[proofplan] We argue by contradiction. Assuming $f$ has no fixed point, we construct a retraction $r: D^n \to S^{n-1} = \partial D^n$ by sending each $x \in D^n$ to the point where the ray from $f(x)$ through $x$ meets $S^{n-1}$. The composition $S^{n-1} \hookrightarrow D^n \xrightarrow{r} S^{n-1}$ is the identity, making $r$ a retraction. Applying the functor $H_{n-1}$ produces a factorisation of the identity on $H_{n-1}(S^{n-1}) \cong \mathbb{Z}$ through $H_{n-1}(D^n) = 0$, which is impossible. [/proofplan] [step:Construct a retraction $r: D^n \to S^{n-1}$ assuming $f$ has no fixed point] Suppose for contradiction that $f(x) \neq x$ for all $x \in D^n$. For each $x \in D^n$, consider the ray starting at $f(x)$ and passing through $x$. Parametrise this ray as \begin{align*} \ell_x(t) := (1 - t) f(x) + t \, x, \quad t \geq 0. \end{align*} At $t = 0$, $\ell_x(0) = f(x) \in D^n$. As $t \to \infty$, $|\ell_x(t)| \to \infty$. Since $|\ell_x(0)| \leq 1$ and $t \mapsto |\ell_x(t)|$ is a continuous function that tends to $\infty$, by the intermediate value theorem there exists a unique $t_x \geq 0$ such that $|\ell_x(t_x)| = 1$ and $t_x$ is the largest such value (i.e., the ray exits the closed ball). Define \begin{align*} r: D^n &\to S^{n-1} \\ x &\mapsto \ell_x(t_x) = (1 - t_x) f(x) + t_x \, x. \end{align*} [guided] Why does the ray from $f(x)$ through $x$ hit $S^{n-1}$? The ray starts at $f(x) \in D^n$ (since $f: D^n \to D^n$) and proceeds in the direction $x - f(x)$, which is nonzero by the assumption $f(x) \neq x$. As $t$ increases, $\ell_x(t)$ moves away from $f(x)$ toward and past $x$, eventually leaving $D^n$. To find the exit point, we solve $|\ell_x(t)|^2 = 1$ for $t \geq 0$. Expanding: \begin{align*} |(1 - t) f(x) + t x|^2 &= (1 - t)^2 |f(x)|^2 + 2t(1 - t) f(x) \cdot x + t^2 |x|^2 = 1. \end{align*} This is a quadratic in $t$ with leading coefficient $|x - f(x)|^2 > 0$ (since $x \neq f(x)$). Since the quadratic takes the value $|f(x)|^2 \leq 1$ at $t = 0$ and tends to $+\infty$, there is a largest root $t_x$, which depends continuously on $x$ (by the quadratic formula and the continuous dependence of the coefficients on $x$). Why is $r$ a retraction? If $x \in S^{n-1}$, then $|x| = 1$ already, and the largest $t$ with $|\ell_x(t)| = 1$ is $t_x = 1$ (at which point $\ell_x(1) = x$). So $r(x) = x$ for all $x \in S^{n-1}$. [/guided] [/step] [step:Verify that $r$ is continuous and satisfies $r|_{S^{n-1}} = \operatorname{id}_{S^{n-1}}$] **Continuity.** The function $x \mapsto t_x$ is continuous on $D^n$: the coefficients of the quadratic $|(1 - t) f(x) + t x|^2 - 1 = 0$ depend continuously on $x$ (since $f$ is continuous), and the larger root of a quadratic with positive leading coefficient depends continuously on the coefficients (provided the discriminant is non-negative, which it is since $t = 0$ gives $|f(x)|^2 - 1 \leq 0$). Since $r(x) = (1 - t_x) f(x) + t_x \, x$ is a composition of continuous functions, $r$ is continuous. **Retraction property.