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 $\mathbb R^5$...