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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 16 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 12 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 12 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 6 hours ago