Home > Analysis, Fixed Point > Brouwer fixed point theorem

## Brouwer fixed point theorem

Let $D$ be an open subset of $\Bbb{R}^n$ such that $\overline{D}$ is homeomorphic to the closed unit ball $\overline{B}$ ( in $\Bbb{R}^n$). If $\phi \in C(\overline{D})$ (continuous function on $\overline{D}$) and $\phi(\overline{D})\subset\overline{D}$, then $\phi$ has a fixed point in $\overline{D}$.

Proof: First reduce the problem to the case where $D=B$. To do this, consider the homeomorphism $h:\overline{D} \to \overline{B}$. Then $\psi=h \circ \phi \circ h^{-1}$ maps $\overline{B}$ to $\overline{B}$, and $\psi(y)=y,\ y \in \overline{B}$ implies (since there exists $x \in \overline{D}$ such that $h(x)=y$) we have $h(\phi(x))=h(x)$, which means $\phi(x)=x$. This reduces the problem to consider our domain the unit ball in $\Bbb{R}^n$.

Now, if $\psi(x_0)=x_0$ for some $x_0 \in \partial B$ then we are done. Suppose this does not happen. Consider the homotopy $h_t(x)=x-t\psi(x), x \in \overline{B}, 0 \leq t\leq 1$.  It is clear that if $x \in \partial B$ and $0 \leq t <1$, then $t \psi(x) \in B$, hence $h_t(x)\neq 0, \forall x \in \partial B, 0\leq t<1$. By hypothesis, $o \notin h_1(\partial B)$. By invariance of topological degree of a mapping at a point which is not in the image of the boundary of the open set on which the mappings are is defined, under a homotopy, we have $d(I-\psi,B,0)=d(I,B,0)$ ($I$ is the identity mapping). But $d(I,B,0)=1$ since $0 \in B$.

If the topology degree of a mapping $\theta$ at a point is not $0$, then there exists a point $x$ in the domain of $\theta$ such $\theta(x)=0$. Therefore, there exists an element $x \in B$ such that $x-\psi(x)=0$, meaning that $\psi$, and therefore $\phi$ has a fixed point.