# Brouwer's fixed point theorem: algebraic part of proof

Every continuous map $f:D^n\rightarrow D^n$ (closed unit disk in $\mathbb{R}^n$) has a fixed point.

The proof of above theorem is easier with some tools in algebraic topology. While understanding the proof, I came across a point in algebraic part of proof. The proof is broadly as follows.

(i) suppose $f(x)\neq x$ $\forall x$. The line $t\mapsto (1-t)x+tf(x)$ intersects $\partial D^n$ in exactly two points.

(ii) Let $g_1(x)$ and $g_2(x)$ be the two points of above line which are on $\partial D^n$.

(iii) Then $x\mapsto g_1(x)$ or $x\mapsto g_2(x)$ is continuous map and one of them is retraction of $D^n$ to $\partial D^n$.

(iv) Apply homological technique to get contradiction. Hence $f(x)=x$ for some $x$.

Question. Are both $g_1$ and $g_2$ retractions or only one of them is retraction?

Partial answer. $(1-t)x+tf(x)=x-t(x-f(x))$will lie on $\partial D^n$ iff $\| x-t(x-f(x))\|=1$ $$\mbox{ i.e. } t^2.\|x-f(x)\|^2-2t\langle x,x-f(x)\rangle+\|x\|^2=1.\hskip5mm(*)$$ The discriminant of above quadratic equation in $t$ is $$\Delta=4\langle x,x-f(x)\rangle^2-4(\|x\|^2-1).\|x-f(x)\|^2$$ Easy to show: $\Delta>0$ for all $x\in D^n$. Thus for (*), there are two values of $t$ for each $x$ namely $$t_1=\frac{2\langle x,x-f(x)\rangle + \sqrt{\Delta}}{2\|x-f(x)\|^2} \mbox{ and } t_2=\frac{2\langle x,x-f(x)\rangle - \sqrt{\Delta}}{2\|x-f(x)\|^2}.$$ Let $$g_1(x)=(1-t_1)x+t_1f(x) \mbox{ and } g_2(x)=(1-t_2)(x)+t_2f(x).$$ What I saw was that when $x\in \partial D^n$ then $g_2(x)=x$ i.e. $g_2$ is a retraction. I was not getting that $g_1(x)=x$ for $x\in \partial D^n$; am I right? This is exactly the question I asked above before partial answer. Also, algebraically can we simplify some arguments to decide which $g_i$ is retraction? (Topologically or pictorially, Hatcher in his book explains in just 1-2 line how retraction is obtained.)

• If I'm allowed to be nit-picky: Since $g_1(x)$ and $g_2(x)$ are indexed with $1$ and $2$ for each separate $x$, neither of them are necessarily continuous. But there is a function that is sometimes equal to $g_1$ and some times equal to $g_2$ and always equal to one of them which is continuous. – Arthur Jan 10 '18 at 5:55
• You might be happier considering the ray from $f(x)$ through $x$ (and on to $\partial D^n$). So restrict $t \in (-\infty,1]$. – Eric Towers Jan 10 '18 at 6:00

Of course, there can be no such retraction, because the ball (contractible) is simply connected, and the circle isn't. .. or the nth homology, $H^n (S^n)=Z$... please excuse any sloppiness, I haven't done this stuff in a while...