Is $(\Sigma^+\Sigma N)\cup (\Sigma N\times \mathbb R^+)$ homeomorphic to $\mathbb R^5$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T01:42:03Z http://mathoverflow.net/feeds/question/22140 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22140/is-sigma-sigma-n-cup-sigma-n-times-mathbb-r-homeomorphic-to-mathbb Is $(\Sigma^+\Sigma N)\cup (\Sigma N\times \mathbb R^+)$ homeomorphic to $\mathbb R^5$? MG 2010-04-22T02:16:32Z 2010-04-22T02:32:38Z <p>Let $N^3$ be Poincare homology sphere, $\Sigma N$ be the spherical suspension of $N$, and it's known that $\Sigma^2 N$ the double suspension is homeomorphic to $S^5$. Let $\Sigma^+\Sigma N$ be the spherical cone over $\Sigma N$, and denote the gluing $\Sigma N\times \mathbb R^+$ to $\Sigma^+\Sigma N$ along their common boundary by $M^5$. Is $M^5$ homeomorphic to $\mathbb R^5$?</p> <p>The intuition I have is $M^5$ is the double suspension deleting one point, so is $\mathbb R^5$. But I am not sure whether this is a real proof.</p> <p>Other concern is, if we delete more, say a small close ball of that point, the rest is still homeomrophic to $M^5$, since it's nothing but $(\Sigma^+\Sigma N)\cup (\Sigma N\times \mathbb [0, 1))$. But $S^5$ 'should' deleting $D^5$ to get $\mathbb R^5$...</p>