The fundamental group

It converts a geometric, hard-to-pin-down question ("how many essentially different loops are there on a circle?") into ordinary integer arithmetic, and is the single computation that all of algebraic topology's winding-number arguments trace back to.

Step 1 — Set up the covering map. Let $p: \mathbb{R} \to S^1$ be $p(t) = (\cos 2\pi t, \sin 2\pi t)$, a continuous surjection where every point of $S^1$ has an evenly-covered neighborhood (a small arc's preimage under $p$ is a disjoint union of open intervals in $\mathbb{R}$, each mapped homeomorphically onto the arc). Fix basepoint $1 = p(0) \in S^1$.

Step 2 — Path lifting. Given any loop $\gamma: [0,1] \to S^1$ with $\gamma(0)=\gamma(1)=1$, there is a unique continuous lift $\tilde\gamma: [0,1] \to \mathbb{R}$ with $p \circ \tilde\gamma = \gamma$ and $\tilde\gamma(0) = 0$ (construct it by covering $[0,1]$ with finitely many subintervals on which $\gamma$ stays inside one evenly-covered neighborhood, and lift piece by piece, each time choosing the branch of $p^{-1}$ continuing from the previous endpoint). Define $\deg(\gamma) = \tilde\gamma(1) \in \mathbb{Z}$ (an integer since $p(\tilde\gamma(1)) = \gamma(1) = 1 = p(0)$ forces $\tilde\gamma(1) \in \mathbb{Z}$).

Step 3 — Homotopy invariance. If $\gamma \simeq_p \gamma'$ via homotopy $H$, the homotopy lifting property (proved the same way as path lifting, one strip at a time) gives a continuous lift $\tilde H$ with $\tilde H(s,0)=\tilde\gamma(s)$, and $\tilde H(0,t)=0$ for all $t$ (constant, since it lifts the constant basepoint loop). Then $t \mapsto \tilde H(1,t)$ is a continuous integer-valued function of $t$ (by the same argument as Step 2), hence constant; so $\deg(\gamma) = \tilde H(1,0) = \tilde H(1,1) = \deg(\gamma')$. Thus $\deg$ is well-defined on homotopy classes $[\gamma] \mapsto \deg(\gamma)$.

Step 4 — Homomorphism property. For loops $\gamma, \delta$, lifting $\gamma * \delta$ by first lifting $\gamma$ to end at $\deg(\gamma)$, then lifting $\delta$ starting from $\deg(\gamma)$ (a shifted copy of $\delta$'s lift, valid since $p$ is invariant under integer translation) shows $\deg(\gamma * \delta) = \deg(\gamma) + \deg(\delta)$. So $[\gamma] \mapsto \deg(\gamma)$ is a group homomorphism $\pi_1(S^1,1) \to \mathbb{Z}$.

Step 5 — Bijectivity. Surjective: for any $n \in \mathbb{Z}$, $\gamma_n(s) = p(ns)$ is a loop with $\deg(\gamma_n)=n$. Injective: if $\deg(\gamma)=0$, the lift $\tilde\gamma$ is a loop in $\mathbb{R}$ based at $0$ (since $\tilde\gamma(1)=0=\tilde\gamma(0)$); $\mathbb{R}$ is convex so the straight-line homotopy $\tilde H(s,t) = (1-t)\tilde\gamma(s)$ contracts $\tilde\gamma$ to the constant path at $0$ rel endpoints, and composing with $p$ gives a path-homotopy from $\gamma$ to the constant loop, so $[\gamma]$ is trivial. Hence $\deg$ is a bijective homomorphism, i.e. an isomorphism $\pi_1(S^1,1) \cong \mathbb{Z}$.

It guarantees solutions in an astonishing range of settings (Nash equilibria, differential equations, economic equilibrium) purely from the shape of $D^2$, with no formula for $f$ needed at all — and the proof is a beautiful example of translating a hard analysis question into an easy algebra question about $\pi_1$.

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.

It is the main computational tool of the subject: it lets you build the fundamental group of a complicated space out of the (often much simpler) fundamental groups of overlapping pieces, which is exactly how the free group $F_2$ for the figure-eight is computed (two circles overlapping in a point).

A full proof requires careful combinatorial bookkeeping; here is the structural outline. Step 1 — Generators. Any loop $\gamma$ in $X$ based at $x_0$ can be subdivided into finitely many sub-paths, each lying entirely in $U$ or entirely in $V$ (using a Lebesgue number argument on the open cover $\{\gamma^{-1}(U), \gamma^{-1}(V)\}$ of the compact interval $[0,1]$), so $[\gamma]$ can be written as a product of loops each representing an element of $\pi_1(U,x_0)$ or $\pi_1(V,x_0)$ (after connecting sub-path endpoints back to $x_0$ through chosen paths in $U \cap V$, which is path-connected). This shows $\pi_1(X,x_0)$ is generated by the images of $\pi_1(U,x_0)$ and $\pi_1(V,x_0)$.

Step 2 — Relations. Any loop $\delta$ lying in $U \cap V$ represents, a priori, two possibly-different elements: $i_U(\delta) \in \pi_1(U,x_0)$ (viewing $\delta$ as a loop in $U$) and $i_V(\delta) \in \pi_1(V,x_0)$ (viewing it as a loop in $V$). Since $\delta$ is literally the same loop in $X$, its images under $\pi_1(U) \to \pi_1(X)$ and $\pi_1(V) \to \pi_1(X)$ must agree; this forces exactly the amalgamation relations $i_U(\delta) = i_V(\delta)$ for every $[\delta] \in \pi_1(U \cap V, x_0)$.

Step 3 — No further relations. A more delicate argument (subdividing homotopies $H: [0,1]\times[0,1] \to X$ the same way, using compactness of the square to get a finite grid where each cell lies in $U$ or $V$) shows that these amalgamation relations are the only relations needed: any two words in the generators representing the same element of $\pi_1(X,x_0)$ can be related by a finite sequence of moves each justified by an amalgamation relation. This identifies $\pi_1(X,x_0)$ exactly with the amalgamated free product.