Meanwhile, let $\tilde X$ be a geometric cone above $\mathbb S^{n-1}$, i.e. $$\tilde X=\left\{(1-t)\mathbf x +t\mathbf v_X\; ;\; t\in(-\infty,1], \mathbf x\in\mathbb S^{n-1}\times \{0\}\right\}$$

where $\mathbf v_X=(\mathbf 0, 1)$ is the peak of the geometric cone. It can be proved that the map $g_X\colon X\rightarrow \tilde X$ given by $g_X(\mathbf x,t)=(1-t)\mathbf x +t\mathbf v_X$ induces a homeomorphism

$$X/\left(\mathbb S^{n-1}\times \{1\}\right)\cong \tilde X$$ and similarly $$Y/\left(\mathbb S^{n-1}\times \{1\}\right)\cong \tilde Y$$ with the peak $\mathbf v_Y$ of the geometric cone now being any point with $(n+1)$-th coordinate equal to 1.

Finally, the geometric cones $\tilde X$ and $\tilde Y$ are homeomorphic to $\mathbb R^n$ -- for example, the projection map of $\tilde X$ onto the first $n$ coordinates gives us a homeomorphism $H_{\tilde X}\colon \tilde X \rightarrow \mathbb R^n$. If we choose $\mathbb v_Y$ to be somewhere "inside" the sphere $f(\mathbb S^{n-1})$, we can also use this same projection as a homeomorphism $H_{\tilde Y}\colon\tilde Y \rightarrow \mathbb R^n$. (Even better, we could choose this point in the beginning of the proof to be the origin without loss of generality and demand $f(\mathbb S^{n-1})$ to enclose it -- omiting this demand probably only gives rise to some trivial complications with translations)

The wanted homeomorphism is then given by the composition $$\mathbb R^n\rightarrow \tilde X \rightarrow X/\left(\mathbb S^{n-1}\times \{1\}\right) \rightarrow Y/\left(\mathbb S^{n-1}\times \{1\}\right) \rightarrow \tilde Y\rightarrow \mathbb R^n.$$

If anyone cares enough to read it, I would be happy to hear where I got it wrong.

I suspected that I was too hasty about the maps $g_X$ and $g_Y$ being quotient maps (if they aren't, the proof breaks down because we can't guarantee that it induces a homeomorphism) -- but one can prove that the map is closed; for $W\subset X$ closed, let $W=\cup_{n\leq 0} W_n$ where $W_n=\mathbb S^{n-1}\times [n,n+1]$. Then $g_X(W)=\cup_{n\leq 0}g_X(W_n)$ is a countable union of locally finite family $\left\{g_X(W_n)\; ; \; n\leq 0\right\}$ of closed sets. They are indeed closed, because $g_X|_{W_n}$ maps from a compact space into a Hausdorff space.

To make sure that this proof is correct, one should consider if this holds for pathological topological spheres like the Alexander horned sphere. I suspect that it doesn't precisely in cases like those, where my intuition couldn't guide me.

• The short answer is that your suspect in the end is well-founded, because you are trying to prove Jordan-Schoenflies theorem for $n\ge3$. This is known to be false, and Alexander's horned sphere is in fact a counterexample (specifically, the unbounded component of its complement is not simply connected, and thus the embedding $f$ does not extend to a homeomorphism of $\Bbb R^3$). – Saucy O'Path Jun 25 '18 at 23:47
• I must say that I've never proved that the components of the complement of the horned sphere aren't homeomorphic to the components of the complement of the sphere. If we assume this for a fact, than it implies that I must've made an error in the proof. If you happen to see it, please let me know. The question is then whether or not the proof works for $n=2$ and if not, if it includes the same pathological step as for $n>2.$ – suhogrozdje Jun 26 '18 at 9:36
• Also, when trying to visualize the situation for $n=3$, it helps to use time as the 4th coordinate (in the proof, $t$ is used suggestively). The infinite cone can then be visualized as a 3-sphere that is shrinking as $t$ gets bigger (proportionally) and finally becomes a point at $t=1$. – suhogrozdje Jun 26 '18 at 10:16

What you are trying to do amounts to proving the Schoenflies Theorem for all topological embeddings $f : \mathbb S^{n-1} \to \mathbb R^n$. If, as you say...

... we are able to construct (or otherwise prove the existence of) a homeomorphism $F\colon \mathbb R^n\rightarrow \mathbb R^n$ such that $F|_{\mathbb S^{n-1}}\equiv f$ for a given embedding $f\colon\mathbb S^{n-1}\rightarrow \mathbb R^n$...

... then you have proved the Schoenflies theorem, as @SaucyO'Path says.

The place where your proof breaks down is that your projection map does not give the homeomorphism $\tilde Y \to \mathbb R^n$ that you assert it to give, because there is no guarantee that this map is one-to-one. In fact, this projection map can fail to be one-to-one even in cases where we already know that the Schönflies Theorem is true.

For example, let's suppose, as you suggest, that $f(\mathbb S^{n-1})$ encloses the origin. Here's the problem: the origin might not be a star point for $f(\mathbb S^{n-1})$. In other words, there might be a vector $\vec u$ in $\mathbb R^n$ such that the ray $R(\vec u) = \{t \vec u \mid t \in [0,\infty)\}$ intersects $f(\mathbb S^{n-1})$ multiple times. If that is the case, then the projection map $\tilde Y \mapsto \mathbb R^n$ is not one-to-one.

And it does no good to assume that there is a star point, because in many examples there isn't one. Tracing the outline of a fattened letter S, for example, gives a Jordan curve in the plane with no star point, so your proof breaks down in this case, even though the Schoenflies Theorem is true in this case. Certainly the Alexander horned sphere has no star point, and that's a good thing because, as @SaucyO'Path points out, the Schönflies Theorem is false in that case.

• Thanks. I see now how the projection fails, even for simple cases for $n=2$. The geometric cone is, however, still homeomorphic to $\mathbb R^n$ (this can still be intuitively seen for $n=2$, even for the mentioned outlines of fattened letters), but the homeomorphism that we need for this job must preserve $f(\mathbb S^{n-1})\times\{0\}$ (i.e. is identity there), which is very difficult (even impossible as seen for horned sphere) to construct -- but it may be that for locally flat $f(S^{n-1})$, or at least in $n=2$, one could construct it. For now I'll wait for my course in algebraic topology. – suhogrozdje Jun 26 '18 at 13:23