# proof of Brouwer fixed point theorem

Proof of the Brouwer fixed point theorem^{}:

Assume that there does exist a map from $f:{B}^{n}\to {B}^{n}$ with no fixed point^{}. Then let
$g(x)$ be the following map: Start at $f(x)$, draw the ray going through $x$ and then let $g(x)$ be
the first intersection^{} of that line with the sphere. This map is continuous^{} and well defined only
because $f$ fixes no point. Also, it is not hard to see that it must be the identity^{} on the boundary
sphere. Thus we have a map $g:{B}^{n}\to {S}^{n-1}$, which is the identity on
${S}^{n-1}=\partial {B}^{n}$, that is, a retraction^{}. Now, if $i:{S}^{n-1}\to {B}^{n}$ is the inclusion
map^{}, $g\circ i={\mathrm{id}}_{{S}^{n-1}}$. Applying the reduced homology functor^{}, we find that
${g}_{*}\circ {i}_{*}={\mathrm{id}}_{{\stackrel{~}{H}}_{n-1}({S}^{n-1})}$, where ${}_{*}$ indicates the induced map on homology^{}.

But, it is a well-known fact that ${\stackrel{~}{H}}_{n-1}({B}^{n})=0$ (since ${B}^{n}$ is contractible^{}), and that
${\stackrel{~}{H}}_{n-1}({S}^{n-1})=\mathbb{Z}$. Thus we have an isomorphism^{} of a non-zero group onto itself
factoring through a trivial group, which is clearly impossible. Thus we have a contradiction^{},
and no such map $f$ exists.

Title | proof of Brouwer fixed point theorem^{} |
---|---|

Canonical name | ProofOfBrouwerFixedPointTheorem |

Date of creation | 2013-03-22 13:11:24 |

Last modified on | 2013-03-22 13:11:24 |

Owner | bwebste (988) |

Last modified by | bwebste (988) |

Numerical id | 6 |

Author | bwebste (988) |

Entry type | Proof |

Classification | msc 47H10 |

Classification | msc 54H25 |

Classification | msc 55M20 |