# the sphere is indecomposable as a topological space

Proposition. If for any topological spaces $X$ and $Y$ the $n$-dimensional sphere $\mathbb{S}^{n}$ 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 $\mathbb{S}^{n}$ is homeomorphic to $X\times Y$. Now $\pi_{n}(\mathbb{S}^{n})\simeq\mathbb{Z}$ and thus we have:

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

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 $\mathbb{S}^{n}$ 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. $\square$

Title the sphere is indecomposable as a topological space TheSphereIsIndecomposableAsATopologicalSpace 2013-03-22 18:31:45 2013-03-22 18:31:45 joking (16130) joking (16130) 8 joking (16130) Theorem msc 54F99