Characteristic classes, Möbius strip, and the cylinder

Question

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).

Answer 1

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.

Answer 2

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.

[Commenters have asked why the one-point compactifications are as I propose. To see this, argue that the spaces described are compact and that removing a single point gives the cylinder or the Möbius strip as appropriate. Thus:

For the cylinder, let $N$ and $S$ be the north and south poles in $S^2$ and remove $[\{N, S\}]$ from the identification space $S^2/R$ where $R$ is the equivalence relation that identifies $N$ and $S$; the result is homeomorphic to $S^2 \setminus \{N, S\}$, which is homeomorphic to the open cylinder.

For the Möbius strip. View the real projective plane $\mathrm{RP}^2$ as the unit disc $D^2 \subseteq \Bbb{C}$ with the boundary arc from $i$ to $-i$ on the left identified with the boundary arc from $-i$ to $i$ on the right. Now remove the origin and note that the resulting space is homeomorphic to what you would get if you removed a small closed disc centred at the origin. Now cut what you have in half along the imaginary axis and glue the two boundary arcs together. What you now have is homeomorphic to the square $[0, 1] \times (0, 1)$ with the left and right edges identified north-south to south-north, i.e., an open Möbius strip.]

Answer 3

The Mobius strip is homotopy equivalent to the cylinder, as both spaces deformation retract onto a circle.

We prove the two are not homeomorphic using local homology; you are right ordinary homology cannot distinguish the two as it is homotopy invariant. The local homology of a point on the Mobius strip is $\cong\Bbb Z$ for all points in the "interior" and is trivial (zero) exactly when the point lies on the boundary circle. No surprises there; this holds for all manifolds with boundary. Computing the local homology of the cylinder $S^1\times I$ we find it "is" $\Bbb Z$ on $S^1\times(0,1)$ and is trivial on the boundary $S^1\times\partial I$; again, no surprises.

However the key observation is that the homological boundary (points with trivial local homology) is path connected (a circle, even) for the Mobius strip whereas it is disconnected for the cylinder; homological boundaries are homeomorphism invariant, so are done. Or use any one of several methods to prove $S^1\times\{0,1\}$ and $S^1$ are not homeomorphic.