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.
Best Answer
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.