Book 7A

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Course 7Book 7A: Topology and the Fundamental GroupChapter 4

The Fundamental Group

Loops up to deformation, and π₁ of the circle.

22 min read · Updated Oct 3, 2026

Read with Lee, Introduction to Topological Manifolds, chapters 7 (homotopy and the fundamental group) and 8 (the circle), or Hatcher, Algebraic Topology, section 1.1. Choose one as your main text for this chapter and the next two, and use the other for examples.

In this chapter · 7 sections
  1. 4.1A robot on a leash
  2. 4.2Homotopy
  3. 4.3The fundamental group
  4. 4.4The fundamental group of the circle
  5. 4.5Consequences
  6. 4.6History
  7. 4.7Exercises

The hypothesis of the Poincaré conjecture is that every loop in the manifold can be shrunk to a point. This chapter turns that sentence into mathematics. Loops are continuous maps of a circle; "shrunk" means continuously deformed, a homotopy; and the loops at a point, up to homotopy, form a group, the fundamental group π1(X)\pi_1(X), which Poincaré introduced in 1895 precisely to tell spaces apart. A space whose fundamental group is trivial is simply connected, and the conjecture says that the 3-sphere is the only closed simply connected three-manifold.

The first real computation is for the circle: π1(S1)=Z\pi_1(S^1) = \mathbb{Z}, the integer being the winding number, how many times a loop goes around. The proof lifts loops from the circle to the line, which is the first instance of a covering space (7A.6 Covering Spaces). The winding number then proves two classical theorems almost for free: Brouwer's fixed point theorem in the disc and the fundamental theorem of algebra.

By the end of this chapter you will be able to:

  • define homotopies of paths and maps, and the fundamental group, and prove it is a group;
  • explain change of basepoint, induced homomorphisms, homotopy equivalence and contractibility;
  • prove π1(S1)≅Z\pi_1(S^1) \cong \mathbb{Z} by path lifting, and compute winding numbers;
  • prove that the circle is not a retract of the disc, and deduce Brouwer's fixed point theorem in dimension 22 and the fundamental theorem of algebra;
  • prove that SnS^n is simply connected for n≥2n \geq 2.

A robot on a leash

In the world In use Tethered robots

A robot connected to a base station by a cable, for power or communication, as is common for underwater inspection vehicles and some robots used in disaster sites, can't move as freely as an untethered one. Two routes to the same goal that pass on different sides of an obstacle leave the cable wrapped differently, and the cable can't be unwrapped without retracing. What matters about a route is not just where it ends but its homotopy class: which routes can be deformed into each other without passing through obstacles. A planner must search for the best route within each class, or avoid classes that would leave the cable snagged or too long.

Robotics researchers have built path planners on exactly this idea: Subhrajit Bhattacharya, Maxim Likhachev and Vijay Kumar ("Topological constraints in search-based robot path planning", Autonomous Robots, 2012) computed invariants that distinguish classes of trajectories around obstacles and planned within prescribed classes, and Soonkyum Kim, Bhattacharya and Kumar applied this to tethered robots ("Path planning for a tethered mobile robot", ICRA 2014). In the plane with obstacles, the relevant invariant is the fundamental group of the obstacle-free region, and for a single obstacle it is the winding number of this chapter (Figure 4.1).

Homotopy

Let XX and YY be topological spaces and I=[0,1]I = [0, 1].

Definition 4.1 Homotopy

Two continuous maps f0,f1:X→Yf_0, f_1 : X \to Y are homotopic, f0≃f1f_0 \simeq f_1, if there is a continuous H:X×I→YH : X\times I \to Y with H(⋅,0)=f0H(\cdot, 0) = f_0 and H(⋅,1)=f1H(\cdot, 1) = f_1. Two paths γ0,γ1:I→Y\gamma_0, \gamma_1 : I \to Y with the same endpoints are path homotopic if there is such an HH with H(0,s)H(0, s) and H(1,s)H(1, s) fixed for all ss.

