Theorem. No Continuous Embedding of the Klein Bottle into ℝ³ [local-0]

This is a standard result from algebraic topology. Using Alexander duality to compute homology, we produce a contradiction.

Proof. [local-1]

Given a Jordan curve $J$ , the set of unordered and unequal pairs of points in $J$ , denoted $S$ , is an open Möbius band. Because every unequal unordered pair of points in the circle determines a unique point in $\mathbb{R}P^2$ : Take the two tangent lines to the circle at these points and intersect them. This will map to points of the real projective plane where we remove the boundary disk, hence an open Möbius strip.

Consider a map $\phi : S \to \mathbb{R}^3$ defined by

\begin{aligned} \phi : S &\to \mathbb{R}^3 \\ \phi(a, b) &= \left( \frac{a+b}{2}, |a - b| \right) \end{aligned}

Geometrically, $\phi$ maps the ordered pair to a point encoding the midpoint of the segment $\overline{ab}$ (2D) and the length of the segment (1D).

If $\phi(a_1, b_1) = \phi(a_2, b_2)$ , then these four points form an inscribed rectangle. Hence we just need to prove that the map $\phi$ is not injective!

Consider $K = \phi(S) \cup \phi(\partial S) \cup \rho(\phi(S))$ where $\rho$ reflects the shape in the XY-plane. This is a Klein bottle by gluing two Möbius strips at their boundary.

If $\phi$ is injective, then $K$ can be embedded into $\mathbb{R}^3$ , which contradicts the theorem that no continuous embedding of the Klein bottle into $\mathbb{R}^3$ exists.