# equivalent statements to statement that sphere is not contractible

Let $V$ be a normed space. Recall the definition of the sphere and the ball in $V$:

 $\mathbb{S}=\{v\in V;\ \ \|v\|=1\};\ \ \ \mathbb{B}=\{v\in V;\ \ \|v\|\leq 1\}.$

The following are equivalent      :

$\mathrm{(1)}$ $\mathbb{S}$ is not contractible;

$\mathrm{(2)}$ for each continous map $F:\mathbb{B}\rightarrow\mathbb{B}$ there exists $x\in\mathbb{B}$ such that $F(x)=x$;

$\mathrm{(3)}$ there is no retraction  from $\mathbb{B}$ onto $\mathbb{S}$.

Proof. The proof of this proposition probably can be found in some books about topology  . I present here the proof from my lecture due to Prof. $\mathrm{G\acute{o}rniewicz}$.

$\mathrm{(1)}\Rightarrow\mathrm{(2)}$ Assume there exists a continous map $F:\mathbb{B}\rightarrow\mathbb{B}$ such that for each $x\in\mathbb{B}$ we have $F(x)\neq x$. Define a map $H:\mathbb{S}\times[0,1]\rightarrow\mathbb{S}$ as follows:

 $H(x,t)=\left\{\begin{matrix}\frac{x-2tF(x)}{\|x-2tF(x)\|},&\mbox{if }0\leq t% \leq\frac{1}{2}\\ &\\ \frac{(2-2t)x-F((2-2t)x)}{\|(2-2t)x-F((2-2t)x)\|}&\mbox{if }\frac{1}{2}\leq t% \leq 1\end{matrix}\right.$

Thanks to the condition $F(x)\neq x$ this map is well defined and it is easy to check that this is a homotopy  from the identity map to constant map. But $\mathbb{S}$ is not contractible. Contradiction   .

$\mathrm{(2)}\Rightarrow\mathrm{(3)}$ Assume there exists a retraction $r:\mathbb{B}\rightarrow\mathbb{S}$. Define a map $F:\mathbb{B}\rightarrow\mathbb{B}$ by the formula   $F(x)=-r(x)$. This map has no fixed point  . Contradiction.

$\mathrm{(3)}\Rightarrow\mathrm{(1)}$ Assume that $\mathbb{S}$ is contractible and take any homotopy $H:\mathbb{S}\times[0,1]\rightarrow\mathbb{S}$ from constant map to identity map, i.e. for all $x\in\mathbb{S}$ we have $H(x,0)=x_{0}$ (for some $x_{0}\in\mathbb{S}$) and $H(x,1)=x$. Define a map $r:\mathbb{B}\rightarrow\mathbb{S}$ as follows:

 $r(x)=\left\{\begin{matrix}x_{0},&\mbox{if }\|x\|\leq\frac{1}{2}\\ &\\ H(\frac{x}{\|x\|},2\|x\|-1)&\mbox{if }\|x\|\geq\frac{1}{2}\end{matrix}\right.$

It is easy to see that this formula defines a retraction from $\mathbb{B}$ onto $\mathbb{S}$. Contradiction. $\square$

Note that this proposition does not state that any of the conditions $\mathrm{(1)},\mathrm{(2)},\mathrm{(3)}$ hold. It only states that they are equivalent. It is well known that all of them are true if $V$ is finite dimensional and all are false if $V$ is infinite dimensional.

Title equivalent statements to statement that sphere is not contractible EquivalentStatementsToStatementThatSphereIsNotContractible 2013-03-22 18:07:53 2013-03-22 18:07:53 joking (16130) joking (16130) 10 joking (16130) Theorem msc 55P99