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Let $M$ be a compact three-dimensional manifold with corners, which is a cobordism of the two-dimensional annulus. In particular, the codimension one boundary of $M$ consists of two copies of the annulus and two copies of the cylinder $[0,1]\times S^1$. Additionally, $M$ has a non-vanishing vector field which is

  • tangent to the boundary components $[0,1] \times S^1$,
  • is transverse and incoming into one of the boundary annulus,
  • transverse and outgoing from the other boundary annulus.

Is it possible to conclude that $M$ is the product cobordism?

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  • $\begingroup$ I would guess that it's more likely you can use the ideas of the Poincare-Hopf theorem to show that every cobordism has such a vector field. $\endgroup$
    – mme
    Commented Nov 11, 2018 at 16:41
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    $\begingroup$ In contact geometry, it is easy to see that any oriented cobordism $M$ has such a vector field. Any such $M$ has a cooriented contact structure which is `standard' near the boundary, so that the Reeb field satisfies the properties desired. $\endgroup$
    – KSackel
    Commented Nov 11, 2018 at 19:25
  • $\begingroup$ I get what your idea was, I think: you wanted to flow from the "incoming" boundary face to the "outgoing" boundary face. But there's no reason to believe that the integral curves actually make it to the other side. If they do, then sure, you have a product cobordism. $\endgroup$
    – mme
    Commented Nov 11, 2018 at 19:40

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