# Diffeomorphism coming from the s-cobordism theorem

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 $F|_{M_0\times1}=f$?

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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.
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. – Misha Dec 21 '12 at 6:37
@Tom Goodwillie: If one adds the condition that $f$ is homotopic to the identity, is the answer pretty trivially no again?... – Mauricio Dec 21 '12 at 16:41