Decomposing the sphere as a product

Solution 1:

Suppose $S^n = X \times Y$, where $X$ and $Y$ are arbitrary topological spaces (that are homotopy-equivalent to CW complexes). With integer coefficients, there is the (unnaturally) split Kunneth sexseq

$0 \rightarrow \bigoplus_i (H_i(X) \otimes H_{m-i}(Y)) \rightarrow H_m(X \times Y) \rightarrow \bigoplus_i \mbox{Tor}(H_i(X),H_{m-i-1}(Y)) \rightarrow 0$.

Clearly $X$ and $Y$ must be path-connected since $\pi_0$ takes products to products (and obviously we should assume $n \geq 1$), so $H_0(X) = H_0(Y)=\mathbb{Z}$. Since $-\otimes \mathbb{Z}$ does nothing to an abelian group, this means that we're getting a copy of $H_*(X)$ in $H_*(X\times Y)$ from the inclusion above when it's tensored against $H_0(Y)$, and similarly for $H_*(Y)$. Moreover, all the homology of $S^n$ must come from the inclusions in the above sexseq, since $\mbox{Tor}$ always consists entirely of torsion. So without loss of generality, $H_*(X) \cong H_*(S^n)$ and $H_*(Y) \cong H_*(\mbox{pt})$. Since $\pi_1$ takes products to products, both $X$ and $Y$ are simply-connected. So the projection $S^n =X \times Y \rightarrow X$ is a homology isomorphism of simply-connected spaces and hence is a (weak) homotopy equivalence, while $Y$ is a simply-connected space with trivial integral homology so it must be (weakly) contractible. Thus the factorization $S^n = X \times Y$ is trivial.

Solution 2:

Assuming that everything is a connected manifold.

Let $X$, $Y$ such that $X \times Y = S^n$. Before we apply the Kunneth theorem lets note that the dimensions of $X$ and $Y$ must add up to to $n$. $$\bigoplus_{i+j=k} H^i(X)\otimes H^j(Y)=H^k(S^n) $$

Suppose $\dim X = i < n$. Then $H^i(X)=\mathbb{Z}=H^{n-i}(Y)$ as these are the only possible non-zero cohomology groups that add up to $n$ which is required by the Kunneth theorem. But therefore $H^i(S^n)=\bigoplus H^k(X)\oplus H^j(Y) \neq 0 $ as the summation contains $H^i(X) \oplus H^0(Y)=\mathbb{Z}$ which is a contradiction. Therefore, wlog, $i=n$ and $Y$ is a point.