Homotopy is an equivalence relation (Exercise 4.9). A homotopy is a continuous family of maps fs=H(⋅,s)f_s = H(\cdot, s) deforming f0f_0 into f1f_1. In a convex subset of Rn\mathbb{R}^n, any two paths with the same endpoints are path homotopic by the straight-line homotopy H(t,s)=(1−s)γ0(t)+sγ1(t)H(t, s) = (1 - s)\gamma_0(t) + s\gamma_1(t). In the plane minus the origin, the two paths around the two sides of the origin are not, as this chapter will prove.

The fundamental group

A loop at x0x_0 is a path γ:I→X\gamma : I \to X with γ(0)=γ(1)=x0\gamma(0) = \gamma(1) = x_0. Two loops can be concatenated: γ⋅δ\gamma\cdot\delta runs through γ\gamma at double speed and then through δ\delta,

(γ⋅δ)(t)={γ(2t),0≤t≤12,δ(2t−1),12≤t≤1.(\gamma\cdot\delta)(t) = \begin{cases}\gamma(2t), & 0 \leq t \leq \frac12,\\ \delta(2t - 1), & \frac12 \leq t \leq 1.\end{cases}
Theorem 4.2 The fundamental group

The set π1(X,x0)\pi_1(X, x_0) of path-homotopy classes of loops at x0x_0 is a group under [γ][δ]=[γ⋅δ][\gamma][\delta] = [\gamma\cdot\delta], with identity the class of the constant loop cx0c_{x_0} and inverse [γ]−1=[γˉ][\gamma]^{-1} = [\bar\gamma], where γˉ(t)=γ(1−t)\bar\gamma(t) = \gamma(1 - t).

Proof. Well defined. If HH is a path homotopy from γ\gamma to γ′\gamma' and KK one from δ\delta to δ′\delta', then running HH and KK side by side is a path homotopy from γ⋅δ\gamma\cdot\delta to γ′⋅δ′\gamma'\cdot\delta'.

The axioms. All three are instances of one principle: reparametrising a path doesn't change its class. If φ:I→I\varphi : I \to I is continuous with φ(0)=0\varphi(0) = 0 and φ(1)=1\varphi(1) = 1, then γ∘φ≃γ\gamma\circ\varphi \simeq \gamma by H(t,s)=γ((1−s)φ(t)+st)H(t, s) = \gamma\big((1 - s)\varphi(t) + st\big). Now (γ⋅δ)⋅ε(\gamma\cdot\delta)\cdot\varepsilon and γ⋅(δ⋅ε)\gamma\cdot(\delta\cdot\varepsilon) traverse the same three loops at different speeds, so they differ by a reparametrisation: associativity. c⋅γc\cdot\gamma is γ\gamma reparametrised (waiting at x0x_0 for half the time): identity. For inverses, H(t,s)=γ(2t(1−s))H(t, s) = \gamma(2t(1 - s)) for t≤12t \leq \frac12 and γ((2−2t)(1−s))\gamma((2 - 2t)(1 - s)) for t≥12t \geq \frac12 shrinks γ⋅γˉ\gamma\cdot\bar\gamma to the constant loop by going out along γ\gamma only as far as time 1−s1 - s and coming back.

The group is usually not abelian (7A.5 Computing π₁): loops around two different holes in a twice-punctured plane don't commute.

Basepoints. If α\alpha is a path from x0x_0 to x1x_1, then [γ]↦[αˉ⋅γ⋅α][\gamma] \mapsto [\bar\alpha\cdot\gamma\cdot\alpha] is an isomorphism π1(X,x0)→π1(X,x1)\pi_1(X, x_0) \to \pi_1(X, x_1). So for path-connected XX the group is independent of the basepoint up to isomorphism, and one writes π1(X)\pi_1(X). The isomorphism depends on α\alpha, up to conjugation.

Induced homomorphisms. A continuous f:X→Yf : X \to Y with f(x0)=y0f(x_0) = y_0 induces f∗:π1(X,x0)→π1(Y,y0)f_* : \pi_1(X, x_0) \to \pi_1(Y, y_0), f∗[γ]=[f∘γ]f_*[\gamma] = [f\circ\gamma]. It is a homomorphism, (g∘f)∗=g∗∘f∗(g\circ f)_* = g_*\circ f_* and id⁡∗=id⁡\operatorname{id}_* = \operatorname{id} (functoriality). So homeomorphic spaces have isomorphic fundamental groups: π1\pi_1 is a topological invariant.

