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A band (or a disk with a hole in it) can be created by gluing two edges of a square in the same direction, while a Möbius band can be created by gluing two edges of a square in opposite directions. These two spaces are not homeomorphic (right?), but proving they are different seems much more difficult. They are homotopy equivalent, both connected -- in general they seem to share the obvious properties.

What is the method of proving these two spaces are different?

I've worked my way through a good chunk of point-set topology but I don't have any foundations in algebraic topology. Does this problem require heavy algebraic topology to solve, or can I do it with point set topology?

Malachi Holden
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  • The boundary of the band is the disjoint union of two circles. The boundary of the Möbius band is a single circle. You ought to be able to prove this without any terribly advanced tool. Considering each space as a quotient of a square with some identifications is likely the easiest way to get an intuitive way into the proof. – Xander Henderson Oct 09 '18 at 04:18
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    I agree with @XanderHenderson, but I would caution that identifying the boundary through topological means is not trivial. One method is to say the boundary is the set of points which don't have a neighborhood homeomorphic to an open disk. However, it is not so easy to prove that the points you want to be the boundary actually have this property! (Homology does the trick.) – Cheerful Parsnip Oct 09 '18 at 04:20
  • This feels like a silly question, but do you mean the boundary when considering these spaces as embedded in R? – Malachi Holden Oct 09 '18 at 04:22
  • No, not the topological boundary as a subset of euclidean space. That would be the whole space! We're talking about the 1 dimensional boundary formed by the two edges of the square that aren't glued. The problem is that it's not easy to show that a homeomorphism has to map this boundary to itself. – Cheerful Parsnip Oct 09 '18 at 04:31
  • @CheerfulParsnip That is often the tradeoff---you can prove that the two spaces are not homeomorphic using elementary techniques, but the advanced techniques are more powerful tools that give you much more elegant and straight-forward proofs. The problem is that you have to learn how to use the big tools! That is, I might add, why I didn't post an answer, and only a comment. $\ddot\smile$ – Xander Henderson Oct 09 '18 at 04:32
  • @CheerfulParsnip Is this the same definition of boundary as usual (in terms of closures and interiors)? If so, in what context? As you said, in R3 the boundary would be the whole space. – Malachi Holden Oct 09 '18 at 04:38
  • @MalachiHolden: No, it's the manifold boundary. – Cheerful Parsnip Oct 09 '18 at 04:42

2 Answers2

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Here is a result with a bit of point set.

Point set fact: if $X,Y$ are hausdorff and homeomorphic, then they have homeomorphic one point compactifications.

With this, you can show that the one point compactification of the mobius band is hoeomorphic to $\mathbb RP^2$. This can be seen explicitly by considering the punctured projective plane.

The one point compactification of the cyllinder is a sphere with two points identified.

You can either stop here if you are satisfied with orientability (or embeddings into $\mathbb R^3$.)

Otherwise, note that the latter space is homotopic to $S^2 \vee S^1$.

Now, you have the job of distinguishing $S^2 \vee S^1$ and $\mathbb RP^2$ up to homotopy is not so bad. For example, what are the possible maps $\pi_1(\mathbb RP^2) \to \pi_1(S^2 \vee S^1)$? Or said differently, there is a loop of order $2$ in $\mathbb RP^2$, is there anything like that in $S^2 \vee S^1$? So on, and so forth.

Andres Mejia
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Let's talk about the "edge" of a strip rather than its boundary. I hope everyone agrees that the edge of the untwisted strip consists of two circles, and that of the Mobius strip consists of one circle, so if we can characterise the edge topologically, it will show these two spaces are not homeomorphic.

Let $X$ be one of the two spaces, and let $x$ be a edge point. Then $x$ has a neighbourhood basis consisting of sets $U$ having the property that $U\setminus\{x\}$ is simply connected. Think of a family of "D-shaped" subsets shrinking towards $x$. But if $x$ is a non-edge points, then this is not possible. Each neighbourhood basis of $x$ contains sets homeomorphic to open subsets of $\Bbb R^2$. Deleting a point from such a set will leave a non-simply connected set (consider a circle centred at the deleted point).

Angina Seng
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