# 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:

$\forall f \in C^\infty \left({B^n \to B^n}\right): \exists x \in B^n: f \left({x}\right) = x$

## Proof

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

Since $f \left({x}\right) \ne x$, the two points $x$ and $f \left({x}\right)$ are distinct and there is a unique straight line on which they both lie.

Call this line $L$ and let $h \left({x}\right) = \partial B^n \cap L$.

If $x \in \partial B^n$, then $h \left({x}\right) = x$ and $h$ restricts to the identity on $\partial B^n$.

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

Hence:

$h \left({x}\right) = t x + \left({1 - t}\right) f \left({x}\right)$

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 $\left|{h \left({x}\right)}\right| = 1$:

$t^2 \left|{x - f \left({x}\right)}\right|^2 + 2 t f \left({x}\right) \cdot \left({x - f \left({x}\right)}\right) + \left|{f \left({x}\right)}\right|^2 - 1 = 0$

$\displaystyle t = \frac {f \left({x}\right) \cdot \left({f \left({x}\right) - x}\right)} {\left|{x - f \left({x}\right)}\right|^2}$
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.
$\blacksquare$