Is It Always Possible to Cross a Surface Exactly Once?

Yesterday, in my physics class, the following question arose:

Is there a closed surface embedded in $\mathbb R^3$ dividing space into two connected components such that all paths from one component to the other cross the surface multiple times?

Note that one can safely replace "closed surface" with "topological sphere" if they like, since this question concerns only the local properties of the surface, not the global structure. You can see that by noting that, if a path with exactly one crossing exists, then it can be restricted to some neighborhood of the crossing and still be a path from one component to the other, crossing the surface once (that is, the path should being within the surface and end outside, and have exactly one point on the surface).

I haven't much idea how to answer such a question; I note that the Jordan-Schönflies theorem implies a negative result in two dimensions, but this doesn't extend to three dimensions.

Solution 1:

Edit: As pointed out in the comments, there is an error here. I believe it is salvageable and am not going to make too much effort to correct it.

Yes, this is always possible. Let $\Sigma$ be the surface and $U_1, U_2$ be the two sides it divides $\Bbb R^3$ into.

Invariance of domain (one of the two times we actually use the topology of the surface) implies that $\Sigma$ is nowhere-dense, so any point $x \in \Sigma$ has points in one of the $U_i$ arbitrarily close to it. In addition, $\partial U_i \subset \Sigma$. These sets are, of course, closed (since they're given by $\overline U_i \cap \Sigma$), so we see that $\partial U_1 \cup \partial U_2$ is a covering of $\Sigma$ by two closed sets; because $\Sigma$ is connected, we see that $\partial U_1 \cap \partial U_2$ is nonempty. Pick a point $x$ in this set. (Actually, it should be true that $\partial U_1 \cap \partial U_2 = \Sigma$, but I haven't been able to prove this. Thanks to Milo Brandt for suggesting this way of avoiding needing to prove it.)

Now pick a sequence of points $x_n \to x$, $x_n \in U_i$ ($i$ fixed), and demand that $d(x_n,x) < 1/n$. I claim that for large enough $n$, $U_i \cap B(x,1/n)$ is path-connected. (This is the second place we use the topology of the surface.) This is because $\Sigma \cap B(x,1/n)$ is homeomorphic to an open subset of $\Bbb R^2$ for large $n$. Pass to the one-point compactification. Here $S = \overline{\Sigma \cap B(x,1/n)}$ is homeomorphic to the one-point compactification of the aforementioned open planar surface. I believe one can calculate that it has $\check H^2(S) = \Bbb Z$, and hence by Alexander duality the complement of $S$ in $S^3$ is two path-connected open sets; but $S^3 \setminus S = B(x,1/n) \setminus \left(\Sigma \cap B(x,1/n)\right)$. Hence $B(x,1/n) \setminus \Sigma$ has two path-connected components; and since I know one of them belongs to $U_1$ and the other to $U_2$ I have the desired claim.

So, supposing $n>N$, $N$ large enough that the above applies, pick a path $f_n: [n,n+1] \to U_i \cap B(x,1/n)$, with $f_n(0) = x_n$ and $f_n(1) = x_{n+1}$. gives a map $f: [N,\infty) \to U_i$, and by construction $\lim_{x \to \infty} f(x) = x$ (taking the limit inside $\overline U_i$). Compactifying we obtain a map $f': [0,1] \to \overline U_i$ such that $f'^{-1}(\Sigma) = 1$. Doing the exact same thing on the other side, we've constructed a path with the desired property.

Don't let the fact that this worked out fool you - topological manifold usually behave terribly. As an example of something like the above with the dimensions swapped, Bing constructed a simple closed curve in $\Bbb R^3$ such that there is no disc that intersects it precisely once (and such that the boundary of the disc links with the curve). See here.

It's also worth seeing where we can weaken the assumption that $\Sigma$ is a surface. We still need some sort of invariance of domain and a way to use Alexander duality. It is a straightforward modification of the above to prove that you can go between two regions in space, separated by a finite connected 2-dimensional polyhedron, while crossing only once; and maybe one can weaken this to "2-dimensional finite CW complex". (In these cases we don't have the property that $\overline U_i \subset \Sigma$, which the original version of this answer tried to prove; but by Milo's argument we don't really need it.)