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I've started studying cobordism and I found difficult formalising the abelian group structure of the oriented bordism group $\Omega_n$.

The main problem I encountered is dealing with the orientation at the boundaries, and so I did some sketches to help me visualise the problem.

To prove reflexivity of the relation it's enough to consider the cylinder over a closed oriented manifold $M$, as depicted in this picture. reflexivity

And to prove existence of the inverse element for any $[M]$ (denoted with $-[M]$) under the operation, we said " consider the cylinder over $M$ and bend one end over the other." From what I understood (not english mother tongue:( ) that is the situation:

inverse

But I'm not convinced due to the fact that I don't see what's changed from the first situation to the second situation, only a little bending cannot change the orientation behaviour of the boundary. Nevertheless I tried depicting the situation.

The convention I used on the orientation of the boundary is the one saying that an oriented basis of $T_p(\partial W)$ followed by an outward pointing vector is an oriented basis of $T_pW$ where $W$ is the manifold of the bordism.

Can someone clarify these passages in the proof of the existence of inverse element?

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    $\begingroup$ What is the exact definition of $\Omega_n$ you are working with? $\endgroup$ – Eric Wofsey Sep 14 '15 at 9:49
  • $\begingroup$ @EricWofsey Let $M_1,M_2$ two $n$-dim closed oriented mfld. We say $M_1$ and $M_2$ are equivalent in $\Omega_n$ if exists a $n+1$ oriented manifold $W$ with boundary such that $\partial W$ is diffeomorphic as an oriented manifold to $M_1\sqcup \bar{M_2}$. By $\bar{M}$ I mean, $M$ with the opposite orientation. The orientation of $\partial W$ is the one induced by the outward pointing vector (as said above) $\endgroup$ – Luigi M Sep 14 '15 at 10:01
  • $\begingroup$ clearly the operation is disjoint union $\endgroup$ – Luigi M Sep 14 '15 at 10:03
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    $\begingroup$ The bending is in the sense of this picture: thomhujanen.files.wordpress.com/2015/06/… In there you would on the one side of the bordism the two circles (with each different orientation) and on the other side the empty set. $\endgroup$ – Daniel Valenzuela Sep 14 '15 at 15:00
  • $\begingroup$ Dear @Dan, thanks for the picture. There is something unclear to me in this picture tough. It seems that your picture lead to the fact that $M \sqcup M \simeq \emptyset$ which cannot be true in general. I'll upload a picture depicting what I think your picture lead to in a moment $\endgroup$ – Luigi M Sep 14 '15 at 20:25
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You don't actually have to bend anything. The definition of a bordism from $M_0$ to $M_1$ is just a manifold $B$ together with an oriented diffeomorphism $\partial B\cong M_0\coprod -M_1$ (where $-M_1$ is $M_1$ with the opposite orientation). Now fix a manifold $M$ and consider $B=M\times[0,1]$; then $\partial B\cong M\coprod -M$. We can split this up in two different ways. First, we can let $M_0=M_1=M$, and find that $B$ is a bordism from $M$ to itself. But we can also set $M_0=M\coprod -M$ and $M_1=\emptyset$, and find that $B$ is a bordism from $M\coprod -M$ to the empty manifold. So between your two pictures, nothing has changed about $B$ or the orientation of its boundary; you've just changed which components of the boundary you're calling $M_0$ and which ones you're calling $M_1$.

The only reason to "bend" in the picture is if you want to visualize $M_0$ being at the bottom of the picture and $M_1$ being at the top: then you have to bend the area around $M\times\{1\}$ down so that it's at the bottom. But this is just to provide intuition; you don't actually need to change what $B$ is.

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  • $\begingroup$ Ok, it's clear for me your first answer's paragraph. But the second one raised a doubt to me. If we bend the cylinder (which, according to me, is the same as my second picture), we would have $M \sqcup M \simeq \emptyset$ because the two ends seems to have the same orientation... but this shouldn't be the case. so where is the fault here? $\endgroup$ – Luigi M Sep 14 '15 at 20:19
  • $\begingroup$ "Bending" doesn't actually change the orientation of anything. The second end still has the opposite orientation from $M$, even if you've bent it over. (This may be difficult to visualize because in the pictures people usually draw $M$ is a circle, and in that case $M$ happens to be oriented-diffeomorphic to $-M$.) $\endgroup$ – Eric Wofsey Sep 14 '15 at 20:31
  • $\begingroup$ Ok, I feared that the problem was that I had a bad picture in mind :), anyway thanks for the clarification, I accept the answer:) $\endgroup$ – Luigi M Sep 14 '15 at 20:32
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The cylinder $M\times I$ induces an orientation on the boundary, the appending outward pointing normal vector being different on both ends. Hence the cylinder gives a bordism between $M$ and $M$, because the boundary is $M\sqcup \bar M$ this implies reflexivity.

For the existence of an inverse the picture of "bending over" should visualize only the partition of the boundary to $(M\sqcup \bar M ) \sqcup \emptyset$. The bending over should only be a homeomorphism for improving the intuition, hence changes nothing about the orientation. The intuition or visualization should just be: a cobordism is an equivalence between the left hand side, and the right hand side. Hence if we want to show that something is cobordant to the empty set (the zero element) we put this something to the left hand side (by bending the cylinder both $M$ and $\bar M$ end up on the LHS).

Hence we showed: $M \sim M$ and $M\sqcup \bar M \sim \emptyset$.

To summarize: The change in both situation is on which side of the bordism we see the boundary parts on (only serves visualization). Both situations are identical in terms of oriented diffeomorphism. It just makes renaming objects easier.

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