The Möbius strip is obtained from a closed band by fixing one end, taking a half twist of the other end and gluing the ends together.

I wonder what happens, if one doesn't just take a half twist but multiple (half) twists of the one end.


  1. Is the resulting space homeomorphic to the classical Möbius strip ?
  2. If they are not homeomorphic, are they homotopy equivalent ?
  • 1
    $\begingroup$ 1. No. Two half-twists makes the resulting band orientable, thus not homeomorphic to the classical one. 2. Yes, because they are all homotopically equivalent to a circle. $\endgroup$
    – Aloizio Macedo
    Sep 24 '16 at 17:29

Any space constructed in this fashion deformation retracts to the core circle, so they are all homotopy equivalent to $S^1$.

I think the study of homeomorphism classes of these objects is an interesting exercise, so I'll leave it as one for you with the following hints as a guide:

  1. Homeomorphism doesn't depend on any embeddings; it's an intrinsic property. So "number of twists," which is definitely an extrinsic property (in this case, it comes from the relationship between the space and its immersion in $\mathbb{R}^3$) isn't going to be as helpful as you might want in classifying up to homeomorphism.

  2. Instead, it may be helpful to think of these spaces intrinsically, in particular as quotients of the square $[0,1]\times [0,1]$. Do this by thinking about how you would "glue" the square to produce these multiply-twisted strips.

  3. Notice that the square does not "know" how many times you've twisted it. All it knows is how you glue it. Can you formalize this into a proof about homeomorphisms between differently-twisted strips?

  4. Based on the result of (3), can you classify all such twisted bands into homeomorphism types?


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