Question

Where can I find the proof of the following fact: If $M$ is a contractible manifold of dimension $n\ge 5$, then the direct product of $M$ and $\mathbb{R}^{n+1}$ is homeomorphic to $R^{2n+1}$ ?

Answer 1

This was proved in the PL-setting in:

McMillan, D. R.; Zeeman, E. C. On contractible open manifolds. Proc. Cambridge Philos. Soc. 58 1962 221–224.

From MathReviews:

"An open manifold is defined to mean a non-compact space that is triangulable by a countable complex which is a combinatorial manifold without boundary. The main theorem is that if $M^n$ is a contractible open manifold, then $M^n\times E^2$ is piecewise linearly homeomorphic to $E^{n+2}$. The principal tool used in the proof is the theorem due to Zeeman which implies that if $M^n$ is a contractible open manifold and $X$ is a subcomplex of codimension $\geq 3$, then $X$ lies in an $n$-cell in $M^n$. The results of this paper have been improved by Stallings for $n>3$ [same Proc. 58 (1962), 481--488] to show that $M^n\times E^1$ is piecewise linearly homeomorphic to $E^{n+1}$. Lemma 3 of the present paper is of independent interest and has been used to study cellularity of sets in products." (Reviewed by M. L. Curtis)

Answer 2

Expanding a little on Andreas Thom's answer, the easiest way to prove this is to use Stallings's theorem that says that a high-dimensional contractible PL manifold that is simply connected at infinity is homeomorphic to $\mathbb{R}^n$ (the key point being that if $M$ is a contractible PL manifold, then $M \times \mathbb{R}$ is clearly simply connected at infinity). There is a really beautiful account of this theorem in chapter 10 of Steve Ferry's notes on geometric topology, available here.