As I understand it the h-cobordism theorem says that if $M, N$ are closed $n$-manifolds and $W$ an $(n+1)$-manifold such that $\partial W = M \coprod \bar N $ then if W, N, M are mutually homotopy equivalent, simply-connected, and smooth then $M$ is homeomorphic to $N$ ($n \ge 5$).
If I've understood that correctly then it would seem as a corollary one could prove that for closed, smooth, simply-connected, $(n \ge 5)$-manifolds homotopy-equivalence implies homeomorphism. This seems absurd since homotopy-equivalence is such a coarse relation.
My argument is the following: let $f:M \longrightarrow N$ be a homotopy equivalence with $g$ the homotopy inverse of $f$. Then $W = M_f = N \times [0,1] \coprod_f M$ is the mapping cylinder, and since $f$ is a homotopy equivalence we know $W$ deformation retracts onto both of its ends. Since $M,N,[0,1]$ are all simply-connected we know $W$ is simply-connected. Thus we can find a smooth manifold (no corners) in the homotopy type of $W = M_f$. Here we can apply the h-cobordism theorem to $W'$ and conclude $M$ is homeomorphic to $N$.
Obviously there must be something wrong with that argument, but I'm not immediately seeing it. Also wikipedia says the h-cobordism theorem can be stated in the DIFF category too, in which case a similar argument would say that isotopy-equivalence implies diffeomorphism for this class of manifolds?