In Lurie's "On the Classification of Topological Field Theories" (and certainly other places) he defines the category $\mathbf{Cob}(n)$ who objects are oriented $(n-1)$ manifolds. Given $M,N\in\mathrm{Cob}(n)$ a morphism $M\to N$ is an $n$-dimensional manifold $B$ equipped with an orientation preserving diffeomorphism $\partial B\simeq \overline{M}\coprod N$ where $\overline{M}$ denotes the manifold $M$ equipped with the opposite orientation.

What is the necessity of having one part of the boundary have the reverse orientation? If we define it so that the above equation is simply $M\coprod N$, do we run into problems?



Yes, it's essential. The boundary of an oriented manifolds inherits an orientation, the determination of which involves (depending on the definition of orientation), e.g. the inner normal at the boundary.

Consider two copies of $S^1$ making up the boundary of an annulus. (In general consider $M\times [0,1]$) The induced orientation (in this case representable by a nonzero one form, hence by an vector field on which this form is $1$) on the first $S^1$ will be pointing into a 'different direction' than the other one, because in both cases you are using the inner normal to determine it.

In order to have $S^1$ -- $M$ in general -- being bordant to itself (what you always want, it's supposed to be an equivalence relation) you have to revert the orientation of one of the copies.

  • $\begingroup$ Ah yes that makes sense. Basically if your arrows are going to be pointing outwards, the boundary circles have to be oriented oppositely. Haha, that's a really infantile way of describing it, but I'm pretty sure I get what you're saying. $\endgroup$ – Jonathan Beardsley Dec 31 '11 at 16:25
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    $\begingroup$ @JBeardz Why infantile? Of course a rigouros proof would need additional reasioning, but the geometric picture is what is really behind it, and in case of $S^1$ it's possible to easily visualize it. I'd claim it's absolutely to the point. $\endgroup$ – user20266 Dec 31 '11 at 17:15

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