Homotopy invariance. More is true: maps that are homotopic (through basepoint-preserving homotopies) induce the same homomorphism. Two spaces XX and YY are homotopy equivalent if there are maps f:X→Yf : X \to Y and g:Y→Xg : Y \to X with g∘f≃id⁡Xg\circ f \simeq \operatorname{id}_X and f∘g≃id⁡Yf\circ g \simeq \operatorname{id}_Y; then π1(X)≅π1(Y)\pi_1(X) \cong \pi_1(Y). For instance the punctured plane R2∖{0}\mathbb{R}^2\setminus\{0\} is homotopy equivalent to the circle (retract each ray onto its point of the unit circle), and so is the annulus and the solid torus D2×S1D^2\times S^1. A space homotopy equivalent to a point is contractible: Rn\mathbb{R}^n, any convex set, any star-shaped set. Contractible spaces have trivial fundamental group.

Definition 4.3 Simply connected

A space is simply connected if it is path-connected and its fundamental group is trivial: every loop can be shrunk to a point through loops.

Figure 4.1. Loops in an annulus, based at the same point. The small loop can be shrunk to a point; the loop that goes once around the hole cannot, and neither can the one that goes around twice, and no two of these three are homotopic. The fundamental group of the annulus is Z\mathbb{Z}, generated by the loop that goes around once (Theorem 4.5).

The fundamental group of the circle

Let p:R→S1p : \mathbb{R} \to S^1, p(s)=(cos⁡2πs,sin⁡2πs)p(s) = (\cos2\pi s, \sin2\pi s), which wraps the line around the circle like a helix projected onto its axis (Figure 4.2). Each small arc of the circle has a preimage that is a disjoint union of copies of that arc, one in each interval (k−12,k+12)(k - \frac12, k + \frac12); this is what makes pp a covering map (7A.6 Covering Spaces).

Lemma 4.4 Path and homotopy lifting
  1. For every path γ:I→S1\gamma : I \to S^1 and every s0∈p−1(γ(0))s_0 \in p^{-1}(\gamma(0)), there is a unique path γ~:I→R\tilde\gamma : I \to \mathbb{R} with p∘γ~=γp\circ\tilde\gamma = \gamma and γ~(0)=s0\tilde\gamma(0) = s_0.
  2. Likewise every homotopy H:I×I→S1H : I\times I \to S^1 has a unique lift H~\tilde H with a prescribed value at (0,0)(0, 0).

Proof. (1) By uniform continuity (2B.3 Compactness), divide II into intervals [tj−1,tj][t_{j-1}, t_j] so small that γ\gamma maps each into an arc AjA_j of length less than half the circle. On such an arc, pp has continuous local inverses, one for each sheet of p−1(Aj)p^{-1}(A_j). Lift on the first interval using the inverse whose sheet contains s0s_0, then on the next interval using the sheet containing the endpoint already reached, and so on. Uniqueness: two lifts with the same start differ by an integer-valued continuous function, which is constant on the connected II (2B.4 Connectedness). (2) The same argument on a fine grid of small squares of I×II\times I.

Figure 4.2. The covering p:R→S1p : \mathbb{R} \to S^1 pictured as a helix over the circle. A loop going once around the circle lifts to a path on the helix that climbs one full turn: from 00 to 11. The end of the lift, an integer, is the loop's winding number.
Theorem 4.5 π1(S1)≅Z\pi_1(S^1) \cong \mathbb{Z}

For a loop γ\gamma at (1,0)(1, 0), let γ~\tilde\gamma be its lift with γ~(0)=0\tilde\gamma(0) = 0, and define the winding number deg⁡γ=γ~(1)∈Z\deg\gamma = \tilde\gamma(1) \in \mathbb{Z}. Then [γ]↦deg⁡γ[\gamma] \mapsto \deg\gamma is an isomorphism π1(S1)→Z\pi_1(S^1) \to \mathbb{Z}.

