Compact $n$-manifold has same integral cohomology as $S^n$?

Question

Let $M$ be a compact connected $n$-manifold without boundary, where $n \ge 2$. Suppose that $M$ is homotopy equivalent to $\Sigma Y$ for some connected based space $Y$. Does $M$ have the same integral homology as $S^n$?

Answer 1

Yes. First, recall that a suspension has no nontrivial cup products. By Poincaré duality, it follows that the integral cohomology of $M$ is torsion in degrees $1$ through $n-1$, and that the same is true of the integral homology. From universal coefficients we know that $H_{n-1}$ is torsion-free, which here implies that it's trivial. By Poincaré duality again, $H^1$ is trivial, so by universal coefficients again, $H_1$ is trivial, and by Poincaré duality again, $H^{n-1}$ is trivial. Now we can repeat the cycle: universal coefficients now implies that $H_{n-2}$ is trivial, etc.