s-cobordism theorem

Remarks

The theorem is due independently to Barden, Mazur and Stallings. The same statements hold for PL manifolds, and for topological manifolds by work of Kirby and Siebenmann. Idea of the proof: as for the h-cobordism theorem, take a handle decomposition and cancel handles. Now the handle chain complex is a free based complex over $\Z[\pi]$, and the obstruction to cancelling all handles is exactly its torsion in $\operatorname{Wh}(\pi)$.

The theorem at work: $\operatorname{Wh}(1) = 0$, which gives back the h-cobordism theorem, and $\operatorname{Wh}(\Z^k) = 0$ by the computation for free abelian groups. So every h-cobordism of dimension at least $6$ on a torus is a product. For $\pi = \Z/5$ the Whitehead group is infinite cyclic, so a closed $5$-manifold with this fundamental group carries h-cobordisms that are not products.

Going deeper

For a student

The theorem explains why surgery theory works with simple homotopy equivalences: an h-cobordism with zero torsion is an s-cobordism, and s-cobordant manifolds of dimension at least $5$ are diffeomorphic. It is one input of the surgery exact sequence, where the structure set is formed up to this relation. Whitehead torsion is computed from algebraic K-theory, since $\operatorname{Wh}(\pi)$ is a quotient of $K_1(\Z[\pi])$. Classical examples come from lens spaces, whose distinction up to simple homotopy uses Reidemeister torsion.