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The question is given below:

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I have made a Mobius band with a paper and twisted it 3-times but I could not describe what I see it may be a 3 knot shape, could anyone give me a hint for solving that question please?

EDIT:

The answer to my previous question is here:

Topologically distinguishing Mobius Strips based on the number of half-twists

But still I do not know how to prove that the boundary of the cylinder is 2 copies of $S^1$ and that of the Mobius band is 1 copy of $S^1$, could anyone help me in doing so?

Intuition
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1 Answers1

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Internally, you cannot tell a manifold with one even number of twists from one with some other even number. They are all orientable. The ones with an odd number of twists are non-orientable, but you can't tell one odd number from another. The boundary is $2$ copies of $S^1$ for the even case and $1$ copy of $S^1$ for the odd.

Ross Millikan
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  • How can I prove rigorously that the boundary of the cylinder is 2 copies of $S^1$? – Intuition Nov 21 '18 at 19:24
  • You parameterize the side of the square from $0$ to $1$ and note that the gluing identified $0$ and $1$. – Ross Millikan Nov 21 '18 at 19:26
  • Could you provide more details please about this in your answer ? Also How can I prove rigorously that the boundary of the mobius band is one copy of $S^1$? – Intuition Nov 21 '18 at 19:31
  • And I think that the answer will be divided according to if n is even or odd, so I think if I twist odd number of times the result wil lbe isomorphic to the mobius band and if I twisted even number of times the result will be isomorphic to the cylinder ..... am I correct? – Intuition Nov 21 '18 at 19:37
  • Yes, that is correct. – Ross Millikan Nov 21 '18 at 19:38
  • but what isomorphism should I use? – Intuition Nov 21 '18 at 19:39
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    You can just use the isomorphism that maps each point on the untwisted square to the corresponding point on the twisted square. The neighborhood of each point is the same as long as the parity of the number of twists is the same. The only difference is the embedding in $\Bbb R^3$, which is not a property of the surface. If the parity is different this isomorphism does not work for points on the edge. – Ross Millikan Nov 21 '18 at 19:46
  • The answer to some of my questions to you is here https://math.stackexchange.com/questions/1379300/topologically-distinguishing-mobius-strips-based-on-the-number-of-half-twists – Intuition Nov 22 '18 at 19:21
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    @hopefully You can find out on origami from a web search. All you need is paper and scissors. I hope you can find these. – Ronnie Brown Nov 25 '18 at 22:27