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Consider the closed curve on the mobius strip that covers both sides of the fundamental polygon: https://thumb9.shutterstock.com/display_pic_with_logo/1663882/462955276/stock-vector-blue-moebius-strip-or-moebius-band-with-centerline-surface-with-only-one-side-and-one-boundary-462955276.jpg

Is this curve viewed as a graph (with one vertex?) non planar since it must intersect itself in the plane? (help me make this more rigorous)

How do I more rigorously prove that certain graph in the mobius strip is non planar? Do I need to use some planarity criterion like Kuratowski's?

user352102
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1 Answers1

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I think you are confusing some concepts.

A graph can be crossing free in some embeddings and not in others - think of a circle embedded as figure 8. A graph is planar if it has a crossing free embedding.

The bounding curve of your Mobius strip is just a circle. What you've noticed is that any plane projection of that strip will lead to an embedding of that circle that has crossings. But the circle is still a planar graph (when "viewed as a graph").

Ethan Bolker
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  • $K_{3,3}$ can be embedded in the mobius strip though, right? https://cornellmath.wordpress.com/2007/07/01/graph-minor-theory-part-2/. and we know that $K_{3,3}$ is non planar. – user352102 Jun 06 '17 at 18:22
  • Now I understand your question. Searching "planar mobius strip" found this answer: https://math.stackexchange.com/questions/75874/planar-graphs-on-m%C3%B6bius-strips . I will flag your nice question as a duplicate. – Ethan Bolker Jun 06 '17 at 19:21