the sphere is indecomposable as a topological space

Proposition. If for any topological spaces $X$ and $Y$ the $n$-dimensional sphere ${𝕊}^n$ is homeomorphic to $X\times Y$, then either $X$ has exactly one point or $Y$ has exactly one point.

Proof. Recall that the homotopy group functor is additive, i.e. ${\pi }_n(X\times Y)\simeq {\pi }_n(X)\oplus {\pi }_n(Y)$. Assume that ${𝕊}^n$ is homeomorphic to $X\times Y$. Now ${\pi }_n({𝕊}^n)\simeq \mathbb{Z}$ and thus we have:

$$\mathbb{Z}\simeq {\pi }_n({𝕊}^n)\simeq {\pi }_n(X\times Y)\simeq {\pi }_n(X)\oplus {\pi }_n(Y).$$

Since $\mathbb{Z}$ is an indecomposable group, then either ${\pi }_n(X)\simeq 0$ or ${\pi }_n(Y)\simeq 0$.

Assume that ${\pi }_n(Y)\simeq 0$. Consider the map $p:X\times Y\to Y$ such that $p(x,y)=y$. Since $X\times Y$ is homeomorphic to ${𝕊}^n$ and ${\pi }_n(Y)\simeq 0$, then $p$ is homotopic to some constant map. Let $y_0\in Y$ and $H:\mathrm{I}\times X\times Y\to Y$ be such that

$$H(0,x,y)=p(x,y)=y;$$

$$H(1,x,y)=y_0.$$

Consider the map $F:\mathrm{I}\times X\times Y\to X\times Y$ defined by the formula

$$F(t,x,y)=(x,H(t,x,y)).$$

Note that $F(0,x,y)=(x,y)$ and $F(1,x,y)=(x,y_0)$ and thus $X\times \{y_0\}$ is a deformation retract of $X\times Y$. But $X\times Y$ is a sphere and spheres do not have proper deformation retracts (please see this entry (http://planetmath.org/EveryMapIntoSphereWhichIsNotOntoIsNullhomotopic) for more details). Therefore $X\times \{y_0\}=X\times Y$, so $Y=\{y_0\}$ has exactly one point. $\mathrm{\square }$