Homotopy Poincaré conjecture - no map inducing the isomorphism on homology

Question

$\newcommand{\Z}{\mathbb{Z}}$ In Terence Tao's notes on page 18, concerning the Poincaré conjecture, he gave the following sketchy proof of the homotopy Poincaré conjecture.

Given $M^3$ a 3-manifold with $\pi_1(M^3)=0$, then automatically $0=H_1(X,\Z)=H^2(X,\Z)=H_2(X,\Z)$. Moreover the action of $\pi_1(M_3) \curvearrowright H^{n-1}(S^{n-1},\Z)$ where $S^{n-1} \hookrightarrow Sp(T(M^3)) \to M^3$ is the fiber of the sphere bundle over the basepoint of $M$, is trivial since $\pi_1(M^3)=0$, so that $M^3$ is orientable. Therefore $H^3(M^3,\Z)=\Z$ and $H^i(M^3,\Z)=0$ for $i>3$ because $M^3$ is a 3 dimensional manifold (easy argument from excision).

Terence Tao then says that it follows from the Whitehead theorem that $M^3 \to S^3$ is a homotopy equivalence. It does not follow since there is no map inducing the isomorphisms on homology between $S^3$ and $M^3$.

Is there a Postnikov decomposition trick to show that we don't need the map to induce the isomorphism or is there an obvious map that I am missing?

Answer 1

There are a couple of ways to produce such a map. One is the approach given by Mike Miller. Another is using two applications of the Hurewicz theorem: one shows that $\pi_2 = 0$, and the second shows that $\pi_3 \cong H_3 \cong \mathbb{Z}$. The map $f : S^3 \to M$ given by a generator of $\pi_3$ then induces an isomorphism on homology, and hence (by homology Whitehead) is a homotopy equivalence.

Answer 2

If $M$ is a connected $n$-manifold, pick any closed embedded disc $D$ and consider the canonical map $M \to M/(M \setminus \text{int}(D)) \cong S^n$. The degree of this map is one, and hence if $M$ is a homology sphere, this is a homology isomorphism. The assumption that $\pi_1 = 0$ then implies it's a homotopy equivalence by Hurewicz and Whitehead.