** For $x \in S^{n-1}$, we have $|x| = 1$, so $t = 1$ satisfies $|\ell_x(1)| = |x| = 1$. For $t > 1$, $\ell_x(t) = f(x) + t(x - f(x))$ moves further from the origin (since $|x| = 1$ and the ray is heading outward). Therefore $t_x = 1$ and $r(x) = \ell_x(1) = x$. Hence $r|_{S^{n-1}} = \operatorname{id}_{S^{n-1}}$. [/step] [step:Derive a contradiction from the induced maps on homology] The retraction $r: D^n \to S^{n-1}$ satisfies $r \circ \iota = \operatorname{id}_{S^{n-1}}$, where $\iota: S^{n-1} \hookrightarrow D^n$ is the inclusion. Applying the functor $H_{n-1}$: \begin{align*} r_* \circ \iota_* = (\operatorname{id}_{S^{n-1}})_* = \operatorname{id}_{H_{n-1}(S^{n-1})}. \end{align*} This means the identity on $H_{n-1}(S^{n-1})$ factors through $H_{n-1}(D^n)$: \begin{align*} H_{n-1}(S^{n-1}) \xrightarrow{\;\iota_*\;} H_{n-1}(D^n) \xrightarrow{\;r_*\;} H_{n-1}(S^{n-1}). \end{align*} By the [Homology of $S^n$](/theorems/2242), $H_{n-1}(S^{n-1}) \cong \mathbb{Z}$ (for $n \geq 2$; for $n = 1$, $H_0(S^0) \cong \mathbb{Z}^2$). Since $D^n$ is contractible (it deformation retracts to the origin via $F(t, x) = (1-t)x$), $H_{n-1}(D^n) = 0$ for $n - 1 \geq 1$, i.e., for $n \geq 2$. The identity on a nonzero group cannot factor through the zero group: if $H_{n-1}(D^n) = 0$, then $\iota_*$ has image $\{0\}$, so $r_* \circ \iota_*$ is the zero map, which contradicts $r_* \circ \iota_* = \operatorname{id}_{H_{n-1}(S^{n-1})} \neq 0$. For $n = 1$: the disk $D^1 = [-1, 1]$ is path-connected, so $H_0(D^1) \cong \mathbb{Z}$, while $S^0 = \{-1, 1\}$ has $H_0(S^0) \cong \mathbb{Z}^2$. A retraction $r: D^1 \to S^0$ would induce a surjection $r_*: H_0(D^1) \cong \mathbb{Z} \twoheadrightarrow H_0(S^0) \cong \mathbb{Z}^2$ (since $r_* \circ \iota_* = \operatorname{id}$, the map $r_*$ is a left inverse to $\iota_*$ and hence surjective). But there is no surjection from $\mathbb{Z}$ to $\mathbb{Z}^2$ (since $\mathbb{Z}^2$ is not cyclic), giving the same contradiction. Therefore the assumption that $f$ has no fixed point leads to a contradiction, completing the proof. [guided] The logical structure of this proof is: "no retraction" $\implies$ "fixed point." The contrapositive is: if $f$ has no fixed point, then one can construct a retraction $r: D^n \to S^{n-1}$, which is impossible for algebraic-topological reasons. Why is there no retraction from $D^n$ to $S^{n-1}$? A retraction $r$ gives a factorisation \begin{align*} \operatorname{id}: H_{n-1}(S^{n-1}) \xrightarrow{\iota_*} H_{n-1}(D^n) \xrightarrow{r_*} H_{n-1}(S^{n-1}). \end{align*} The middle group is $H_{n-1}(D^n) = 0$ (since $D^n$ is contractible), while the outer groups are $H_{n-1}(S^{n-1}) \cong \mathbb{Z} \neq 0$. The identity on a nonzero group cannot factor through zero, because any homomorphism through the zero group is the zero map. This "no retraction" theorem is the topological content; Brouwer's fixed point theorem is a corollary via the geometric construction of $r$ from a fixed-point-free map. The $n = 1$ case is worth noting separately. Here $D^1 = [-1, 1]$ and $S^0 = \{-1, 1\}$. A continuous map $f: [-1, 1] \to [-1, 1]$ with $f(-1) \neq -1$ and $f(1) \neq 1$ would have $f(-1) > -1$ and $f(1) < 1$. The function $g(x) = f(x) - x$ satisfies $g(-1) > 0$ and $g(1) < 0$, so by the intermediate value theorem, $g(c) = 0$ for some $c \in (-1, 1)$, i.e., $f(c) = c$. This elementary argument does not use homology, but the homological proof works uniformly in all dimensions. [/guided] [/step]