Show that a closed simply-connected 3-manifold is homotopy equivalent to \(S^3\).
Solution
Let \(M\) be a closed simply-connected \(3\)-manifold. Poincaré duality gives \[ H_i(M;\mathbb Z)\cong \begin{cases} \mathbb Z,&i=0,3,\\ 0,&i=1,2. \end{cases} \] Thus \(M\) is a homology \(3\)-sphere.
Since \(M\) is simply connected and \(H_2(M)=0\), the Hurewicz theorem implies \(\pi_2(M)=0\): otherwise \(\pi_2\) would be the first nonzero higher homotopy group and would map isomorphically to \(H_2\). Hence \(M\) is \(2\)-connected. Hurewicz now gives \[ \pi_3(M)\xrightarrow{\cong}H_3(M)\cong\mathbb Z. \] Choose \[ f:S^3\to M \] representing a generator. Then \(f_*\) is an isomorphism on \(H_3\), and it is automatically an isomorphism on all other homology groups as well. Both spaces are simply connected CW complexes (or have CW type), so the homology Whitehead theorem implies \[ \boxed{M\simeq S^3.} \]