Suppose one has a closed $n$-manifold M, with $\pi_i(M)=0$ for $i=1,\ldots n-1$. Then does $M$ have to be homeomorphic to a sphere?
Note that this does not follow immediately from the generalised Poincare conjecture which states that if all homotopy groups (including of degree higher than $n$) are equal to that of a sphere then $M$ is homeomorphic to a sphere.
I am guessing it will follow from the GPC after applying some trick to show $\pi_{k}(M)=\pi_{k}(S^n)$ for $k>n$.
Note, however that under are assumptions we get $\pi_{n}(M) \cong \mathbb{Z}$ by the Hurewicz theorem.
The assumptions imply that $M$ is an integral homology sphere, whether that helps or not I am unsure.
Edit: I think I found a reference. On page 79 we have the quote". But any
simply connected integral homology sphere is of the homotopy type of the sphere (of
the same dimension), by the theorems of Hurewicz and Whitehead." Page 79 of SPACES OF TOPOLOGICAL COMPLEXITY ONE
MARK GRANT, GREGORY LUPTON and JOHN OPREA.
But I would stil like to know how the argument works (how do we apply Whiteheads theorem here?).
I am guessing we use $\pi_{n}(M) \neq 0$ to construct a non-trival map $S^n \rightarrow M$, but we need to know that this map preserves all homotopy groups?
Best Answer
You can prove this using the homology version of Whitehead's theorem (see Corollary 4.33 of Algebraic Topology by Hatcher):
If $M$ is an $(n-1)$-connected closed $n$-manifold, then $\pi_n(M) \cong \mathbb{Z}$ by the Hurewicz theorem. Let $f : S^n \to M$ be a representative of a generator of $\pi_n(M)$. Then $f_* : H_n(S^n; \mathbb{Z}) \to H_n(M; \mathbb{Z})$ is an isomorphism. It follows from the above theorem that $f$ is a homotopy equivalence. Now by the topological Poincaré conjecture, $M$ is homeomorphic to $S^n$.