# Brouwer fixed point theorem

• May 15th 2011, 09:40 PM
jackie
Brouwer fixed point theorem
I just learned the Brouwer Fixed Point Theorem which states that $\displaystyle D^2$, the 2-sphere, has the fixed point property. I understand most part of the proof by contradiction by assuming that for any map $\displaystyle f: D^2 \rightarrow D^2$, then $\displaystyle f(x)\neq x$ for every $\displaystyle x\in D^2$, but I got stuck on why the function $\displaystyle r(x): D^2 \rightarrow S^1$ defined by $\displaystyle r(x)=p_x$ where $\displaystyle p_x$ is the intersection of the ray starting at $\displaystyle f(x)$ passing through $\displaystyle x$ and leaves $\displaystyle D^2$, is a continuous function. The author (Hatcher) gives an explanation for it, but I can't really justify it with a rigorous proof. He said that "continuity of $\displaystyle r$ is clear since small pertubations of x produces small pertubations of $\displaystyle f(x)$, hence also small pertubations of the ray through these two points" How can I turn this into an $\displaystyle \epsilon-\delta$ proof?
• May 15th 2011, 10:18 PM
Jose27
Take $\displaystyle 1=|(1-s)x+sf(x)|$, solve for $\displaystyle s\in \mathbb{R}$ and pick the nonnegative root ($\displaystyle |.|$ denotes euclidean distance).
• May 15th 2011, 10:33 PM
Drexel28
Also, you only need to prove that the map which takes a point of the disc to its corresponding circle point is cont. since your map is the composition of that with $\displaystyle f$. But, that map, call it $\displaystyle R$, is given by $\displaystyle R:ce^{i\theta}\mapsto e^{i\theta}$ where $\displaystyle ce^{i\theta}$ is the unique representation of an element of $\displaystyle \mathbb{D}^2$ as the product of something in $\displaystyle c\in(0,1)$ and $\displaystyle e^{i\theta}\in\mathbb{S}^1$. From there it's clear how to proceed.
• May 15th 2011, 11:24 PM
jackie
Quote:

Originally Posted by Jose27
Take $\displaystyle 1=|(1-s)x+sf(x)|$, solve for $\displaystyle s\in \mathbb{R}$ and pick the nonnegative root ($\displaystyle |.|$ denotes euclidean distance).

Thank you very much for your help, Jose, but I don't really follow the hint (Sorry. I'm slow on this). So you write the equation of the ray passing through $\displaystyle x$ and $\displaystyle f(x)$, and set its Euclidean distance equals to 1. So, I have $\displaystyle 1=\mid s(f(x)-x)+x\mid$. I tempted to say that $\displaystyle s=\frac{1-x}{f(x)-x}$, but that does not seem right since the expression is Euclidean distance.

Thanks a lot for your help too, Drexel. I'll try to work on your hint tomorrow morning. My brain is not working now.(Worried)