h-cobordism theorem

Remarks

The theorem is due to Smale. Idea of the proof: a Morse function on $W$ gives a handle decomposition of $W$ on $M_0$. The homotopy equivalence condition makes the handle chain complex acyclic, and handles are traded and cancelled in pairs until none remain. Cancelling a pair needs the Whitney trick to remove extra intersection points of spheres in a level set. The Whitney trick needs simple connectivity and level sets of dimension at least $5$. A cobordism with no handles is a product.

The theorem at work: let $\Sigma$ be a smooth closed $m$-manifold homotopy equivalent to $S^m$ with $m \ge 6$. Removing two disjoint open balls from $\Sigma$ gives an h-cobordism between two copies of $S^{m-1}$, which is therefore a product. It follows that $\Sigma$ is homeomorphic to $S^m$, the generalized Poincaré conjecture in these dimensions.

Both extra hypotheses are needed. Without simple connectivity there are h-cobordisms that are not products; the obstruction is the Whitehead torsion, see the s-cobordism theorem. For $n = 4$ the smooth statement fails: Donaldson found smooth $5$-dimensional h-cobordisms between simply connected $4$-manifolds that are not products.

Going deeper

For a student

The h-cobordism theorem converts a homotopy-theoretic condition into a diffeomorphism. It is the base of the classification of high-dimensional manifolds through surgery theory: surgery produces cobordisms, and the theorem shows that suitable cobordisms are products. Its proof is the model application of Morse theory and handle cancellation. The s-cobordism theorem extends it to manifolds with a nontrivial fundamental group. In dimension $4$ the topological version holds by Freedman's work, but the smooth version fails.