Open ball does not have fixed point

Question

How we can prove that the open ball in $R^n$ does not have fixed point property (by algebraic topology concepts)?
I know $D^n$ -closed ball in $R^n$- has fixed point property by Brouwer's theorem, but can't see how to find the function which has not a fixed point in the aforementioned case.

Would be grateful for your help.

Answer 1

$\mathbb R^n$ does not have fixed point property (think of translation). The open ball in $\mathbb R^n$ is homeomorphic to $\mathbb R^n$, so it does not have fixed point property either.

Answer 2

Consider the open ball of radius $\frac 1 2$ around $(\frac 1 2, 0, 0, \ldots, 0)$, and the function $f$ which takes the $(n-1)$-dimensional cross-section at first coordinate $x$ to that of first coordinate $x^2$.