Brouwer Fixed-Point Theorem

Imagine you have two sheets of graph paper of the same size, one atop the other. Your friend takes one piece, crumples it into a ball, and tossies it onto the other sheet so that no piece of the ball extends beyond the edge of the bottom paper. Brouwer's Fixed-Point theorem tells us that there is atleast one point in the ball that lies directly above its corresponding point on the flat piece, in other words, there is atleast one point that is directly above its original position.

The theorem works in other dimensions too. Imagine a bowl of punch. The theorem insists that when your friend stirs the punch, atleast one liquid particle is in the same position as it was before stirring.

In more precise language, a continuous function from an $n$-ball to an $n$-ball (where $n$>0 is the dimension) must have a fixed point.

Short of a proof, it is easier to understand why it works if you think of a one dimensional line. To be concrete, suppose we map the real number line $[0,1]$ continuously to $[\frac{1}{3},\frac{5}{6}]$ via $f(x) = \frac{1}{2}x + \frac{1}{3}$. We get the following map:

0 _________________________ 1 $\rightarrow$

           $\frac{1}{3}$________________ $\frac{5}{6}$

We notice that $\frac{1}{3} = f(0) > 0$ and that $f(x)$ continues to be greater than $x$ for a while. Since $\frac{5}{6} = f(1) < 1$, it must flip, and thereafter $f(x) < x$. In this case, Brouwer's Fixed-Point theorem, that there must exist a point with $f(x)=x$, follows directly from the Intermediate Value theorem. In fact, it is easy to check that there is a fixed point at $x=\frac{2}{3}$.



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