# Are invertibly cobordant manifolds diffeomorphic

cobordismdifferential-geometrydifferential-topology

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?

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.