[Math] Simply connected 3-manifolds are homotopy equivalent to 3-spheres

algebraic-topologyhomotopy-theorymanifoldsspheres

Let $M$ be a simply connected $3$-dimensional manifold (smooth, closed, connected). How to prove that $M$ has a homotopy type of a $3$-sphere?
I would like to use somehow Whitehead theorem: so the problem reduces to show the weak homotopy equivalence (i.e. the existence of a map $f:M \to S^3$ inducing isomorphisms on every homotopy group).

Best Answer

By Poincaré duality, $H_2(M)=0$ (use universal coefficient theorem). Applying Hurewicz, we get that $M$ is $2$-connected and that the Hurewicz map $h:\pi_3(M)\to H_3(M)\cong\mathbb{Z}$ is an isomorphism. Letting $f:S^3\to M$ represent a generator of $\pi_3(M)$, we see that $f$ is a homotopy equivalence by Whitehead (using the fact that manifolds are homotopy equivalent to CW complexes).

This is exercise 4.2.15 in Hatcher.

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