contractible manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T05:23:17Z http://mathoverflow.net/feeds/question/60113 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60113/contractible-manifolds contractible manifolds Mark Sapir 2011-03-30T19:45:42Z 2011-03-30T20:19:24Z <p>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 $R^{2n+1}$ ? </p> http://mathoverflow.net/questions/60113/contractible-manifolds/60117#60117 Answer by Andreas Thom for contractible manifolds Andreas Thom 2011-03-30T20:05:30Z 2011-03-30T20:05:30Z <p>This was proved in the PL-setting in:</p> <p>McMillan, D. R.; Zeeman, E. C. <em>On contractible open manifolds.</em> Proc. Cambridge Philos. Soc. 58 1962 221–224. </p> <p>From MathReviews:</p> <p>"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)</p> http://mathoverflow.net/questions/60113/contractible-manifolds/60118#60118 Answer by Andy Putman for contractible manifolds Andy Putman 2011-03-30T20:19:24Z 2011-03-30T20:19:24Z <p>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 <a href="http://www.math.rutgers.edu/~sferry/ps/geotop.pdf" rel="nofollow">here</a>.</p>