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I was trying to prove that if $S$ is a non-orientable surface embedded in orientable 3-manifold $M$. Then $S$ does not seperate the normal bundle $NS$ and $M$ as well.

It's easy to check that if $S$ is orientable(for example the cylinder in $\Bbb{R}^3$), $S$ may seperate the normal bundle, I have no idea why the seperating property is related to the orientability of the normal bundle ?

Ethan Bolker
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yi li
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  • a surface is non orientable iff it contains a Mobius band, therefore if we move along the middle line the Mobius band you will go from "inside " to the "outside" of $NS \setminus S$ without crossing $S$, however that's not a rigorous proof. – yi li Jun 04 '22 at 16:48
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    You know that $NS$ is non-orientable (since $TS$ is non-orientable and $TM|_S$ is orientable), and hence there is a closed loop in $S$ with an orientation-reversing section of $NS$ on it. The claim about $M$ follows from the tubular neighborhood theorem. – Ted Shifrin Jun 04 '22 at 19:54
  • The same link as at your other question may help here, as well. – Andrew D. Hwang Jun 04 '22 at 20:46
  • hi @Ted Shifting can you elaborate more on the first claim why orientation reversing section implies it does not seperate the normal bundle? – yi li Jun 05 '22 at 03:07
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    Because you have a path in the complement of the zero section from one element of a fiber to its negative. – Ted Shifrin Jun 05 '22 at 03:12

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