Brouwer fixed point theorem

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May 15th 2011, 09:40 PM
jackie

Brouwer fixed point theorem

I just learned the Brouwer Fixed Point Theorem which states that $D^2$ , the 2-sphere, has the fixed point property. I understand most part of the proof by contradiction by assuming that for any map $f: D^2 \rightarrow D^2$ , then $f(x)\neq x$ for every $x\in D^2$ , but I got stuck on why the function $r(x): D^2 \rightarrow S^1$ defined by $r(x)=p_x$ where $p_x$ is the intersection of the ray starting at $f(x)$ passing through $x$ and leaves $D^2$ , 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 $r$ is clear since small pertubations of x produces small pertubations of $f(x)$ , hence also small pertubations of the ray through these two points" How can I turn this into an $\epsilon-\delta$ proof?
May 15th 2011, 10:18 PM
Jose27

Take $1=|(1-s)x+sf(x)|$ , solve for $s\in \mathbb{R}$ and pick the nonnegative root ( $|.|$ 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 $f$ . But, that map, call it $R$ , is given by $R:ce^{i\theta}\mapsto e^{i\theta}$ where $ce^{i\theta}$ is the unique representation of an element of $\mathbb{D}^2$ as the product of something in $c\in(0,1)$ and $e^{i\theta}\in\mathbb{S}^1$ . From there it's clear how to proceed.
May 15th 2011, 11:24 PM
jackie

Quote:

Originally Posted by Jose27

Take $1=|(1-s)x+sf(x)|$ , solve for $s\in \mathbb{R}$ and pick the nonnegative root ( $|.|$ 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 $x$ and $f(x)$ , and set its Euclidean distance equals to 1. So, I have $1=\mid s(f(x)-x)+x\mid$ . I tempted to say that $s=\frac{1-x}{f(x)-x}$ , 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)