Proof:

We assume the contrary, that is to say a continuous function $f$ of the closed unit disc into itself exists which has no fixed points. We produce a retraction from $E^2$ onto $S^1$ which would be a contradiction.

Figure: $E^2$ is not a retract of $S^1$

[width=.4]GKSBook/fig12/fig12.eps

The ray emanating from $f(x) \in E^2$ and passing through $x \in E^2$ namely

$$tx + (1-t)f(x),\quad t\geq 0,$$

meets the circle $S^1$ at a point denoted by $r(x) = t_0x + (1-t_0)f(x)$ where, $t_0$ is a root of the quadratic

$$\langle tx + (1-t)f(x),\; tx + (1-t)f(x)\rangle = 1. \eqno(10.1)$$

We recast this quadratic as

$$t^2({\vert f({\bf x})- {\bf x}\vert}^2) - 2tf({\bf x}) \cdot (f({\bf x}) - {\bf x}) - (1 - {\vert f({\bf x})\vert}^2) = 0. \eqno(10.2)$$

Since the coefficient of $t^2$ is never zero, the roots are continuous functions of ${\bf x}$ and they are real. Moreover the roots differ in sign or one of the roots is zero. Take $t_0$ to be the larger root for constructing $r(x)$. From (10.1) we see that $r$ maps $E^2$ to $S^1$. Note that if $\vert x\vert = 1$ then $t = 1$ satisfies the quadratic and so must be the larger root. Hence we conclude $r(x) = x$ if $\vert x\vert = 1$ and we get a retraction of $E^2$ onto $S^1$ which is a contradiction. $\square$