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Let $Cob^{3}$ denote the cobordism category of $1$ dimensional manifolds i.e the objects are finite disjoint union of circles and morphisms are represented by surfaces.

Is it possible to treat the circles as ordered circles? To be specific, if a surface $S$ represents a morphism from $k$ circles to $m$ circles, starting with an order of $k$ circles can we induce an order on the $m$ circles so that it is well defined i.e respects the composition order.

As we know any $1+1$ dimensional TQFT corresponds to a Frobenius extension over a commutative ring. Is it true for the ordered cobordism category as well?

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  • $\begingroup$ It's not really clear to me what you want this order to satisfy, but you certainly have morphisms which shuffle the circles around. For example, starting with 4 circles which you ordered somehow, I can have a "pair of pants" morphism to 3 circles, which pairs the first and the last circle (and does cylinders on the remaining two circles). Also, there are morphisms from the empty set to any collection of circles, certainly you can't induce a well-defined ordering on the target out of nothing? $\endgroup$ Mar 3 '21 at 16:20
  • $\begingroup$ So for the example you mentioned where first and the fourth circles are paired to a single circle and does nothing to the second and the third circles, can you order the ending 3 circles somehow out of the ordering in your starting 4 circles... for example you may choose to order to be 1<2<3 where 1=the circle merged from 1st and 4th circles, 2= second circle as it is the identity, and 3= third circle... It is kind of ad hoc!! Is it possible to make it well defined? $\endgroup$ Mar 3 '21 at 17:28
  • $\begingroup$ Yes, sounds very ad hoc. I'm afraid I don't know what definition you have in mind (especially for the second example, where all possible orderings look equally compatible) $\endgroup$ Mar 3 '21 at 17:32

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