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I was reminded by question 77569 about something that has bothered me off and on for a while now. Consider the model of the projective plane given by the last diagram on page 37 of these notes (it's just a lens-shape in the plane, with opposite sides identified in a "chasing" way):

Möbius strip

Now if you remove an open disk from it, what's left, call it $M$, is homeomorphic to a Möbius strip. Unlike the projective plane, the Möbius strip can be embedded in ${\bf R}^3$ with no self-intersections. So it's natural to ask whether you can use the model $M$ to construct a physical Möbius strip. That is, does the presence of the hole make it possible to carry out in 3-space the indicated identification of edges, without tearing or self-intersection, provided the material being used is sufficiently elastic.

Experiments with paper have not been encouraging. Is there some obstruction to making a Möbius strip from $M$?

Another plane model for a Möbius strip is a triangle with 2 edges identified "chasing" and one edge unidentified. The 2nd figure on page 40 of those notes will do as an illustration. Again, I haven't been able to carry out the identification with paper in 3-space; is it possible? or is there some obstruction?

I should add that of course I know that one can make a Möbius strip from the usual model, a rectangle with one pair of opposite edges identified chasing. And I know that these other models are homeomorphic to the Möbius strip. The questions are about using the other models to make a Möbius strip.

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    As for your second example of a triangle, that is very close to the usual rectangle model. If you have a trapezoid with two opposite sides of length $1$ and $\epsilon$ unidentified, and the other two opposite sides identified with the "chasing" orientation, then the the triangle model is the limit as $\epsilon\to 0$. The usual strip model happens when $\epsilon=1$.2011-11-01
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    That seems to be a strange direction to go in -- since the model in the notes is not an actual embedding into $\mathbb R^3$ (they say "we’re not too worried about where these surfaces live, we won’t pursue that point"), you couldn't construct it to start with and cut out a disk from it. I suppose you could remove one of the "eardrums" from a [Boy's surface](http://en.wikipedia.org/wiki/Boy%27s_surface) and get a partially self-intersecting Möbius strip, though.2011-11-01
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    @Jim, yes, and now I am reminded that I think I saw a discussion somewhere of how short the unidentified sides of a rectangle could be (relative to the identified pair) and still permit making a Mobius strip. I'll have to see if I can find it. But are you suggesting your comment answers my question?2011-11-01
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    @Henning, we have a failure of commutativity here. I didn't mean, construct the projective plane, then cut out a disk; I meant, cut out a disk (from the diagram), then construct a Mobius strip (if possible).2011-11-01
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    @t.b., thanks for the edit.2011-11-01
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    @GerryMyerson: No I'm just observing. I also tried making a model of the first one (lens minus disk) out of paper and tape, and I couldn't get it to glue up.2011-11-01
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    @GerryMyerson: Have you seen Stephen Barr's book "Experiments in topology." I can view it on Google books. Chapter 3 is about constructing the "shortest Möbius strip."2011-11-01
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    @Jim, yes, I realized yesterday that that was where I had seen the discussion I referred to in my earlier comment. I believe Barr completely solves the problem of the triangle model. If I get some time today to read it and digest it I may post a summary as an answer. Together with Henning's answer to the other part of my question (once I've understood it), that should settle matters.2011-11-01

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