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

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.

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Presumably you mean "they are homotopy-equivalent" not "homotopic". And you're asking if they're simple homotopy-equivalent, or do you mean something more precise by "related by a simple homotopy"? You are using a fair bit of non-standard terminology. The answer is no in general, since Whitehead torsion gets in the way. See the Wikipedia page on the h-cobordism theorem. – Ryan Budney Nov 26 '10 at 20:20
Ryan, I misread your comment before; sorry. – Tom Goodwillie Nov 27 '10 at 4:58
I wouldn't have known if you hadn't told me, Tom. So, no problem. :) – Ryan Budney Nov 27 '10 at 5:31

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 $M_1\times I$.