Is there a version of the 2d cobordism hypothesis for surfaces with non-empty incoming and outgoing boundary?

Question: Is there a condition on an object $x$ of an $(\infty,2)$-category $\mathcal C$ which is equivalent to $x = Z(pt_+)$ for a unique TFT $Z$ from the $(\infty,2)$-category of framed bordisms where we only allow $2$-cobordisms for which the incoming and outgoing boundary of every component is non-empty.

Remark: if I only demanded that the outgoing boundary was non-empty, this is called the non-compact bordism category non-compact -see below, and Defn 4.2.10 in http://www.math.harvard.edu/~lurie/papers/cobordism.pdf for the oriented version.

Motivation: The kind of examples I have in mind are things like string topology for a non-compact oriented manifold (this would be an oriented theory rather than framed, but I want to try to separate out the conditions imposed by giving rise to a framed theory, and the fixed point data for the action of $SO(2)$.

Background (from Jacob Lurie's paper linked above): The cobordism hypothesis in two dimensions states that fully extended 2d framed TFTs

$Bord_2 ^{fr} \to \mathcal C$

are equivalent to fully dualizable objects in $\mathcal C$ (where $\mathcal C$ is some symmetric monoidal $(\infty,2)$-category).

Explicitly, an object $x \in \mathcal C$ is fully dualizable if

1. It is dualizable (with dual $x^\vee$)
2. The evaluation morphism $ev:x \otimes x^\vee \to 1_{\mathcal C}$ has both a right and a left adjoint.

By duality, the adjoints $ev^R$ and $ev^L$ give rise to endomorphisms $S$ and $T$ of $x$ which are inverses of each other ($S$ is called the Serre automorphism).

There is also a non-compact version (as far as I understand): Let $Bord_n ^{fr,nc}$ be the bordism category in which every connected componant of a surface has a non-empty outgoing boundary. A non-compact 2d TFT

$Bord_n^{fr,nc} \to \mathcal C$

is equivalent to a $(1 + 1/2)$-dualizable object in $\mathcal C$. That is, an object $x$ which is

1. Dualizable,
2. The evaluation morphism has a right adjoint,
3. The corresponding endomorphism of $x$ is invertible.

Thoughts: Both full and 1.5 dualizibility are conditions that can be checked on the level of homotopy 2-categories. If an object $x$ is fully dualizable then the dualizing data (evaluation, unit and counit for the adjunction, etc.) are essentially uniquely determined.

The issue for me is that I don't see an obvious way to express the generators and relations for the non-empty incoming and outgoing boundary bordism category in terms of duals and adjoints. I could see a potential answer to my question along the lines of:

1. $x$ is dualizable,
2. $x$ admits an automorphism $S$, giving rise to morphisms $coev^\ast = (S\otimes 1_{x^{\ast}}) \circ coev$ and $coev^! = (S^{-1} \otimes 1_{x^\ast}) \circ coev: 1_{\mathcal C} \to x \otimes x^\ast$,
3. There are 2-morphisms $coev^! \circ ev: \to 1_{x^\ast \otimes x}$ and $1_{x^\ast \otimes x} \to coev^\ast \circ ev$ (corresponding to "saddle" cobordisms).
4. These satisfy some relations (I am picturing something like the identity that relates the comultiplication and multiplication in a Frobenius algebra...)

In any case, if there is an answer along these lines, my question is: if an object $x$ admits such a collection of data, is this collection unique?

I hope this makes some sense...

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I fixed a trivial typo, an $X$ that should have been an $x$. –  David Roberts Nov 1 '12 at 23:41
I was under the impression that the standard notion of a fully extended framed TQFT already incorporates non-empty incoming and outgoing boundary. Am I mistaken about this? –  Qiaochu Yuan Nov 2 '12 at 1:42
@Qiaochu: I meant that (each component of) the cobordism must have non-empty incoming and outgoing boundary. I.e. there are no cups or caps. If I only insisted this for outgoing boundary, this would be called non-compact, as in Defn 4.2.10 of Lurie's paper. I'll edit to clarify. –  Sam Gunningham Nov 2 '12 at 1:56
There will not be a condition which gives you such a TQFT. Such a TQFT will involve a 1-dualizable object (a condition) plus the structure necessary for you to extend the $(\infty,1)$-tqft to framed 1-handles. In the non-compact case you also have 0-handles and, roughly, together the 0-handles and 1-handles are just a property of the object. Alone they are structures. A similar thing happens for framed 2D tqfts where you only allow 0-handles, but not 1-handles. These correspond to 1-dualizable objects together with some structure. –  Chris Schommer-Pries Nov 2 '12 at 11:02
Thanks Chris! That answers my question as far as I am concerned (though I'll have to do some thinking to see why what you say is true). To clarify: the answers to my questions in the first paragraph and the last paragraph are "no". –  Sam Gunningham Nov 2 '12 at 16:20