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 $\mathbb {R} ^{n}$, 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

1. 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 $S^{n-1}$