Proof that the Klein bottle is homeomorphic to $T/S$ where $T$ is the torus of revolution and $S$ is the equivalence relation given by $(x, y, z) \sim (x', y', z')$ if and only if $(x, y, z) = \pm (x', y', z')$.
How do I must do it? Sorry, I am new in this.
[Math] Homeomorphism of Klein Bottle
algebraic-topologygeneral-topologyklein-bottle
Related Solutions
For any $c>0$ the relations $$(x,y)\sim (x, \,y+ k c)\quad(k\in{\mathbb Z}), \qquad (x,y)\sim \bigl(x+\ell,\,(-1)^\ell y\bigr) \quad(\ell\in{\mathbb Z})$$ define a Klein bottle $K_c$ of "length" $1$ and "width" $c$ as a quotient of the $(x,y)$-plane , and with a rectangle $[0,1]\times[0,c]$ as fundamental domain.
The identity map $\,\iota: \,{\mathbb R}^2\to{\mathbb R}^2$ realizes $K_2$ as a double cover of $K_1$: Each point $(x,y)_{\sim1}\in K_1$ has two preimages in $K_2$, namely $(x,y)_{\sim2}$ and $(x,y+1)_{\sim2}$.
Let's directly glue two Möbius strips into a Klein bottle.
- Take a Möbius strip and make its bottom wider:
- Make the rear part of the band like a half of a bottle, also the front part of the band like a tube.
- Take another Möbius strip and repeat steps 1 & 2 for the new one. Then glue new strip with the old one along their boundaries. Now we get a Klein bottle.
Therefore, a Klein bottle is homeomorphic to the glued Möbius strips.
Best Answer
Given that the equivalence relation is a reflection through the origin, you can view the quotient by cutting the torus in half along a plane (say, $x = 0$), and gluing the boundary appropriately together. Since this half-torus is just a cylinder, you should be able to reasonably either draw a picture, or relate this to the usual description of the Klein bottle for the rest.