= Manifold criterion for a suspension
{title2=$SX\text{ is a manifold}\iff X\cong S^d$}
For a nonempty <compact> <topological manifold> $X^d$ without boundary, its <suspension> is a topological <manifold> without boundary exactly when $X$ is homeomorphic to $S^d$. The forward implication is a <local homology from a link> calculation at a suspension vertex: $\widetilde H_j(X)$ is $\mathbb Z$ in degree $d$ and zero elsewhere. For $d\geq2$, put a coordinate ball between two nested cone neighbourhoods. The inclusion of their punctures is a <homotopy equivalence> but factors through the <simply connected> punctured ball, forcing $\pi_1(X)=0$. Thus $X$ is a <homotopy sphere>, and the <topological generalized Poincare theorem> gives the result. The dimensions zero and one follow from the finite-set and <circle> classifications. Conversely $S(S^d)\cong S^{d+1}$. If manifolds with boundary are allowed, the interval and its disk suspension show why this formulation needs the boundary restriction.
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