every map into sphere which is not onto is nullhomotopic

. 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 $y_{0}\not\in\mathrm{im}(f)$. It is well known that there is a homeomorphism $\phi:\mathbb{S}^{n}\setminus\{y_{0}\}\to\mathbb{R}^{n}$. Then we have an induced map

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

Since $\mathbb{R}^{n}$ is contractible, then there is $c\in\mathbb{R}^{n}$ such that $\phi\circ f$ 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 $\psi(x)=\phi^{-1}(x)$ (note that $\psi$ is not the inverse of $\phi$ because $\psi$ is not onto) and take any homotopy $H:\mathrm{I}\times X\to\mathbb{R}^{n}$ from $\phi\circ f$ 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

 $F(0,x)=\psi(H(0,x))=\psi(\phi(f(x)))=f(x);$
 $F(1,x)=\psi(H(1,x))=\psi(c)\in\mathbb{S}^{n}.$

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

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

Title every map into sphere which is not onto is nullhomotopic EveryMapIntoSphereWhichIsNotOntoIsNullhomotopic 2013-03-22 18:31:41 2013-03-22 18:31:41 joking (16130) joking (16130) 8 joking (16130) Theorem msc 55P99