Proof. γ~(1)\tilde\gamma(1) is an integer because p(γ~(1))=γ(1)=(1,0)p(\tilde\gamma(1)) = \gamma(1) = (1, 0). Well defined: a path homotopy HH from γ\gamma to γ′\gamma' lifts to H~\tilde H; the lift's right edge H~(1,⋅)\tilde H(1, \cdot) lies in the discrete set p−1(1,0)=Zp^{-1}(1, 0) = \mathbb{Z} and is continuous, hence constant, so γ~(1)=γ~′(1)\tilde\gamma(1) = \tilde\gamma'(1). Homomorphism: the lift of γ⋅δ\gamma\cdot\delta is γ~\tilde\gamma followed by δ~+γ~(1)\tilde\delta + \tilde\gamma(1), which ends at γ~(1)+δ~(1)\tilde\gamma(1) + \tilde\delta(1). Onto: the loop ωn(t)=p(nt)\omega_n(t) = p(nt) has degree nn. Injective: if deg⁡γ=0\deg\gamma = 0, the lift is a loop in R\mathbb{R}, which is contractible by a straight-line homotopy H~\tilde H; then p∘H~p\circ\tilde H contracts γ\gamma.

For a loop in the punctured plane, γ(t)∈C∖{0}\gamma(t) \in \mathbb{C}\setminus\{0\}, the winding number around 00 is the degree of γ/∣γ∣\gamma/|\gamma|, and for piecewise smooth loops it equals 12πi∮γdzz\frac{1}{2\pi i}\oint_\gamma\frac{dz}{z}, the integral of 5A.2 Cauchy’s Theorem and Its Consequences and the argument principle of 5A.3 Residues and Fourier Transforms: the integral measures the total change of the angle, which is 2π2\pi times the change of the lift.

In the world Model The dog on a leash

A dog on a leash walks around a lamppost and back to its owner. Whether the leash can be freed without unclipping is decided by the winding number of the dog's path around the post, measured relative to the owner: if it is zero, the dog can walk the leash free; if not, the leash is wrapped, and no amount of wandering that keeps the leash on the same side of the post will undo it. The same integer counts the turns of a garden hose around a tree and of a tether around a pillar.

Consequences

Theorem 4.6 No retraction of the disc onto its boundary

There is no continuous map r:D2→S1r : D^2 \to S^1 with r(x)=xr(x) = x for all x∈S1x \in S^1.

Proof. If there were, then with i:S1→D2i : S^1 \to D^2 the inclusion, r∘i=id⁡S1r\circ i = \operatorname{id}_{S^1}, so r∗∘i∗=id⁡r_*\circ i_* = \operatorname{id} on π1(S1)=Z\pi_1(S^1) = \mathbb{Z}. But i∗i_* maps into π1(D2)=0\pi_1(D^2) = 0, since the disc is convex. The identity of Z\mathbb{Z} cannot factor through the trivial group.

Corollary 4.7 Brouwer's fixed point theorem in dimension 2

Every continuous map f:D2→D2f : D^2 \to D^2 has a fixed point.

Proof. If f(x)≠xf(x) \neq x for all xx, send each xx to the point where the ray from f(x)f(x) through xx leaves the disc. This is continuous and fixes the boundary circle: a retraction, which doesn't exist.

In the world Analogy Stirring coffee

Stir a cup of coffee gently, without splashing, and let it settle. If the surface is idealised as a disc and the stirring as a continuous map of the disc to itself, Brouwer's theorem says some point of the surface ends up exactly where it started.

Where the picture breaks Where the picture breaks

Real stirring is a three-dimensional, often turbulent flow, and the surface layer isn't moved by a single continuous map of a disc to itself: fluid moves between the surface and the depths, and "the same point of coffee" is not well defined at the molecular scale. The theorem applies to the idealisation, not to the cup. What the picture does convey correctly is that the fixed point is forced by topology alone, with no information about how the stirring was done.

The fundamental theorem of algebra, again. If p(z)=zn+an−1zn−1+⋯+a0p(z) = z^n + a_{n-1}z^{n-1} + \dots + a_0 had no root, then for each r≥0r \geq 0 the loop γr(t)=p(re2πit)/∣p(re2πit)∣\gamma_r(t) = p(re^{2\pi it})/|p(re^{2\pi it})| in S1S^1 would be defined, and the family would be a homotopy from γ0\gamma_0 (constant, winding number 00) to γR\gamma_R for large RR, which winds nn times because p(z)≈znp(z) \approx z^n on the large circle (Exercise 4.12). So n=0n = 0. This is the topological core of the complex-analytic proofs in 5A.2 Cauchy’s Theorem and Its Consequences and 5A.3 Residues and Fourier Transforms.

