Question: Determine whether the Klein bottle is homeomorphic to the union of two Mobius bands attached along their boundary circles.

The Klein bottle is the quotient space $$ K=I^2 /{\sim}, \quad (x,0)\sim(x,1), \; (0,y)\sim(1,1-y), \; \forall x,y\in I $$ The Möbius band is the quotient space $$ M=I^2 /{\sim}, \quad (0,y)\sim(1,1-y) $$

What would be a good way to approach this question? I have not had any success constructing a map between spaces

Edit: I remember that homeomorphism must preserve orientability. So this could be used to disprove a homeomorphism.

The mobius band is non-orientable, as is the klein bottle. What I am not sure about is if we take the union of two non-orientable Mobius bands and attach their boundary circles, do we still get a non-orientable surface.

I think the gluing the boundary step may switch the orientabilty. So we have an orientable surface which therefore cannot be homeomorphic to the non-orientable Klein bottle.

I am unsure how to prove this in a formal way (with equations and notation etc)

Would appreciate your help

  • $\begingroup$ You cannot embed a non-orientable surface into an orientable one for obvious reasons. So the strips stay non-orientable. Non-orientability doesn't cancel, it's more like cancer. $\endgroup$ – user326572 Apr 11 '16 at 23:47
  • $\begingroup$ The way to prove it in a formal way is to first produce a good intuitive picture, such as in the answer of @user326572, and then describe the features of that picture precisely in coordinates. $\endgroup$ – Lee Mosher Apr 12 '16 at 1:57
  • $\begingroup$ @LeeMosher thanks. I am not sure how to construct these pictures $\endgroup$ – amiz9 Apr 12 '16 at 9:54
  • $\begingroup$ @amiz9 What is there to construct? The square is $[0,1]^2$ and one of the strips is $\{(x,y-x/2)|x\in[0,1],y\in[1/2,1]\}$. If you don't understand how subspaces and quotient spaces interact, I suggest you work it out in detail yourself. $\endgroup$ – user326572 Apr 12 '16 at 16:01

This is a diagram of the Klein bottle, note that the diagonal lines divide it into 2 Möbius strips sharing a boundary:

So the answer is yes.

  • $\begingroup$ please can you explain how you arrived at this diagram, as I would not be able to do this myself? $\endgroup$ – amiz9 Apr 12 '16 at 9:53
  • 1
    $\begingroup$ @amiz9 At first I tried dividing it horizontally by a line in the middle. But that doesn't quite work since the left side of the upper strip gets identified with the right side of the bottom strip. This is a modification of that idea. $\endgroup$ – user326572 Apr 12 '16 at 10:34

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