Brouwer's Fixed Point Theorem

Recall from one of the results on The Induced Mapping from the Fundamental Groups of Two Topological Spaces page that if $S^1$ is not a retract of $D^2$. We will use this result to prove the famous Brouwer's fixed point theorem.

Theorem 1 (Brouwer's Fixed Point Theorem): Every continuous function from the closed unit disk onto itself has a fixed point. That is, if $f : D^2 \to D^2$ is a continuous then there exists a point $(x, y) \in D^2$ such that $f(x, y) = (x, y)$.

• Proof: Let $f : D^2 \to D^2$ be a continuous function and suppose that no fixed point exists. That is, for every $(x, y) \in D^2$ we have that $f(x, y) \neq (x, y)$.

• Since $f(x, y) \neq (x, y)$ there must exist a unique line starting at $f(x, y)$, passing through $(x, y)$, and then hitting the boundary of $D^2$ (the unit circle $S^1$. Define a function $g : D^2 \to S^1$ for each $(x, y) \in D^2$ by $g(x, y)$ will be the unique point on the boundary of $S^1$.

• Then $g$ is a clearly a continuous function, and $g \circ \mathrm{in} = \mathrm{id}_{S^1}$. So $g$ is a retraction of the unit circle $S^1$ with the closed unit disk $D^2$.

• But this is impossible from the result stated at the top of the page. So there exists and $(x, y) \in D^2$ such that $f(x, y) = (x, y)$. $\blacksquare$

Theorem 2 (Brouwer's Fixed Point Theorem for $D^n$): Every continuous function $f : D^n \to D^n$ has a fixed point.

Here, $D^1$ is the closed unit interval, $D^2$ is of course the closed unit disk, $D^3$ is the closed unit ball, etc…

The proof of Theorem 2 is analogous to that of Theorem 1, so we will omit it.

Theorem 3 (General Brouwer's Fixed Point Theorem): If $A \subset \mathbb{R}^n$ is homeomorphic to $D^n$ then every continuous function $f : A \to A$ has a fixed point.