# Brouwer's Fixed Point Theorem/Smooth Mapping

## Theorem

A smooth mapping $f$ of the closed unit ball $B^n \subset \R^n$ into itself has a fixed point:

$\forall f \in \map {C^\infty} {B^n \to B^n}: \exists x \in B^n: \map f x = x$

## Proof

Suppose there exists such a mapping $f$ of the unit ball to itself without fixed points.

Since $\map f x \ne x$, the two points $x$ and $\map f x$ are distinct and there is a unique ray from $x$ to $\map f x$ on which they both lie.

Call this line $L$ and let $\map h x = \partial B^n \cap L$.

If $x \in \partial B^n$, then $\map h x = x$ and $h$ restricts to the identity on $\partial B^n$.

Since $x$ is in the line segment between $\map f x$ and $\map h x$, one may write the vector $\map h x - \map f x$ as a multiple $t$ times the vector $x - \map f x$, where $t \ge 1$.

Hence:

$\map h x = t x + \paren {1 - t} \map f x$

Since $f$ is smooth, the smoothness of $t$ with respect to $x$ implies the smoothness of $h$.

Taking the dot product of both sides of this formula and noting that $\size {\map h x} = 1$:

$t^2 \size {x - \map f x}^2 + 2 t \map f x \cdot \paren {x - \map f x} + \size {\map f x}^2 - 1 = 0$

$\displaystyle t = \frac {\map f x \cdot \paren {\map f x - x} } {\size {x - \map f x}^2}$
an expression for $t$ in smooth terms of $x$.
Hence $h$ is a smooth retract of a compact manifold onto its boundary, which contradicts the Retraction Theorem.
$\blacksquare$