Let $M$ and $N$ be oriented, closed, $n-1$ manifolds and $F$ a cobordism from $M$ to $N$ and $G$ a cobordism from $N$ to $M$ such that the composite cobordism $G\circ F\cong M\times I$ and $F\circ G\cong N\times I$. Does the existence of an "inverse" cobordism imply that $M$ is diffeomorphic to $N$?

My first, naive strategy is to take the following composite: if $j: N\to F$ is the inclusion of $N$ into $F$, $\psi: G\circ F\to M\times I$ the diffeomorphism, and $\pi: M\times I\to M$ the projection on the first factor, then the composite $\pi \circ \psi \circ j$ is a map from $N\to M$. However, it is easy to foresee that this map may fail to be injective or surjective.

Is there reason to believe this statement to be true?

## Best Answer

This statement is false for general $n$ (I'll write $n$ where you've written $n-1$).

For $n\geq5$ we can disprove the statement using a rather common trick. Suppose we are given two (closed) manifolds $M$ and $N$ of dimension $n\geq5$, and suppose that the manifolds are $h$-cobordant (recall that this means that their inclusions $M \hookrightarrow W$ and $N \hookrightarrow W$ into the cobordism $W$ are homotopy equivalences). The Whitehead torsion of this cobordism is an element $\tau = \tau(W,M) \in \operatorname{Wh}(\pi_1(M))$. The $s$-cobordism theorem says that $\tau$ vanishes if and only if the cobordism is a cylinder (i.e., iff $M$ and $N$ are diffeomorphic).

Onto either end of $W$ it is possible to glue an $h$-cobordism whose torsion is $-\tau$; the composite cobordism then has Whitehead torsion $0$ (these facts follow from properties of Whitehead torsion, plus a part of the $s$-cobordism theorem that says you can realize every element of the Whitehead group as the torsion of some $h$-cobordism). Our composite cobordism is hence a cylinder by the $s$-cobordism theorem.

This means that it's actually always possible to get the setup you describe in your question; you get the "inverse" of $W$ by this construction and it's not difficult to see that this will actually be a $2$-sided inverse to $W$. But if the statement you hope to be true actually holds, it would follow that all $h$-cobordant manifolds are diffeomorphic, contradicting the $s$-cobordism theorem.