Does Thom's J-equivalence imply Whitehead's simple homotopy?

Question

Rene Thom came up with the idea of J-equivalence:

Let $M_1$ and $M_2$ be manifolds that are oriented, compact and smooth. Then they are J-equivalent if there is a smooth manifold $X$ with boundary $M_1-M_2$ (we reverse the orientation of $M_2$) and both $M_1$ and $M_2$ are deformation retracts of $X$.

Thus if $M_1$ and $M_2$ are J-equivalent, they are homotopic. Are they related by a simple homotopy?

Note: Asking if they are instead diffeomorphic is, I believe, an open question. Although scattered results exists, eg this theorem of Smale: If $M_1$ and $M_2$ are J-equivalent homotopy spheres of dimension $2m-1$ with $m>2$, then they are diffeomorphic.

Answer 1

This "J-equivalence" is usually called h-cobordism. The results on it are not scattered! They are quite complete, except in low dimensions.

Given any $M_1$, and given an element $\tau$ of the Whitehead group of (the fundamental group of) $M_1$, there is always an h-cobordism $X$ such that the torsion of the inclusion map $M_1\to X$ is $\tau$. In fact, up to diffeomorphism fixed on $M_1$ there is a unique such $X$. In particular, in the case when $M_1\to X$ is a simple homotopy equivalence the only possible $X$ is $M_1\times I$.

If the fundamental group is trivial, or more generally if the Whitehead group is trivial (which it is for many and conjecturally all torsion-free fundamental groups), then h-cobordism implies diffeomorphism.