contractible manifolds - MathOverflow

contractible manifolds

2011-03-30T19:45:42Z

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 by Andreas Thom for contractible manifolds

2011-03-30T20:05:30Z

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 by Andy Putman for contractible manifolds

2011-03-30T20:19:24Z

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.