Products. π1(X×Y)≅π1(X)×π1(Y)\pi_1(X\times Y) \cong \pi_1(X)\times\pi_1(Y), since a loop in a product is a pair of loops and a homotopy is a pair of homotopies. So π1(T2)=π1(S1×S1)≅Z2\pi_1(T^2) = \pi_1(S^1\times S^1) \cong \mathbb{Z}^2, and the torus is not simply connected: the 2-dimensional Poincaré theorem's first test case (7A.3 Manifolds and Surfaces).

Theorem 4.8 Spheres of dimension at least 2 are simply connected

For n≥2n \geq 2, π1(Sn)=0\pi_1(S^n) = 0.

Proof. Let γ\gamma be a loop in SnS^n. By uniform continuity, divide II into intervals on each of which γ\gamma stays within an open hemisphere. Within an open hemisphere, homotope the piece of γ\gamma, keeping its endpoints, to the shortest great-circle arc between them (the hemisphere is homeomorphic to a convex set by projection, so a straight-line homotopy there works). The new loop is a finite union of great-circle arcs, a set of zero area in SnS^n for n≥2n \geq 2, so it misses some point qq. But Sn∖{q}≅RnS^n\setminus\{q\} \cong \mathbb{R}^n by stereographic projection (7A.2 Compactness and Compactification), which is contractible, so the loop can be shrunk to a point in Sn∖{q}S^n\setminus\{q\}.

In particular S3S^3 is simply connected: it satisfies the hypothesis of the Poincaré conjecture, which says it is the only closed three-manifold that does (7A.9 The Poincaré Conjecture, Precisely).

Where this goes Loops, spheres and higher homotopy groups

The fundamental group is the first of the homotopy groups πk(X)\pi_k(X), the classes of maps of the kk-sphere into XX. Perelman's finite-extinction argument for simply connected three-manifolds uses π2\pi_2 and π3\pi_3: a simply connected closed three-manifold has non-trivial π3\pi_3, so it contains a sphere of spheres that can't be shrunk, and the area of the best such family is forced to zero by the Ricci flow in finite time (12C.2 Finite Extinction). Sweepouts and their width are the subject of 10A.8 Min–Max and Width.

History

Henri Poincaré defined the fundamental group in Analysis Situs (1895), as the group of loops up to deformation, and used it to distinguish three-manifolds with the same homology (7A.8 Homology in Brief). Luitzen Brouwer proved his fixed point theorem in 1910–12 (in dimension 3 it had been proved by Piers Bohl in 1904). The winding number goes back to Gauss and Cauchy, as the change of the argument along a closed curve. Camille Jordan studied closed curves and their deformations in the 1860s.

Recall Where we stand

Homotopy is continuous deformation; loops at a point up to path homotopy form the fundamental group π1(X,x0)\pi_1(X, x_0), with concatenation as product. For path-connected spaces it is independent of the basepoint, it is functorial, and homotopy equivalent spaces have isomorphic fundamental groups; contractible spaces are simply connected. Lifting paths and homotopies through p:R→S1p : \mathbb{R} \to S^1 proves π1(S1)=Z\pi_1(S^1) = \mathbb{Z}, the winding number. Hence the circle is not a retract of the disc, Brouwer's theorem holds in dimension 22, and polynomials have roots. π1(T2)=Z2\pi_1(T^2) = \mathbb{Z}^2, and SnS^n is simply connected for n≥2n \geq 2. 7A.5 Computing π₁ computes fundamental groups of spaces built by gluing.

Exercises

Exercise 4.9 Homotopy is an equivalence relation

Show that path homotopy is reflexive, symmetric (reverse the homotopy parameter) and transitive (run two homotopies one after the other, each at double speed in ss). Where is continuity of the combined homotopy checked?

