Brouwer fixed-point theorem

This article describes a theorem about spheres

Statement

In the language of spheres

Any continuous map from a disc to itself must have a fixed point. In other words, for any natural number $n$ , if $D^{n}$ denotes the spherical disc in $R^{n}$ , any continuous map $f : D^{n} \to D^{n}$ must have a point $x$ such that $f (x) = x$ .

In the language of simplices

Any continuous map from the standard $n$ -simplex, to itself has a fixed point.

Particular cases

Case $n = 1$

This says that any continuous map from the closed unit interval $[0, 1]$ to itself has a fixed point. This particular case is often proved as a consequence of the intermediate value theorem for continuous real-valued functions. Specifically, if $f : [0, 1] \to [0, 1]$ is the function, then the function $g (x) : = f (x) - x$ crosses over from a non-positive to a nonnegative function and hence must be zero for some intermediate value of $x$ .

Facts used

No-retraction theorem: This states that there does not exist a continuous retraction from $D^{n}$ to $S^{n - 1}$ , i.e., there is no continuous map from $D^{n}$ to $S^{n - 1} = \partial D^{n}$ that restricts to the identity map on $S^{n - 1}$ .

Proof

The Brouwer fixed-point theorem follows easily from the no-retraction theorem. Suppose $f : D^{n} \to D^{n}$ is a continuous map with no fixed points. Define a map $g : D^{n} \to S^{n - 1}$ , that sends $x \in D^{n}$ to the unique point on $S^{n - 1}$ that is colllinear with $x$ and $f (x)$ in such a way that $x$ lies between that point and $f (x)$ . We can see that:

Since $f (x)$ is never equal to $x$ , and $x$ is inside the unit disc, $g$ is well-defined throughout $D^{n}$
$g$ is continuous
$g$ is a retraction because it fixes every point on $S^{n - 1}$