**Proof plan.** We argue by contradiction. Assuming $f$ has no fixed point, the map $x \mapsto x - f(x)$ is everywhere nonzero, which allows us to construct a ray from $f(x)$ through $x$ and follow it until it hits the boundary $\partial \bar{B}$. This defines a map $r: \bar{B} \to \partial \bar{B}$. The main work is showing that $r$ is continuous and fixes every boundary point (Claims 1–3), making it a continuous retraction. The No Retraction Theorem — the statement that no continuous retraction from $\bar{B}$ onto $\partial \bar{B}$ exists — then provides the contradiction. (The No Retraction Theorem is itself a consequence of either the non-existence of a smooth retraction via Stokes' theorem, or of the machinery of singular homology; we take it as given here.)

**Step 1 (Setup and the retraction candidate).**

Assume for contradiction that $f(x) \neq x$ for every $x \in \bar{B}$. For each $x \in \bar{B}$, the vector $x - f(x) \in \mathbb{R}^n$ is nonzero, so the ray \begin{align*} t &\mapsto x + t(x - f(x)), \quad t \geq 0 \end{align*} is well-defined and starts at $x$ (when $t = 0$) heading away from $f(x)$. Define $r(x)$ to be the point where this ray exits $\bar{B}$, i.e. $r(x) := x + \lambda(x)(x - f(x))$ where $\lambda(x) \geq 0$ is determined by the condition $\|r(x)\| = 1$.

**Step 2 (Explicit formula for $\lambda$).**

[claim: Existence And Uniqueness Of Lambda] For each $x \in \bar{B}$ with $f(x) \neq x$, there exists a unique $\lambda(x) \geq 0$ such that \begin{align*} \|x + \lambda(x)(x - f(x))\|^2 &= 1, \end{align*} and it is given by \begin{align*} \lambda(x) &= \frac{-\langle x,\, x - f(x) \rangle + \sqrt{\langle x,\, x - f(x) \rangle^2 + \|x - f(x)\|^2(1 - \|x\|^2)}}{\|x - f(x)\|^2}. \end{align*} [/claim]

[proof] Set $d(x) := x - f(x)$. Expanding the norm-squared condition gives \begin{align*} \|x + \lambda \, d(x)\|^2 &= \|x\|^2 + 2\lambda \langle x, d(x) \rangle + \lambda^2 \|d(x)\|^2 = 1. \end{align*} This is a quadratic in $\lambda$ with leading coefficient $a := \|d(x)\|^2 > 0$ (since $d(x) \neq 0$), linear coefficient $b := 2\langle x, d(x) \rangle$, and constant term $c := \|x\|^2 - 1 \leq 0$ (since $\|x\| \leq 1$). The discriminant is \begin{align*} b^2 - 4ac &= 4\langle x, d(x) \rangle^2 - 4\|d(x)\|^2(\|x\|^2 - 1) \\ &= 4\langle x, d(x) \rangle^2 + 4\|d(x)\|^2(1 - \|x\|^2) \\ &\geq 4\|d(x)\|^2(1 - \|x\|^2) \geq 0, \end{align*} so the quadratic has real roots. The two roots are \begin{align*} \lambda_\pm &= \frac{-\langle x, d(x) \rangle \pm \sqrt{\langle x, d(x) \rangle^2 + \|d(x)\|^2(1 - \|x\|^2)}}{\|d(x)\|^2}. \end{align*} Since $\sqrt{\langle x, d(x) \rangle^2 + \|d(x)\|^2(1 - \|x\|^2)} \geq |\langle x, d(x) \rangle|$, the root $\lambda_+$ satisfies $\lambda_+ \geq 0$, while $\lambda_- \leq 0$. The unique non-negative root is therefore $\lambda(x) := \lambda_+$. [/proof]

**Step 3 (Continuity of $r$).**

[claim: Continuity Of The Retraction] The map $r: \bar{B} \to \partial \bar{B}$ defined by $r(x) := x + \lambda(x)(x - f(x))$ is continuous. [/claim]

[proof] It suffices to show that $\lambda: \bar{B} \to [0, \infty)$ is continuous, since $r$ is then a composition of continuous maps. The formula for $\lambda(x)$ from Claim 1 is \begin{align*} \lambda(x) &= \frac{-\langle x,\, x - f(x) \rangle + \sqrt{\langle x,\, x - f(x) \rangle^2 + \|x - f(x)\|^2(1 - \|x\|^2)}}{\|x - f(x)\|^2}. \end{align*} The numerator is a composition of continuous functions: the inner product $\langle \cdot, \cdot \rangle$ and the norm $\|\cdot\|$ are continuous on $\mathbb{R}^n$, $f$ is continuous by hypothesis, and $t \mapsto \sqrt{t}$ is continuous on $[0, \infty)$. The expression under the square root is non-negative (as shown in Claim 1). The denominator $\|x - f(x)\|^2$ is continuous and strictly positive on $\bar{B}$ because $f(x) \neq x$ for every $x \in \bar{B}$ by assumption, and $\bar{B}$ is compact, so $\inf_{x \in \bar{B}} \|x - f(x)\| > 0$ (a continuous positive function on a compact set attains its infimum, which must be positive). Therefore $\lambda$ is a quotient of continuous functions with a strictly positive denominator, hence continuous. [/proof]

**Step 4 ($r$ fixes the boundary).**

[claim: Boundary Identity] For every $x \in \partial \bar{B}$, we have $r(x) = x$. [/claim]

[proof] Let $x \in \partial \bar{B}$, so $\|x\| = 1$. The formula for $\lambda(x)$ gives \begin{align*} \lambda(x) &= \frac{-\langle x,\, x - f(x) \rangle + \sqrt{\langle x,\, x - f(x) \rangle^2 + \|x - f(x)\|^2 \cdot 0}}{\|x - f(x)\|^2} \\ &= \frac{-\langle x,\, x - f(x) \rangle + |\langle x,\, x - f(x) \rangle|}{\|x - f(x)\|^2}. \end{align*} Now $\langle x, x - f(x) \rangle = \|x\|^2 - \langle x, f(x) \rangle = 1 - \langle x, f(x) \rangle$. Since $f(x) \in \bar{B}$ and $\|x\| = 1$, the Cauchy–Schwarz inequality gives $\langle x, f(x) \rangle \leq \|x\| \|f(x)\| \leq 1$, so $\langle x, x - f(x) \rangle \geq 0$. Therefore $|\langle x, x - f(x) \rangle| = \langle x, x - f(x) \rangle$, and the numerator becomes \begin{align*} -\langle x, x - f(x) \rangle + \langle x, x - f(x) \rangle &= 0. \end{align*} Hence $\lambda(x) = 0$, and $r(x) = x + 0 \cdot (x - f(x)) = x$. [/proof]

**Step 5 (Contradiction).**

Claims 2 and 3 establish that $r: \bar{B} \to \partial \bar{B}$ is a continuous retraction: it is continuous and satisfies $r(x) = x$ for all $x \in \partial \bar{B}$. However, the No Retraction Theorem states that there is no continuous retraction from $\bar{B}$ onto $\partial \bar{B}$ (the proof, via either Stokes' theorem or singular homology, shows that the existence of such a retraction would contradict topological properties of the sphere). This contradicts the existence of $r$, so the assumption that $f$ has no fixed point is false. There exists $\bar{x} \in \bar{B}$ with $f(\bar{x}) = \bar{x}$.