Moebius band not homeomorphic to Cylinder.

Question

I have been trying to think of a rather basic way of proving this, but it seems a bit elusive. In the case with boundary (looking at them as quotients in $ [0,1] \times [0,1] $), they can be distinguished from the connectedness of the boundary (thanks Stefan), but I'm interested in the open case (looking at them as quotients in $ [0,1] \times (0,1) $, i.e. as manifolds without boundary).

If there is not such basic way to do this, it would be interesting to read about some advanced methods you guys know about.

Answer 1

Here is the most elementary argument I can come up with. Let $C$ denote the open cylinder and $M$ denote the open Moebius band. Note that there exists a compact subset $K\subset C$ such that for any compact $K'$ such that $K\subseteq K'\subset C$, $C\setminus K'$ is disconnected (for instance, you can take $K$ to be the "equator" of $C$). On the other hand, I claim that no compact subset $K\subset M$ has this property. Indeed, let $K\subset M$ be compact; identifying $M$ as a quotient of $[0,1]\times(0,1)$ in the standard way, there is then some $\epsilon>0$ such that $K$ is contained in the image of $[0,1]\times[\epsilon,1-\epsilon]$ under the quotient map (for instance, because the images of sets of the form $[0, 1] \times (\epsilon, 1-\epsilon)$ are open and cover $K$). Let $K'$ be this image; then $K'$ is compact, $K\subseteq K'\subset M$, and $M\setminus K'$ is connected.

Answer 2

Let's be clear we have $X = [0,1]\times (0,1)$ with two different identification maps:

• $f:(0,x) \in X \mapsto (1,x) \in X$ this gives a cylinder $Y$
• $f:(0,x) \in X \mapsto (1,1-x) \in X$ this gives a Möbius band $Z$

Even though $Y$ and $Z$ are open, we can take their closure and $\partial \overline{Y} = S^1 \cup S^1$ while $\partial \overline{Z} = S^1 $. See Chapter 4 of the notes of Allen Hatcher on Point Set Topology for a discussion on quotient spaces.

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#1 Consider the subspace $X_0= (0,1) \times [\frac{1}{4}, \frac{3}{4}]$ common to both spaces. Let $f:Y\to Z$ be a homeomorphism and the closure $\overline{X_0}\subset Y$ is a cylinder while $\overline{f(X_0)} \subset Z$ is a Mobius band.

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#2 Cylinder is orientable. Mobius band isn't.

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#3 Remove the meridian circle $ [0,1] \times \frac{1}{2} $ from a cylinder - it is not connected. Remove a meridian circle from a Möbius band, it remains connected. Jordan curve is not necessary, merely that if a continuous map $\phi:[0,1] \to \mathbb{R}$ has positive and negative values, then $\phi(x) = 0$ somewhere.

Answer 3

An open-ended cylinder has the Jordan curve property - remove any subset homemorphic with a circle and you get a space which is not path connected. There is a circle you can remove from any Möbius strip which leaves the space connected.

Of course, the Jordan curve theorem is notoriously tricky and technical to prove, but at least it is intuitive what is going on.

Answer 4

One-point compactification of an open cylinder is homotopy equivalent to $S^2\vee S^1$; and one-point compactification of Moebius band is just $\mathbb RP^2$. One-point compactifications of homeomorphic spaces have to be homeomorphic too, but two obtained spaces have different $\pi_1$.

Answer 5

The open Moebius band contains the closed Moebius band as a subspace, the cyclinder (or puncutred plane) does not. Though a rigorous justification of the latter may be a bit tricky.

Answer 6

Okay, I'm throwing my hat in the ring here. The cylinder is orientable while the Möbius band is not. It's easy to see that the Möbius band isn't, since you can find an explicit orientation reversing path. To show that the cylinder is orientable is slightly trickier, but is still easy since one has a method to define a consistent orientation across the entire cylinder. For example, thinking of the cylinder as embedded in space, take an outward pointing normal and use the right hand rule to induce an orientation of the cylinder.

Answer 7

As with most questions where we want to keep from getting into deep arguements, it is a questions of how deep can we go? Here is what I came up with for a intuitive arguement:

First, we notice that we can embed (Put in so it doesn't really change) the cylinder $C$ into $\mathbb{R}^2$. Just think of sitting the cylinder on the table, putting your hands inside and pushing the sides down and out away from the center.

If the cylinder and Möbius strip were homeomoprhic, we could also embed it in the plane. But we cannot, and here is why:

Any map from the Möbius strip into the plane will not be injective, which means that (at least) two points will be sent to the same place, no matter what. Here is a Möbius strip, with the two sides identified.

And here is the same Möbius strip with two circles that only intersect once in the Möbius strip. (This is the important part.)

If we wanted to embed the Möbius strip, we would also embed the circles. So lets put the red circle in first, standardly, and the try to put in the blue circle.

We can see that blue circle should be on both sides of the red circle, but for that to happen, they must intersect each other twice! So, we can't embed the Möbius strip into the plane, hence, the cylinder and the Möbius strip are not homeomorphic.