## Teorema del punto fijo de Brouwer

Nos damos un pequeño break de la teoría de juegos para hablar un poco del teorema del punto fijo de Brouwer. Como vimos en el post sobre teoría de juegos y el teorema del punto fijo éste es un teorema que fundamenta la teoría de juegos y un teorema muy importante en topología y topología algebraica. Veremos el teorema y su prueba en inglés.

The theorem states the following: Let ${f:D^2\rightarrow D^2}$ be a continuous map, where

$\displaystyle D^2 = \{(x, y)\in\mathbb{R}^2 : x^2 + y^2\leq 1\}$

Then ${f}$ has a fixed point, i.e., there is some point ${(x, y)\in D^2}$ with the property that ${f(x, y) = (x, y)}$

Proof Suppose that ${f:D^2\rightarrow D^2}$ does not have a fixed point, so that ${f(x, y) \neq (x, y)}$ for all ${(x, y) \in D^2}$. So, for each point ${(x, y) \in D^2}$ we get two points ${(x, y)}$ and ${f(x, y)}$, and we can draw a line through them both. Extend this line beyond ${(x, y)}$ until it meets the boundary of ${D^2}$ (i.e., ${\mathbb{S}^1}$), and let ${g(x, y)}$ be the point where this happens. So we get a function ${g : D^2\rightarrow \mathbb{S}^1}$ as in the picture.

This map ${g}$ is continuous, essentially because if ${(x', y')}$ is sufficiently close to ${(x, y)}$, then ${f(x', y')}$ will be close to ${f(x, y)}$ (since ${f}$ is continuous) and, hence, ${g(x', y')}$ will be reasonably close to ${g(x, y)}$. More rigorously, if ${A}$ is an open arc around ${g(x, y)}$, then there is some radius ${r}$ such that whenever ${(x', y')}$ is in the open ball ${B_r(x, y)}$ and ${f(x', y')}$ is in the open ball ${B_r(f(x, y))}$, then ${g(x', y')}$ is in ${A}$, as depicted below, where ${A}$ is indicated by a bold line, and the balls around ${(x, y)}$ and ${f(x, y)}$ are indicated by the dotted circles of their perimeters. Any straight line which passes through both balls will hit the circle in the region ${A}$.

Since ${f}$ is continuous, there is some radius ${\delta}$ such that ${f(x', y')\in B_r(f(x, y))}$ whenever ${(x', y')\in B_\delta(x, y)}$. Hence the preimage ${g^{-1}(A)}$ contains ${B_{\min(\delta,r)}(x, y)}$. The same argument can be applied to any point in the preimage, so ${g^{-1}(A)}$ is open, i.e., ${g}$ is continuous. If ${(x, y)}$ is on the boundary of ${D^2}$, then ${g(x, y) = (x, y)}$ no matter what ${f(x, y)}$ is. Now define a map

$\displaystyle F : \mathbb{S}^1\times I \rightarrow \mathbb{S}^1$

by ${F((x, y), t) = g(tx, ty)}$. This map ${F}$ is continuous, so we can think of it as a homotopy between the map ${h : \mathbb{S}^1\rightarrow \mathbb{S}^1}$ defined by ${ h(x, y) = F((x, y), 0)}$ and ${j : \mathbb{S}^1\rightarrow \mathbb{S}^1}$ defined by ${j(x, y) = F((x, y), 1)}$. Now ${h(x, y) = g(0, 0)}$ for all ${(x, y)}$, so ${h}$ is the constant map and thus ${\deg(h) = 0}$. On the other hand, however, ${j(x, y) = g(x, y) = (x, y)}$ for all ${(x, y)}$, so ${j}$ is the identity map and ${\deg(j) = 1}$. If ${F}$ is a homotopy between ${h}$ and ${j}$, then these degrees must be equal. Since they are not, the map ${F}$ cannot exist. Hence nor can ${g}$, showing in turn that the map ${f}$ must have had a fixed point in the first place.

El teorema de Brouwer es un teorema de punto fijo que dice que una aplicación continua de un conjunto convexo y compacto en si mismo tiene un punto fijo. Schauder y Tychonoff ademas probaron que el teorema sigue siendo valido para espacios normados; y también para espacios localmente compactos. Luitzen Egbertus Jan Brouwer fue el principal teórico del Intuicionismo Matemático y el fundador de la topologia moderna.

Referencias:
-Mathematics – The Harper Collins Dictionary. Borowski & Borwein 1991
-Essential Topology – Martin Crossley 2005
Brouwer fixed point theorem by Palmieri
The Brouwer fixed point theorem and the game of Hex

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### One Response to Teorema del punto fijo de Brouwer

1. varasdemate says:

Se me olvido mencionar que lo de homotopia se puede ver en cualquier libro de topologia. El mencionado como referencia es muy bueno.