# Fill in the blanks: “1Cob is the free ____ category on a ____”

This is probably straightforward, but I'm having trouble writing down a precise statement. "Everyone knows" that the cobordism category $\text{2Cob}$ (all manifolds compact and oriented) is the free symmetric monoidal category on a commutative Frobenius object. What is the analogous statement for $\text{1Cob}$?

It looks something like the free symmetric monoidal category on an object with a (left and right) dual, but I'm not sure if I'm interpreting the orientation on points correctly.

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How is it symmetric monoidal? i don't think it there is a cobordism from 2 points to one point that is not a "projection". – Sean Tilson Feb 6 '11 at 4:53
Sean, it's symmetric monoidal under disjoint union. The cobordism you're talking about is part of "Frobenius" for the 2cob story, not "symmetric monoidal." – Gabriel C. Drummond-Cole Feb 6 '11 at 5:04
thanks GCDC! (blank space) – Sean Tilson Feb 6 '11 at 5:18
I would have said "the free symmetric monoidal 1-category on a dualizable object", but you seem to think that this isn't quite right? How's your site-specific google-fu? Have you tried searching the TWF archives? See also ncatlab.org/nlab/show/cobordism+hypothesis . For the unoriented category, you might say it's "the free symmetric monoidal 1-category on a self-dual object". As you move up the ladder, you'll also have to start thinking about other questions, like framed versus oriented versus ... – Theo Johnson-Freyd Feb 6 '11 at 6:20
@Theo: ah. I wasn't sure enough about all the definitions on that page, but I think I know what "stable" means now. – Qiaochu Yuan Feb 6 '11 at 12:05

You might want to think of morphisms $S \to T$ as in bijection with morphisms $I \to -S + T$, where $I$ is the unit (represented by the empty 0-manifold). Then all the components or edges of the latter morphism do go from $-$ to $+$. The general rule is that an edge between points on opposite sides of the morphism go from $+$ to $+$ or $-$ to $-$, and go between points of opposite orientation if they are on the same side of the morphism. – Todd Trimble Feb 6 '11 at 13:34