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I have found a proposition who says: A manifold M is not orientable if it contains a Moebius band. How can I prove this?

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  • $\begingroup$ Interesting question: the converse implication! $\endgroup$ – Martín-Blas Pérez Pinilla Jan 31 '14 at 8:54
  • $\begingroup$ @Martín-BlasPérezPinilla interesting remark! Is it true? $\endgroup$ – andreasvr Jan 31 '14 at 8:58
  • $\begingroup$ More hypothesis required. Can you check the the exact wording? $\endgroup$ – Martín-Blas Pérez Pinilla Jan 31 '14 at 9:25
  • $\begingroup$ If a Manifold isn't orientable then it contains an open subset diffeomorphic to the Moebius band.. Is this the right claim? $\endgroup$ – andreasvr Jan 31 '14 at 9:31
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    $\begingroup$ Open is impossible if ${\rm dim M}>2$. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 31 '14 at 9:34
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We have:

1) The mobius strip is not orientable.

2) let $f:M\rightarrow$ N be is a diffeo. M is orientable iff N is orientable.

3) Every open subset of a orientable manifold is a orientable manifold.

Let M be the mobius strip, U subset of N and $f:U\rightarrow$ M diffeomorphism.

Suppose that N is orientable. Then by 3) we have U is a orientable manifold.

Since $f:U \rightarrow$ M is a diffeo by 2) M is orientable. Contradition.

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If $M$ were orientable, the Moebius band will inherit the orientation of $M$.

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  • $\begingroup$ Martin thanks for your answer. If i want to prove that the Moebius band isn't orientable, without using the vector normal, only with the definition of oriented atlas, what i might to do? I've seen there is a similar post but it don't answered completely this question. Thanks. $\endgroup$ – andreasvr Jan 31 '14 at 8:56
  • $\begingroup$ Thank you for your comment! $\endgroup$ – user89987 Jan 31 '14 at 9:05
  • $\begingroup$ See math.stackexchange.com/questions/15602/…. And maybe Differential Forms and Applications (do Carmo) has another argument that I can't remember. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 31 '14 at 9:06
  • $\begingroup$ I think we can´t use this argument because the mobius strip is not open. We have: If M is a orientable manifold, then every open subset of M is a orientable manifold. $\endgroup$ – Luiz Apr 23 '16 at 14:48
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The claim is incorrect. Euclidean space $M=\mathbb{R}^3$ contains the Mobius band but $M$ is orientable.

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  • $\begingroup$ Is a submanifold? $\endgroup$ – Martín-Blas Pérez Pinilla Jan 31 '14 at 9:07
  • $\begingroup$ The usual picture of the Mobius band in Euclidean space represents it as a submanifold. A parametrisation can be found in most elementary differential geometry textbooks. Of course if $M$ is 2-dimensional the claim is true. $\endgroup$ – Mikhail Katz Jan 31 '14 at 9:09
  • $\begingroup$ This is the exercice: I have the Moebius band M, the quotient of the square $[0,1]x\times [0,1]$ with the projection map $\pi \colon [0,1]\times [0,1]\to M$. After prove that the Moebius band isn't orientable conseider $M^°=\pi([0,1]\times (0,1)$. Show that $M^°$ and every manifold which contains an open set diffeomorphic to $M^°$ is not orientable. $\endgroup$ – andreasvr Jan 31 '14 at 9:13
  • $\begingroup$ So @user72694 the claim is true only in 2-dimensional manifolds? How we can generalize it for prove that a manifold isn't orientable? thanks $\endgroup$ – andreasvr Jan 31 '14 at 9:16
  • $\begingroup$ @user72694 the Moebius band is contained in ${\Bbb R}^3$ like ${\Bbb R}^2$ in ${\Bbb R}^3$? There are several definitions of submanifolds with adjectives. In any case, you are right, the simple condition of subset isn't enough. More is required. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 31 '14 at 9:22

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