Cancellation in topological product

Question

I was wondering whether $M\times \mathbb{R}$ is homeomorphic to $N\times \mathbb{R}$ implies $M$ is homeomorphic to $N$, where let us say $M,N$ are smooth manifolds. (They are certainly homotopy equivalent.) More generally what if $\mathbb{R}$ is replaced by some other manifold say $\Sigma$? Any possible approach/idea?

Answer 1

The answer is no.

Let $X$ be the Whitehead manifold which is a contractible three-dimensional manifold. Despite the fact that $X$ is not homeomorphic to $\mathbb{R}^3$, $X\times\mathbb{R}$ is homeomorphic to $\mathbb{R}^4 = \mathbb{R}^3\times\mathbb{R}$.

Note, there is a similar question on MathOverflow that may be of interest.