I am trying to pin down the right category of $ 2 $-dimensional bordisms that lies beneath the equivalence between $ 2 $-dimensional topological quantum field theories and commutative Frobenius algebras (see also my previous question).
The category that suits our needs must be a category of oriented bordisms, but the notion of oriented bordism troubles me.
Freed's notes (or equivalently the pretty good Moore and Saxena's TASI lecture) gives the following definitions.
Definition 1 (unoriented bordisms). Let $\Sigma_0$ and $\Sigma_1$ be the two closed manifolds. An $ n $-dimensional bordism from $\Sigma_0$ to $\Sigma_1$ is a triple $(M, i_0, i_1)$ where
- $M$ is an $ n $-dimensional compact manifold with boundary where we defined a partition of the connected components of its boundary $ \partial M $ into in-boundaries and out-boundaries;
- $ i_0 $ and $ i_1 $ are smooth embeddings $$ i_0 \colon \Sigma_0 \times [0,\varepsilon[ \to M \qquad i_1 \colon \Sigma_1 \times ]1-\varepsilon, 1] \to M $$ that define diffeomorphisms $ \Sigma_0 \cong i_0(\Sigma_0,0)\cong \partial_0M $ and $ \Sigma_1\cong i_1(\Sigma_1,1)\cong \partial_1M $ between the "abstract" in- and out-boundaries $ \Sigma_0 $ and $ \Sigma_1 $, and the (disjoint union of) the chosen in- and out-boundaries $ \partial_0M $ and $ \partial_1M $.
Definition 2 (oriented bordisms). Let $\Sigma_0$ and $\Sigma_1$ be the two oriented closed manifolds. An $ n $-dimensional bordism from $\Sigma_0$ to $\Sigma_1$ is a triple $(M, i_0, i_1)$ where
- $M$ is an $ n $-dimensional oriented compact manifold with boundary, whose boundary $ \partial M $ inherits the boundary orientation and where we defined a partition of the connected components of $ \partial M $ into in-boundaries and out-boundaries;
- $ i_0 $ and $ i_1 $ are smooth orientation preserving embeddings $$ i_0 \colon \Sigma_0 \times [0,\varepsilon[ \to M \qquad i_1 \colon \Sigma_1 \times ]1-\varepsilon, 1] \to M $$ that define (orientation preserving) diffeomorphisms $ \Sigma_0 \cong i_0(\Sigma_0,0)\cong \partial_0M $ and $ \Sigma_1\cong i_1(\Sigma_1,1)\cong \partial_1M $ between the "abstract" in- and out-boundaries $ \Sigma_0 $ and $ \Sigma_1 $, and the (disjoint union of) the chosen in- and out-boundaries $ \partial_0M $ and $ \partial_1M $.
As I understood it, to specify an unoriented bordism we need to provide its in/out components "by hand"; that is, the partition of $ \partial M $ is an external structure we need to put in when explicitly constructing the bordism.
On the other hand, my intuition keeps suggesting that the choice of an orienation must in some way fix the in- and out-boundaries of an oriented bordism [1].
It is indeed undoubtedly possible to 1) take two oriented manifolds $ \Sigma_0^{n-1} $ and $ \Sigma_0^{n-1} $, 2) take an oriented manifold $ M^n $, and 3) let the orientation of $ M $ to specify a partition of the components of $ \partial M $ into "in" and "out" components. If we then also define embeddings $ i_0 $ and $ i_1 $ that orientation-preservingly identify each $ \Sigma_i $ with the corresponding choice of in- and out-components, then according to Definition 2 we get an oriented bordism.
My point now, though, is that Definition 2 allows for oriented bordism $ (M,i_0,i_1) $ where the partition of $ \partial M $ into in- and out-components is not the one induced by the orientation of $ M $.
Question. Should Definition 2 be tweaked as to not to explicitly require the end user to specify a partition of the set of components of $\partial M$?
[1] I can illustrate this idea with a pictorial description. In dimension $ 2 $ the in-boundaries are the ones who (with the induced orientation) are "rotating" so that by the right hand rule we are pointing to the inside of the underlying manifold. The same rule applies mutatis mutandis for the "out" ones.