Exercise 4.10 Changing the basepoint

Show that βα([γ])=[αˉ⋅γ⋅α]\beta_\alpha([\gamma]) = [\bar\alpha\cdot\gamma\cdot\alpha] is a well-defined isomorphism from π1(X,x0)\pi_1(X, x_0) to π1(X,x1)\pi_1(X, x_1), with inverse βαˉ\beta_{\bar\alpha}. If π1\pi_1 is abelian, show βα\beta_\alpha doesn't depend on α\alpha.

Exercise 4.11 Computing winding numbers

Find the winding number around 00 of: (a) γ(t)=(2+cos⁡2πt)e4πit\gamma(t) = (2 + \cos2\pi t)e^{4\pi it}; (b) γ(t)=e2πit+3\gamma(t) = e^{2\pi it} + 3; (c) the boundary of the square [−1,1]2[-1, 1]^2 traversed counterclockwise; (d) γ(t)=e2πit(e2πit−12)\gamma(t) = e^{2\pi it}(e^{2\pi it} - \frac12). (For (d), the winding number of a product is the sum.)

Solution

(a) 22: the modulus stays positive and the argument increases by 4π4\pi. (b) 00: the loop stays in the half-plane Re⁡z>0\operatorname{Re}z > 0, where the argument has a continuous branch. (c) 11. (d) 1+1=21 + 1 = 2, since e2πit−12e^{2\pi it} - \frac12 winds once around 00.

Exercise 4.12 The fundamental theorem of algebra, in detail

Let p(z)=zn+an−1zn−1+⋯+a0p(z) = z^n + a_{n-1}z^{n-1} + \dots + a_0 with n≥1n \geq 1. (a) Show that for R>1+∑∣ak∣R > 1 + \sum|a_k|, the straight-line homotopy from p(Re2πit)p(Re^{2\pi it}) to (Re2πit)n(Re^{2\pi it})^n never passes through 00, so the two loops have the same winding number, nn. (b) Complete the argument in the text.

Solution

(a) ∣p(z)−zn∣≤∑∣ak∣Rk<Rn|p(z) - z^n| \leq \sum|a_k|R^k < R^n for ∣z∣=R|z| = R under the condition, so (1−s)p(z)+szn=zn+(1−s)(p(z)−zn)≠0(1 - s)p(z) + sz^n = z^n + (1 - s)(p(z) - z^n) \neq 0. (b) If pp has no zeros, r↦p(re2πit)r \mapsto p(re^{2\pi it}) for 0≤r≤R0 \leq r \leq R is a homotopy in C∖{0}\mathbb{C}\setminus\{0\} from a constant loop to a loop of winding number nn, so n=0n = 0, a contradiction.

Exercise 4.13 Retracts and their fundamental groups

Show that if A⊆XA \subseteq X is a retract of XX, then i∗:π1(A)→π1(X)i_* : \pi_1(A) \to \pi_1(X) is injective. Use this to show that the circle {∣z∣=2}\{|z| = 2\} is not a retract of the closed disc {∣z∣≤2}\{|z| \leq 2\}, but is a retract of the annulus {1≤∣z∣≤2}\{1 \leq |z| \leq 2\}.

Exercise 4.14 Rehearsal: which three-manifolds are simply connected?

Using the results of this chapter, decide which of the following closed three-manifolds are simply connected: S3S^3; T3=S1×S1×S1T^3 = S^1\times S^1\times S^1; S2×S1S^2\times S^1; the lens space L(p,q)L(p, q) (accept from 7A.6 Covering Spaces that π1(L(p,q))=Z/p\pi_1(L(p, q)) = \mathbb{Z}/p). Which of them could be counterexamples to the Poincaré conjecture, if it were false? (None of them: the ones that are simply connected are spheres. 7A.9 The Poincaré Conjecture, Precisely explains why the remaining candidate, the Poincaré homology sphere, isn't one either.)

Solution

S3S^3: simply connected (Theorem 4.8). T3T^3: π1=Z3\pi_1 = \mathbb{Z}^3. S2×S1S^2\times S^1: π1=π1(S2)×π1(S1)=Z\pi_1 = \pi_1(S^2)\times\pi_1(S^1) = \mathbb{Z}. L(p,q)L(p, q): Z/p≠0\mathbb{Z}/p \neq 0 for p≥2p \geq 2. Only S3S^3 is simply connected, and it is the sphere, so none is a counterexample.

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