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The standart way to prove the theorem is to assume that there are no fixed points for a function $f: D^{n} \rightarrow D^{n}$ and from that obtain a retraction $r: D^{n} \rightarrow \partial D^{n}$, which sends $x$ to the intersection of the ray $x + t(f(x) - x)$ with the boundary $\partial D^{n}$, from which we then can see the contradiction by examining the induced homomorphisms in the homology groups.

My question is: doesn't this actually give us a deformation retraction? If $t_{x}$ is $t \leq 0$, for which $||x + t_{x}(f(x) - x)|| = 1$, we can define $$ F(x, \lambda) = \begin{cases} (1 - \lambda) x + \lambda (x + t_{x}(f(x) - x)), & \text{if } x \in D^{n} \backslash \partial D^{n} \\ x & \text{if } x \in \partial D^{n} \end{cases} $$ which seems to be the required homotopy relative to $\partial D^{n}$, since $t_{x}$ is continuous as a function of $x$, so this is just a composition of continuous functions. Contradiction then is with the fact that homology is different for $D^{n}$ and $S^{n - 1}$, so we still come to the needed conclusion.

By doing a quick search before posting the question, it seems that retraction variant is more educationally useful, since idea can be used in the more general proof as seen in this question, but this seems to also work and I just want to check that I didn't miss something obvious here.

Sergey Rusakov
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  • The condition $x + t_{x}(f(x) - x) = 1$ does not make sense. – Paul Frost May 13 '21 at 16:22
  • @PaulFrost oh, yes, I meant the norm is equal to one, corrected it, thank you! – Sergey Rusakov May 13 '21 at 16:45
  • With the definition $r(x) = x + t_x(f(x) - x)$ you need $t_x \le 0$. In fact, $r(x) = x + t_x(f(x) - x) = x$ iff $t_x = 0$. – Paul Frost May 13 '21 at 22:37
  • @PaulFrost Not necessarily, I think. You can choose either $t_{x} > 0$ or $t_{x} < 0$, it just fixes to which of the two solutions of $||x + t_{x}(f(x) - x)|| = 1$ you are sending $x$ in a consistent way for all $x$ so retraction is continuous. – Sergey Rusakov May 13 '21 at 23:24
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    In fact you get two intersection points: One with $t_x \le 0$ and one with $t_x \ge 1$ (for $0 < t_x < 1$ you end up in the interior of $D^n$). But if you take the positive value $t_x$, you will not get a retraction. For each $x \in S^{n-1}$ you want to have $r(x) = x$ and this is true iff $t_x = 0$. Thus you must not choose $t_x > 0$. This is geometrically clear: Your ray starts at $x$ and it goes in positive direction through $f(x) \ne x$ . Thus the map $r$ will never keep $x$ fixed if you insist on $t_x > 0$. – Paul Frost May 13 '21 at 23:32
  • @PaulFrost Oh right, I see it now, was too concentrated on what happens to points in the interiour fo $D^{n}$, thanks! – Sergey Rusakov May 13 '21 at 23:43

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You have to take the ray $\bar f_x$ starting at $f(x)$ and going through $x$ which has parameterization $\bar f_x(t) = f(x) + t(x - f(x))$ with $t \ge 0$. There exists a unique $\tau(x) \ge 1$ such that $\lVert \bar f_x(\tau(x)) \rVert = 1$. The function $\tau : D^n \to \mathbb R$ turns out to be continuous. Clearly $\tau(x) = 1$ iff $x \in S^{n-1}$. Now we take

$$r(x) = \bar f_x(\tau(x)) = f(x) + \tau(x)(x - f(x)) .$$

Anyway, if we are given a retraction $r : D^n \to S^{n-1}$ (whatever its construction may be), then we can define

$$F : D^n \times I \to D^n, F(x,s) = (1-s)x + sr(x) .$$ This is well-defined because $D^n$ is convex and $x, r(x) \in D^n$. Clearly $F(x,0) = x$, $F(x,1) = r(x)$ and $F(x,t) = x$ for all $x \in S^{n-1}$. Thus $r$ is a strong deformation retraction.

Paul Frost
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