The Klein bottle is homeomorphic to the boundary of the product of the Möbius band with a disk

Question

Can someone please give me a hint or the intuition in how to prove that the Klein bottle $\cong \partial$(Möbius strip $\times D^1 $ ) where $\cong$ means homeomorphic.

Answer 1

The Klein bottle can be defined as the quotient space $$ K=I^2 /{\sim}, \quad (x,0)\sim(x,1), \; (0,y)\sim(1,1-y), \; \forall x,y\in I $$ Let $p:I^2\to K$ be the canonical quotient map. The Möbius band is the quotient space $$ M=I^2 /{\sim}, \quad (0,y)\sim(1,1-y) $$ with a quotient map $q: I^2\to M$. We have a product map $$ q' = q\times 1_I : I^2\times I \to M\times I $$ Consider the map $f:I^2\to I^3$ $$ f(x,y) = \begin{cases} (x,1/2+4y,0) &\text{ if } y\in[0/8,1/8]\\ (x,1,4(y-1/8)) &\text{ if } y\in[1/8,3/8] \\ (x,1-4(y-3/8),1) &\text{ if } y\in[3/8,5/8] \\ (x,0,1-4(y-5/8)) &\text{ if } y\in[5/8,7/8] \\ (x,4(y-7/8),0) &\text{ if } y\in[7/8,8/8] \end{cases} $$ This $f$ induces an $\tilde f: K \to M\times I$ such that $q'f = \tilde fp$. It is easy to show that $\tilde f$ is injective, so it is a homeomorphism onto its image $q'(I\times(I\times\partial I\cup\partial I\times I))$, which is the boundary of $M\times I$.