I have been thinking how to distinguish the (open) Möbius strip from a(n open) cylinder.

What does not work

  • Standard invariants from general topology, as connectedness or compactness,
  • Invariants that depend on the mere homotopy type of the space, as homotopy groups and (co)homology.

What does work

The only invariant I can think of is orientability. My question is therefore:

Is there any other invariant that can be used to show that the Möbius strip and a cylinder are not homeomorphic?

If we regard both spaces as line bundles over the circle, we can show with the Stiefel–Whitney classes that they are non-isomorphic vector bundles, but this seems to be a weaker statement that being non-homeomorphic. (And moreover Stiefel–Whitney can be treated as a reformulation of the orientability argument).


2 Answers 2


You can use the fact that if you cut the open Möbius strip around center the resulting space is connected.

Any homeomorphism from the Möbius strip to a cylinder will induce an isomorphism on fundamental groups, so if a homeomorphism existed it must send the center of the Möbius strip to a curve homotopic to a circle going around the cylinder, and removing anything homotopic to such a curve from the cylinder disconnects it.

  • $\begingroup$ Beautiful argument! $\endgroup$ Jun 8, 2020 at 21:31
  • $\begingroup$ @PawełCzyż If you've never done it I highly endorse making a Möbius strip and cutting it, or getting a non mathematician to do so! Lots of cool stuff can happen. $\endgroup$ Jun 8, 2020 at 21:40
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    $\begingroup$ This needs to be stated a little carefully as one needs to define the "centre" of the Moebius strip. A simple closed path on the strip can disconnect it. For example take a path that is at 1/3 of the width from the edge. Perhaps define the "centre" as a loop that generates the homotopy group. $\endgroup$
    – Kapil
    Jun 9, 2020 at 6:04
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    $\begingroup$ How would you show that such curve disconnects the cylinder? This looks hard, a variation of Jordan curve theorem. $\endgroup$
    – freakish
    Jun 9, 2020 at 6:26
  • $\begingroup$ @JamesCameron Thanks – although I have seen it, I would not come up with your beautiful solution. $\endgroup$ Jun 9, 2020 at 9:50

As an alternative to James's answer, you can look at the one-point compactifications. For the cylinder, you get a space homeomorphic to a sphere with two points identified, which has $\Bbb{Z}$ for its fundamental group. For the Möbius strip you get a space homeomorphic to the real projective plane, which has $\Bbb{Z}/2\Bbb{Z}$ for its fundamental group.

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    $\begingroup$ Great solution! I wish there were a possibility to choose more than one "Accepted answer"... $\endgroup$ Jun 8, 2020 at 21:33
  • $\begingroup$ No worries! If there are several possible answers, someone has to get there first. $\endgroup$
    – Rob Arthan
    Jun 8, 2020 at 21:34

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