Embedded submanifold in a cylinder Let $M^n$ be an $n$-dimensional topological closed manifold. Suppose there exists an embedding $i:M \to M \times [0,1]$ such that $i(M)$ is contained in the interior of $M \times [0,1]$ and separates $M \times \{0\}$ and $M \times \{1\}$.
Can we show the region bounded by $M \times \{0\}$ and $i(M)$ is homeomorphic to $M \times [0,1]$?
 A: In fact, there are "nice" counterexamples. There is a notion of an inertial h-cobordism on $M$. It is an h-cobordism with both boundaries homeomorphic to $M$. By the s-cobordism theorem, all h-cobordisms are invertible in the sense that we can stack one on top of the other and get the identity cobordism.
Given any inertial h-cobordism we can extract one of the diagrams you suggest. Namely, stack the inverse h-cobordism on top of it and consider the embedding of $M$ into the middle. If the h-cobordism is nontrivial, than you get a counter example to your question. In fact, if $i:M \rightarrow M \times [0,1]$ is assumed to have a bicollar, these notions are exactly the same.
Page 3 of Whitehead torsion of inertial h-cobordisms talks about how to construct nontrivial examples.
A: I think the answer is no. Identify $S^2\times [0,1]$ with a large ball in $\mathbb R^3$ centered at the origin with a neighborhood of the origin removed. Then I believe the Alexander horned sphere (embedded within the large ball with the small ball ``inside it'') provides the desired separation, but the outside part is not simply connected. See Section 2.B of Hatcher for details.
