Does Klein bottle homeomorphic to a tube whose two end circles each sewed to mobius bands?

Question

The Klein bottle is homeomorphic to the result of cutting out two distinct discs from $S^2$ and sewing in two Möbius bands as in Example 7. We shall show that the Klein bottle is in fact the result of sewing two Möbius bands along their boundaries. From there it is easy to check that the result is the same—up to homeomorphism—to two Möbius bands sewed to a thin band, as in Example 1. Of course, the thin band is homeomorphic to a sphere with two distinct open discs removed (check this).

[Taken from Page 122 of Topology: An Introduction to the Point-set and Algebraic Areas By Donald W. Kahn ]

I've understood how Klein bottle is homeomorphic to the union of two mobius bands whose boundaries are identified. But I could not understand how it could be homeomorphic to the tube whose two end circles identified with mobius bands, respectively, as written in above text.

It would appreciate, if anyone could give me a precise proof or any intuition.

Answer 1

A precise proof: the object you describe is a nonorientable (because it contains a mobius band) compact manifold without boundary. Its Euler characteristic is the same as that of the Klein bottle. So by the classification theorem for compact surfaces, they are homeomorphic.

An alternative intuitive explanation: The typical picture of a Klein bottle (in 3-space) has a plane of symmetry -- the part to the left of that plane is the reflection of the part to the right. Bandsaw the K-bottle along this plane, separate the two pieces, and notice that the "cut line" is just a circle (albeit one that intersects itself in 3-space). Make an extruded copy of this circle, and glue it between the two cut halves. You now have a half-klein, a cylinder, and a half-klein, glued up as described.

The only remaining step is to see that the half-klein is in fact a Mobius band, but that will, I hope, not be too hard for you.