[proofplan] The proof splits into two parts. First, we show that the fixed-point statement reduces to the non-existence of a retraction $D^n \to S^{n-1}$: a continuous map $f: D^n \to D^n$ with no fixed point produces a retraction via a geometric ray construction, so ruling out retractions forces fixed points. Second, we rule out retractions using the functoriality of singular homology: if a retraction $r: D^n \to S^{n-1}$ existed, the composition $r_* \circ i_*$ would factor the identity on $H_{n-1}(S^{n-1}) \cong \mathbb{Z}$ through $H_{n-1}(D^n) = 0$, which is impossible. The case $n = 1$ is handled separately via connectedness. [/proofplan] [step:Reduce the fixed-point statement to non-existence of a retraction] Assume the no-retraction claim (proved in the next step). Let $f: D^n \to D^n$ be continuous; suppose for contradiction that $f(x) \neq x$ for every $x \in D^n$. Define \begin{align*} g: D^n &\to S^{n-1} = \partial D^n \\ x &\mapsto \text{the unique point where the ray from } f(x) \text{ through } x \text{ meets } S^{n-1}. \end{align*} Concretely, set $u(x) = x - f(x) \neq 0$ (nonzero by hypothesis) and let $g(x) = x + t(x)\, u(x)$, where $t(x) \ge 0$ is the unique non-negative value solving $|x + t\, u(x)|^2 = 1$. Expanding: \begin{align*} |x + t\, u(x)|^2 = |x|^2 + 2t\, x \cdot u(x) + t^2 |u(x)|^2 = 1 \end{align*} is a quadratic in $t$ with positive leading coefficient $|u(x)|^2 > 0$ and non-positive constant term $|x|^2 - 1 \le 0$. The discriminant is \begin{align*} \Delta(x) = 4(x \cdot u(x))^2 + 4|u(x)|^2(1 - |x|^2) \ge 0, \end{align*} and the unique non-negative root is \begin{align*} t(x) = \frac{-x \cdot u(x) + \sqrt{(x \cdot u(x))^2 + |u(x)|^2(1 - |x|^2)}}{|u(x)|^2}. \end{align*} For $x \in S^{n-1}$ (so $|x| = 1$), the formula gives $t(x) = 0$ provided $x \cdot u(x) \ge 0$. We verify: $x \cdot u(x) = x \cdot (x - f(x)) = |x|^2 - x \cdot f(x) = 1 - x \cdot f(x)$. By the Cauchy-Schwarz inequality, $x \cdot f(x) \le |x|\,|f(x)| = |f(x)| \le 1$ (since $f(x) \in D^n$), so $x \cdot u(x) \ge 0$. Thus $t(x) = 0$ and $g(x) = x$, so $g|_{S^{n-1}} = \operatorname{id}_{S^{n-1}}$. Continuity of $g$ follows from continuity of $u$, $x \cdot u(x)$, $|u(x)|^2$ (all polynomial in $x$ and $f(x)$), the non-vanishing of $|u(x)|^2$ on $D^n$ (by the fixed-point-free hypothesis), and continuity of the square root on $[0, \infty)$. So $g: D^n \to S^{n-1}$ is a continuous retraction, contradicting the no-retraction claim. Hence $f$ has a fixed point. [guided] The geometric picture: if $f(x) \neq x$ for every $x$, then for each $x \in D^n$ we have two distinct points $f(x)$ and $x$. Draw the ray starting at $f(x)$, passing through $x$, and continuing until it hits the boundary sphere $S^{n-1}$. Call the exit point $g(x)$. This defines a map $g: D^n \to S^{n-1}$. We must verify two properties to obtain a contradiction: **Retraction property ($g|_{S^{n-1}} = \operatorname{id}$).** If $x$ is already on the boundary $S^{n-1}$, the ray from $f(x)$ through $x$ exits $D^n$ at $x$ itself. Algebraically, this reduces to showing $t(x) = 0$ when $|x| = 1$, which requires $x \cdot u(x) \ge 0$. We compute: \begin{align*} x \cdot u(x) = x \cdot (x - f(x)) = 1 - x \cdot f(x) \ge 1 - |x|\,|f(x)| = 1 - |f(x)| \ge 0, \end{align*} where the first inequality is Cauchy-Schwarz and the second uses $f(x) \in D^n$, so $|f(x)| \le 1$. **Continuity.** The formula for $g(x) = x + t(x)\, u(x)$ involves polynomial expressions in $x$ and $f(x)$, division by $|u(x)|^2 > 0$ (nonzero because $f$ has no fixed points), and a square root of a non-negative quantity. Each of these operations is continuous, so $g$ is continuous on $D^n$. Since $g: D^n \to S^{n-1}$ is a continuous retraction, this contradicts the no-retraction claim proved below, so $f$ must have a fixed point. The case $n = 0$: $D^0 = \{0\}$ is a single point, so every self-map is the identity, which has a fixed point. The no-retraction statement is vacuously true since $S^{-1} = \varnothing$. [/guided] [/step] [step:Rule out retractions $r: D^n \to S^{n-1}$ via the $(n-1)$st homology functor] We show that no continuous retraction $r: D^n \to S^{n-1}$ exists for $n \ge 1$. **Case $n = 1$.