Brouwer fixed point theorem problem

Question

the proof for Brouwer fixed point theorem involved creating a function g that mapped the disk $D_n$ into its boundary, where $g(x)$ was obtained by extending a segment from $f(x)$ through $x$ and into the boundary of $D_n$.

Define such $g:D_n\to D_n$ with following properties

1. $g$ is continuous
2. $g$ maps all points of $D_n$ to boundary $D_n$
3. $g$ leaves boundary points fixed

which shows that there is no such $g$ exists.

Now suppose that, instead, we create $g$ in a different way: suppose that $g(x)$ is obtained by extending a segment from $x$ through $f(x)$ and into the boundary of $D_n$. Explain why the argument will fail if this "new" $g$ is used in place of the "original" $g$.

How can I explain this?

Answer 1

If $x$ is in the boundary, it should be mapped to itself. This is not so for your function, since $x \mapsto f(x)$ and extended to the boundary, but the resulting point cannot be $x$.