TITLE: 3-dimensional h-cobordisms
QUESTION [7 upvotes]: Let $W$ be a $3$-dimensional $h$-cobordism of closed surfaces $M_0$ and $M_1$. Can we prove that $W$ is trivial? That is, $W$ is homeomorphic to $M_0 \times [0,1]$.

REPLY [9 votes]: For $M_0=M_1=S^2$, this follows from the 3-dimensional Poincaré conjecture: glueing in 3-balls, you get a simply connected $3$-manifold, that has to be $S^3$, so $W=S^2\times\left[0,1\right]$.
For surfaces of higher genus, the result follows from Waldhausens rigidity theorem, which says that homotopy-equivalent Haken manifolds with homeomorphic boundaries are homeomorphic. (An irreducible $3$-manifold with non-spherical boundary is necessarily a Haken manifold.)