Does it follow that $Y$ is homotopy equivalent to $S^{n-1}$?

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 it follow that $Y$ is homotopy equivalent to $S^{n-1}$?

Answer 1

I'll prove the statement I have in my comment. That any CW-complex with the homology of $S^n$ suspends to a space homotopy equivalent to $S^{n+1}$ for $n\geq 1$. If $X$ is such a CW-complex, by Mayer-Vietoris, $SX$ has the homology of $S^{n+1}$. As $X$ is $(0)$-connected, $\pi_1(SX)=0$ by the Freudenthal suspension theorem.

By applying Hurewicz repeatedly, we have $0=H_k(SX)\cong\pi_k(SX)$ for $1\leq k\leq n+1$, and $H_{n+1}(SX)=\pi_{n+1}(SX)=\Bbb Z$ and the Hurewicz map $$H:\pi_{n+1}(SX)\to H_{n+1}(SX)$$ is an isomorphism. In particular, there exists $\alpha :S^{n+1}\to SX$ with $\alpha_*([S^{n+1}])$ generating $H_{n+1}(SX)$.

Now $\alpha$ is map between simply-connected spaces which induces isomorphism on homology , so it is a weak homotopy equivalence (cf. Spanier 7.6.25) and hence a homotopy equivalence if $X$ is a CW-complex.

In particular if $M$ is $S^4$, and $Y$ is a non-trivial homology 3-sphere (e.g. the Poincaré dodecahedral space), then $S(Y)$ is homotopy equivalent to $S^4$.