# Is $(\Sigma^+\Sigma N)\cup (\Sigma N\times \mathbb R^+)$ homeomorphic to $\mathbb R^5$?

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 $\mathbb R^5$?

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.

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 $\mathbb R^5$...

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The fact that $\Sigma^2 N\cong S^5$ is usually called the Cannon--Edwards Theorem. –  HJRW Apr 22 '10 at 18:27
Yes, $M^5$ is homeomorphic to $\mathbb{R}^5$ and the homeomorphism is easy to write down (use for example the coordinates given by the usual description of cones and suspension as quotients). –  Benoît Kloeckner Apr 28 '10 at 9:23