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Two closed oriented $n-1$ dimenisonal manifolds $A,B$ are cobordant if there is a compact $n$ dimensional manifold $M$ such that $A \sqcup -B$ diffeomorphic(orientation-preserving) to boundry of $M$.

If there is a diffeomorphism between $A$ and $B$ , can we conclude that $A,B$ are cobordant? Intuitively, I think it is true, but I have trouble to prove this claim.

iefjkfdhfure
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  • Is not this is the right definition: Two oriented closed $n-1$ dimensional manifolds $A, B$ are said to cobordant if there is a compact $n$ dimensional oriented manifold $M$ such that there is an orientation preserving diffeomorphism from $\partial M$ to $A\sqcup -B$ – Sumanta May 30 '21 at 08:39
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    how about $A\times [0,1]$? – Tim kinsella May 30 '21 at 08:42
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    @Timkinsella, you are right. If there is an orientation preserving diffeomorphism from $A$ to $B$, then $A\times [0,1]$ is the right choice for $M$. – Sumanta May 30 '21 at 08:45
  • @Sumanta you are right. I will edit my question. – iefjkfdhfure May 30 '21 at 08:50
  • @Timkinsella why $A×[0,1]$ works for any orientation-preserving diffeomorphism? – iefjkfdhfure May 30 '21 at 08:56
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    @iefjkfdhfure The boundary of the $1$-dimensional smooth manifold $[0, 1]$ with the usual orientation, i.e. from $0$ to $1$, is defined as $\big(−{0}\big) \cup {1}$. Also, for any two oriented manifold $X,Y$ with $\partial Y=\varnothing$ we have $$\partial (X\times Y)=(\partial X)\times Y\textbf{ as oriented manifolds.}$$ – Sumanta May 30 '21 at 09:11

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