On pg53, Zeidler gives the Brouwer's Fixed Point Theorem
The continuous operator $A: M \to M$ has a fixed point provided $M$ is a compact, convex and nonempty set in a finite dimensional normed space over $\mathbb{K}$
Then he proceed to prove this using "intuitive proof"
Let $M$ be a closed disk in $\mathbb{R^2}$, and let $A:M\to M$ be a continuous operator.
Suppose $A:M\to M$ is fixed point free such that $Au \neq u$, $\forall u \in M$
Then construct an operator $R: M \to \partial M$ as follows:
For each point $u \in M$ follow the directed line segment from $Au$ to $u$ to intersection with $\partial M$ and let the intersection point be $Ru$
Obviously $R$ is a so called retraction that is $R$ is continuous and $Ru = u$, $ \forall u \in \partial M$
Intuitively such a retraction does not exist, hence the desired contradiction.
Why is it intuitive that such retraction does not exist? $Ru =u$ means that for each $u$ in the boundary of $M$, $Ru$ takes the point to the boundary which is still the same $u$, so it makes total sense to say $Ru =u$.
Secondly, if $Ru \neq u$, then $RAu \neq u$ hence $Au \neq u$...wouldn't this contradict the theorem?
Can someone please guide me through this? Thanks!
Regarding non-existence of the retraction: it can be done with the machinery of an algebraic topology.
Let's consider $f: D^{2} \rightarrow D^{2}$ - a continous map from the 2-dimensional ball to itself. Since $D^{2}$ is contractible, then $\pi_{1}(D^{2}) = \{0 \}$. $\partial D^{2} = S^{1}$, and $\pi_{1}(S^{1}) = \mathbb{Z}$. So, we see that this two spaces are not homotopy equivalent, in particular there is no retraction.
In order to proceed in the case of $D^{n}$, $n>2$, it's sufficient to compute the $n-1$th homology groups of $D^{n}$ and $\partial D^{n}$.