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Let $f:M_0\rightarrow M_1$ be a diffeomorphism between two compact $n$-dimensional manifolds. Let $W^{n+1}$ be an $h$-cobordism between $M_0$ and $M_1$. Assume that the cobordism has no torsion and its dimension is high enough so that, by the $s$-cobordism theorem, there is a diffeomorphism $F:M_0\times I\rightarrow W$. Can $F$ be taken so that $F|_{M_0\times 0}=id_{M_0}$ and $F|_{M_0\times1}=f$?

Thanks in advance!

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The question is unclear, mainly because it's not clear what "can be taken ... up to isotopy" means. Under the only interpretation that seems at all reasonable, the answer is pretty trivially "no".

I'll assume that the given $F$ is assumed to take $M_0\times 1$ to $M_0$ by the identity, and that the question is whether there is necessarily another $F$ that is the same on $M_0\times 0$ and is $f$ on $M_0\times 1$. The answer is no. For example, let $M_0=M_1=S^n$ and let $W$ be $S^n\times I$, with $F$ the identity and $f$ orientation-reversing.

Am I misunderstanding?

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    $\begingroup$ Tom: The question is, indeed, poorly phrased. I think, OP is asking for conditions under which a diffeomorphism $f: M\to M$ is (smoothly) pseudo-isotopic to the identity. The obvious necessary condition, as you observed, is that $f$ is homotopic to the identity. There are diffeomorphisms of $S^6$ which are homotopic but not (smoothly) pseudo-isotopic to the identity (coming from exotic 7-spheres). If $M$ is simply-connected (of dimension $\ge 5$) then Cerf proved that smooth pseudo-isotopy is equivalent to smooth isotopy, which is the best one can hope for. $\endgroup$
    – Misha
    Commented Dec 21, 2012 at 6:37
  • $\begingroup$ @Tom Goodwillie: If one adds the condition that $f$ is homotopic to the identity, is the answer pretty trivially no again?... $\endgroup$
    – Mauricio
    Commented Dec 21, 2012 at 16:41
  • $\begingroup$ @Mauricio: No, the answer is nontrivial for maps homotopic to the identity, see my comment above. (That is, unless you regard, say, existence of exotic spheres and Cerf's theory as trivialities. :-)) $\endgroup$
    – Misha
    Commented Dec 21, 2012 at 17:50

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