# every map into sphere which is not onto is nullhomotopic

Proposition^{}. Let $X$ be a topological space^{} and $f:X\to {\mathbb{S}}^{n}$ a continous map from $X$ to $n$-dimensional sphere which is not onto. Then $f$ is nullhomotopic.

Proof. Assume that there is ${y}_{0}\in {\mathbb{S}}^{n}$ such that ${y}_{0}\notin \mathrm{im}(f)$. It is well known that there is a homeomorphism $\varphi :{\mathbb{S}}^{n}\setminus \{{y}_{0}\}\to {\mathbb{R}}^{n}$. Then we have an induced map

$$\varphi \circ f:X\to {\mathbb{R}}^{n}.$$ |

Since ${\mathbb{R}}^{n}$ is contractible, then there is $c\in {\mathbb{R}}^{n}$ such that $\varphi \circ f$ is homotopic to the constant map in $c$ (denoted with the same symbol $c$). Let $\psi :{\mathbb{R}}^{n}\to {\mathbb{S}}^{n}$ be a map such that $\psi (x)={\varphi}^{-1}(x)$ (note that $\psi $ is not the inverse^{} of $\varphi $ because $\psi $ is not onto) and take any homotopy^{} $H:\mathrm{I}\times X\to {\mathbb{R}}^{n}$ from $\varphi \circ f$ to $c$. Then we have a homotopy $F:\mathrm{I}\times X\to {\mathbb{S}}^{n}$ defined by the formula^{} $F=\psi \circ H$. It is clear that

$$F(0,x)=\psi (H(0,x))=\psi (\varphi (f(x)))=f(x);$$ |

$$F(1,x)=\psi (H(1,x))=\psi (c)\in {\mathbb{S}}^{n}.$$ |

Thus $F$ is a homotopy from $f$ to a constant map. $\mathrm{\square}$

Corollary. If $A\subseteq {\mathbb{S}}^{n}$ is a deformation retract^{} of ${\mathbb{S}}^{n}$, then $A={\mathbb{S}}^{n}$.

Proof. If $A\subseteq X$ then by deformation retraction (associated to $A$) we understand a map $R:\mathrm{I}\times X\to X$ such that $R(0,x)=x$ for all $x\in X$, $R(1,a)=a$ for all $a\in A$ and $R(1,x)\in A$ for all $x\in X$. Thus a deformation retract is a subset $A\subseteq X$ such that there is a deformation retraction $R:\mathrm{I}\times X\to X$ associated to $A$.

Assume that $A$ is a deformation retract of ${\mathbb{S}}^{n}$ and $A\ne {\mathbb{S}}^{n}$. Let $R:\mathrm{I}\times {\mathbb{S}}^{n}\to {\mathbb{S}}^{n}$ be a deformation retraction. Then $r:{\mathbb{S}}^{n}\to {\mathbb{S}}^{n}$ such that $r(x)=R(1,x)$ is homotopic to the identity map (by definition of a deformation retract), but on the other hand it is homotopic to a constant map (it follows from the proposition, since $r$ is not onto, because $A$ is a proper subset^{} of ${\mathbb{S}}^{n}$). Thus the identity map is homotopic to a constant map, so ${\mathbb{S}}^{n}$ is contractible. Contradiction^{}. $\mathrm{\square}$

Title | every map into sphere which is not onto is nullhomotopic |
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Canonical name | EveryMapIntoSphereWhichIsNotOntoIsNullhomotopic |

Date of creation | 2013-03-22 18:31:41 |

Last modified on | 2013-03-22 18:31:41 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 8 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 55P99 |