# Characteristic classes, Möbius strip, and the cylinder

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).

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.

• Beautiful argument! Jun 8, 2020 at 21:31
• @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. Jun 8, 2020 at 21:40
• 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. Jun 9, 2020 at 6:04
• How would you show that such curve disconnects the cylinder? This looks hard, a variation of Jordan curve theorem. Jun 9, 2020 at 6:26
• @JamesCameron Thanks – although I have seen it, I would not come up with your beautiful solution. 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.

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