Brouwer's fixed point theorem

Question

Theorem: If $f:D^n\rightarrow D^n$ is continuous then there is $x \in D^n$ such that $f(x)=x$.

To prove the theorem we assume that $f$ is cts but has no fixed point, that is $f(x)\neq x$ for all $x\in D^n$.

We define $g: D^n\rightarrow D^n$ where $g$ is well-defined if $f$ has no fixed point. Also, $g$ has the following properties:

1. $g$ is cts.
2. $g$ maps $D^n$ into boundary.
3. $g$ leaves boundary points fixed.

Now consider another function $h$ that is obtained by extending a segment from $x$ through $f(x)$ and into the boundary of $D^n$. Note that like $g$, the function $h$ also maps $D^n$ into its boundary.

But then the proof of the Brouwer FPT fails if $g$ is replaced with $h$. Why is it so?

Answer 1

The map $g$ only exists if $f$ with the required properties exist. If $g$ exists then you reach a contradiction and so $g$ can not exist which implies $f$ with the required properties can not exist. Similarly $h$ only exists if such an $f$ exists (as if $f$ fixes a point then you can not extend a segment from $x$ through $f(x)$ in any well-defined continuous way). If $h$ exists then we reach a contradiction in the same way and so such an $h$, and hence such an $f$, cannot exist.