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Freed's notes give the following definition of oriented bordism.

Definition. 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

  1. $M$ 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;
  2. $ 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 $.

Now, take an oriented $ 2 $-dimensional closed manifold $ \Sigma_0 $ and let's try to define a bordism from $ \Sigma_0 \sqcup \overline{\Sigma}_0 $ to the null manifold $ \emptyset^2 $, where $ \bar\Sigma_0 $ is just $ \Sigma_0 $ with its orientation reversed.

Let $ M $ be the oriented manifold $ M = \Sigma_0 \times [0,1] $, where $ [0,1] $ carries its canonical "from left to right" orientation and where the boundaries of $ M $ acquire the induced "Stokes" orientation [1].

To define an oriented bordism the two maps $$ i_0\colon \left(\Sigma_0\sqcup \bar\Sigma_0\right)\times [0,\epsilon[\to M\qquad i_1\colon \emptyset^2\times [1 - \epsilon, 1[\to M $$ must be defined. Here $ (\Sigma_0 \sqcup \bar{\Sigma}_0) \times [0,\varepsilon[ $ is nothing but the disjoint union of two (times an unspecified number of) cylinders. One of them, $ \Sigma_0 \times [0,\varepsilon[ $, has the same orientation as $ M $, hence I can embed it into $ M $ in an orientation preserving way. On the other hand, the remaining one $ \bar\Sigma_0 \times [0,\varepsilon[ $ has the opposite orientation, and of course I cannot.

Is there something I'm missing? I see basically no other way of defining a bordism from $ \Sigma_0\sqcup \bar \Sigma_0 $ to the null manifold (and it's quite clear that one must exist).

[1] We can depict the manifold $ M $ as a cylinder with two inward-pointing arrows stemming from its bases.

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The issue arises from incorrectly identifying the boundary orientation. For the cylinder $M = \Sigma_0 \times [0,1]$ with the product orientation, the boundary orientation is $\partial M = (-\Sigma_0) \sqcup \Sigma_1$, where $\Sigma_1$ is $\Sigma_0$ with the same orientation.

To get a cobordism from $\Sigma_0 \sqcup \bar{\Sigma}_0$ to $\emptyset$, you need $\partial M = (-\Sigma_0) \sqcup (-\bar{\Sigma}_0) = (-\Sigma_0) \sqcup \Sigma_0$. This is achieved by taking $M = \Sigma_0 \times [0,1]$ but declaring both boundary components as in-boundaries. Then the embeddings $i_0$ map $\Sigma_0$ to one boundary (which appears as $-\Sigma_0$) and $\bar{\Sigma}_0$ to the other (which appears as $-\bar{\Sigma}_0 = \Sigma_0$). Both maps are orientation-preserving because the boundary orientation automatically reverses the orientation of incoming boundaries relative to their object orientation. Your desired cobordism $M : \Sigma_0 \sqcup \bar{\Sigma}_0 \to \emptyset$ exists by taking the cylinder with both ends as inputs.

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  • $\begingroup$ Can you clarify what you mean by $ \Sigma_0 $, $ \bar \Sigma_0 $, $ -\Sigma_0 $ and $ -\bar\Sigma_0 $? I suspect $ -\Sigma_0 = \bar\Sigma_0 $ so that $ -\bar\Sigma_0 = \Sigma_0 $, am I right? $\endgroup$ Commented yesterday
  • $\begingroup$ Yes. So $-\Sigma_0$ denotes the manifold $\Sigma_0$ with the orientation reversed by the boundary convention, so $-\Sigma_0 = \bar{\Sigma}_0$ and thus $-\bar{\Sigma}_0 = \Sigma_0$. And in the cylinder $M = \Sigma_0 \times [0,1]$, the boundary orientation give $\partial M = (-\Sigma_0) \sqcup \Sigma_0 = \bar{\Sigma}_0 \sqcup \Sigma_0$. By declaring both ends as in-boundaries, this matches the required $\Sigma_0 \sqcup \bar{\Sigma}_0$ (up to reordering), making it a valid cobordism to $\emptyset$. $\endgroup$ Commented 15 hours ago
  • $\begingroup$ It's definitely more clear now, thank you. But one thing still nags me. Let's call the orientation on $ M = \Sigma \times [0,1] $ that makes $ \partial M = (-\Sigma_0) \sqcup \Sigma_0 $ the positive one, and let's try to to define $ i_0 \colon (\Sigma_0 \sqcup \bar\Sigma_0)\times [0, \varepsilon[ \to M $. Then $$ (\Sigma_0\sqcup \bar\Sigma_0) = (\Sigma_0\times [0, \varepsilon[) \sqcup (\bar\Sigma_0\times [0, \varepsilon[). $$ so that we can actually define two maps and then stick them together using the universal property of the coproduct. (Continues below) $\endgroup$ Commented 11 hours ago
  • $\begingroup$ Clearly, then, $ \Sigma_0 \times [0, \varepsilon[ $ is "positive-oriented" cylinder with boundary equal to $ -\Sigma_0 $ that can be embedded into the "$ -\Sigma_0 $" boundary component of $ M $ by one of these maps. On the other hand, $ \bar\Sigma_0 \times [0, \varepsilon[ $ is a "negative-oriented" cylinder with boundary equal to $ \Sigma_0 $. Where do I embed this? I.e., how do I define an orientation-preserving ("collar") map $ \bar\Sigma_0 \times [0, \varepsilon[\to M $? $\endgroup$ Commented 11 hours ago
  • $\begingroup$ Well, the resolve is that for an in-boundary, the standard collar map is actually orientation-reversing in the interval direction. The correct embedding for the $\bar{\Sigma}_0$ component is $i(x,t) = (x, 1-t)$ which uses a decreasing parameterization of the interval. This ensures the map is orientation-preserving when the boundary is declared as incomning. The "outward normal first" convention for the boundary orientation then forces the collar parameter to run backwards for in-boundaries. $\endgroup$ Commented 5 hours ago

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