The 2D Brouwer fixed-point theorem

Step 1 — Suppose for contradiction $f$ has no fixed point. Then $f(x) \ne x$ for every $x \in D^2$, so the ray from $f(x)$ through $x$ is well-defined; let $r(x) \in S^1 = \partial D^2$ be the point where this ray exits the disk. Since $f(x) \ne x$ varies continuously and never vanishes, $r: D^2 \to S^1$ is continuous.

Step 2 — $r$ is a retraction. For $x \in S^1$ already on the boundary, the ray from $f(x)$ through $x$ exits exactly at $x$ itself (since $x$ is already on the boundary and the ray only crosses the boundary once going outward from inside), so $r(x) = x$ for all $x \in S^1$. Thus $r: D^2 \to S^1$ is a retraction: a continuous map fixing the subspace $S^1$ pointwise.

Step 3 — Retractions are impossible: apply the functor $\pi_1$. Let $i: S^1 \hookrightarrow D^2$ be inclusion. Since $r \circ i = \mathrm{id}_{S^1}$ (by definition of retraction), applying the fundamental group functor gives $\pi_1(r) \circ \pi_1(i) = \pi_1(\mathrm{id}_{S^1}) = \mathrm{id}_{\pi_1(S^1)}$ on the level of groups (functoriality: $\pi_1$ turns composition of continuous maps into composition of homomorphisms). But $\pi_1(S^1) \cong \mathbb{Z}$ while $\pi_1(D^2)$ is trivial (every loop in the convex disk contracts by the straight-line homotopy), so $\pi_1(i): \mathbb{Z} \to \{e\}$ and $\pi_1(r): \{e\} \to \mathbb{Z}$ compose to the zero map, which cannot equal the identity on $\mathbb{Z}$ (a nontrivial group) — contradiction.

Step 4 — Conclude. The contradiction in Step 3 shows no such retraction $r$ can exist, so the assumption in Step 1 (that $f$ has no fixed point) must be false. Hence every continuous $f: D^2 \to D^2$ has a fixed point.