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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?


marked as duplicate by Ethan Bolker, Namaste, Daniel W. Farlow, C. Falcon, dantopa Jun 7 '17 at 0:57

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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").


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