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.)

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    $\begingroup$ 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. $\endgroup$ – Arthur Jan 10 '18 at 5:55
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    $\begingroup$ You might be happier considering the ray from $f(x)$ through $x$ (and on to $\partial D^n$). So restrict $t \in (-\infty,1]$. $\endgroup$ – Eric Towers Jan 10 '18 at 6:00

I never got as far as splitting hairs and worrying which direction, just as long as you go along the line (segment) until you hit the boundary. .. it seems you can go in either direction (the result is continuous, since continuous functions are subtracted)

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...

Incidentally, this proof is due to Mo Hirsch, whom I had the pleasure of studying under at Berkeley. ..


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