# 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. $\mathrm{\square}$

Title | the sphere is indecomposable as a topological space |
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Canonical name | TheSphereIsIndecomposableAsATopologicalSpace |

Date of creation | 2013-03-22 18:31:45 |

Last modified on | 2013-03-22 18:31:45 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 8 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 54F99 |