** The disk $D^1 = [-1, 1]$ is connected, while $S^0 = \{-1, 1\}$ is disconnected (it has two path components). The continuous image of a connected space is connected, so any continuous $r: [-1, 1] \to \{-1, 1\}$ must be constant. But a retraction satisfies $r(-1) = -1$ and $r(1) = 1$, so $r$ is not constant. Contradiction. **Case $n \ge 2$.** Suppose a retraction $r: D^n \to S^{n-1}$ exists, and let $i: S^{n-1} \hookrightarrow D^n$ denote the inclusion. The retraction condition is $r \circ i = \operatorname{id}_{S^{n-1}}$. Applying the $(n-1)$st homology functor, which is functorial by [Continuous Maps Induce Homomorphisms on Homology](/theorems/1939): \begin{align*} H_{n-1}(S^{n-1}) \xrightarrow{i_*} H_{n-1}(D^n) \xrightarrow{r_*} H_{n-1}(S^{n-1}), \end{align*} with $r_* \circ i_* = (r \circ i)_* = (\operatorname{id}_{S^{n-1}})_* = \operatorname{id}_{H_{n-1}(S^{n-1})}$. By [Homology of Spheres](/theorems/1945), $H_{n-1}(S^{n-1}) \cong \mathbb{Z}$ (valid since $n - 1 \ge 1$). The disk $D^n$ is contractible: the straight-line homotopy $F: D^n \times [0,1] \to D^n$, $F(x, t) = (1-t)x$ deforms $\operatorname{id}_{D^n}$ to the constant map $c_0: x \mapsto 0$. By [Homotopic Maps Induce Equal Maps on Homology](/theorems/1944), $(\operatorname{id}_{D^n})_* = (c_0)_*$ on all homology groups. Since $c_0$ factors as $D^n \to \{0\} \to D^n$, and $H_k(\{0\}) = 0$ for $k \ge 1$, we get $(c_0)_* = 0$ in positive degrees. Hence $\operatorname{id}_{H_k(D^n)} = 0$ for $k \ge 1$, forcing $H_k(D^n) = 0$ for $k \ge 1$. In particular, $H_{n-1}(D^n) = 0$. The composition $r_* \circ i_*: \mathbb{Z} \to 0 \to \mathbb{Z}$ factors through the zero group, so $r_* \circ i_* = 0$. But $r_* \circ i_* = \operatorname{id}_{\mathbb{Z}} \neq 0$. This contradiction shows no retraction exists for $n \ge 2$. [guided] The strategy is to find an algebraic invariant that a retraction would have to preserve but cannot. The general principle: if $r: D^n \to S^{n-1}$ is a retraction of the inclusion $i: S^{n-1} \hookrightarrow D^n$, then $r \circ i = \operatorname{id}_{S^{n-1}}$. Applying any functor $F$ to this equation gives $F(r) \circ F(i) = \operatorname{id}_{F(S^{n-1})}$. If $F(D^n)$ is "too small" (e.g., the zero group), the identity on $F(S^{n-1})$ cannot factor through it. **Which functor?** We use $H_{n-1}$, the $(n-1)$st singular homology functor. This is functorial by [Continuous Maps Induce Homomorphisms on Homology](/theorems/1939): a continuous map $f: X \to Y$ induces a group homomorphism $f_*: H_k(X) \to H_k(Y)$, and $(f \circ g)_* = f_* \circ g_*$. **Computing $H_{n-1}(S^{n-1})$.** By [Homology of Spheres](/theorems/1945), $H_k(S^{n-1}) \cong \mathbb{Z}$ when $k = n-1$. We need $n - 1 \ge 1$ for this group to be nonzero, which is why $n \ge 2$ is required. **Computing $H_{n-1}(D^n)$.** The key is contractibility. The straight-line homotopy $F(x, t) = (1-t)x$ deforms $\operatorname{id}_{D^n}$ to the constant map $c_0: x \mapsto 0$. By [Homotopic Maps Induce Equal Maps on Homology](/theorems/1944), $(\operatorname{id}_{D^n})_* = (c_0)_*$. The constant map factors as $D^n \xrightarrow{\pi} \{0\} \xrightarrow{j} D^n$, so $(c_0)_* = j_* \circ \pi_*$. For $k \ge 1$, $H_k(\{0\}) = 0$ (a point has vanishing homology in positive degree), so $(c_0)_* = 0$. Since $(c_0)_* = (\operatorname{id}_{D^n})_* = \operatorname{id}_{H_k(D^n)}$, we get $\operatorname{id}_{H_k(D^n)} = 0$, which forces $H_k(D^n) = 0$ for all $k \ge 1$. **The contradiction.** The composition $r_* \circ i_*: H_{n-1}(S^{n-1}) \to H_{n-1}(D^n) \to H_{n-1}(S^{n-1})$ equals $\operatorname{id}_{\mathbb{Z}}$ (from $r \circ i = \operatorname{id}$) but must factor through $H_{n-1}(D^n) = 0$, so it equals $0$. Since $\operatorname{id}_{\mathbb{Z}} \neq 0$, no retraction can exist. **Why $n = 1$ is separate.** For $n = 1$, $H_0(S^0) = \mathbb{Z}^2$ and $H_0(D^1) = \mathbb{Z}$, so the homology argument does not directly produce a contradiction at the group level. Instead, connectedness suffices: $[-1,1]$ is connected but $\{-1, 1\}$ is not, so no continuous surjection $[-1,1] \to \{-1,1\}$ can fix both points. [/guided] [/step] [step:Combine both assertions to complete the proof] The previous step established that no continuous retraction $D^n \to S^{n-1}$ exists for any $n \ge 1$ (the $n = 0$ case is vacuous). The first step showed that the absence of retractions implies every continuous $f: D^n \to D^n$ has a fixed point. Combining the two completes the proof. [/step]