The Fundamental Group

JWR Feb 6, 2005

1 The Fundamental Group

1.1. When γ : I → X is a path, we denote by [γ] the homotopy class of γ (rel ∂I); here ∂I = {0, 1}. Two paths α, β : I → X with α(1) = β(0) determine a path α · β : I → X via the formula ( 1 α(2t), 0 ≤ t ≤ 2 ; (α · β)(t) := 1 β(2t − 1), 2 ≤ t ≤ 1.

The fundamental group of the pointed space (X, x0) is

π1(X, x0) = {[γ]: γ(0) = γ(1) = x0}.

The group operation is well deﬁned by [α][β] := [α · β]. The identity element is the constant path and the inverse operation is deﬁned by

[γ]−1 := [¯γ], γ¯(t) := γ(1 − t).

Note that any path α : I → X gives an isomorphism

π1(X, x0) → π1(X, x1):[γ] 7→ [¯α · γ · α] where x0 = α(0), x1 = α(1), and (as before)γ ¯(t) := γ(1 − t). In this sense, if X is path connected, the fundamental group is independent of the choice of the base point. Remark 1.2. A groupoid is an algebraic structure consisting of two sets B (the objects) and G (the morphisms) and ﬁve maps s, t : G → B (source and target), e : B → G (identity), m : G s×t G → G (composition), and i : G → G (inverse) satisfying the following axioms:

(Identity Law) g · ex = g, ey · g = g where s(g) = x, t(g) = y.

(Associative Law) g1 · (g1 · g3) = (g1 · g1) · g3 for g1, g2, g3 ∈ G s×t G.

−1 −1 (Inverse Law) g · g = gey, g · g = ex where s(g) = x, t(g) = y.

1 Here we are using the abbreviations

G s×t G := {(g1, g2) ∈ G × G : t(g2) = s(g1)},

−1 ex := e(x), g1 · g2 := m(g1, g2), g := i(g). If the map i and the Inverse Law are dropped we recover the deﬁnition of a category. In a category a morphism which has a (necessarily unique) two-sided inverse is called an isomorphism (or an automorphism if source and target are the same). Thus a groupoid is a category in which every morphism is an isomorphism. For a category (B,G) and x, y ∈ B denote the set of morphisms from x to y by Gx,y := {g ∈ G : s(g) = x, t(g) = y}.

The set Gx := {g ∈ Gx,x : g is an isomorphism} is called the automorphism group of the object x. With these deﬁnitions π1(X, x0) is the automorphism group x0 in the fundamental groupoid deﬁned by

B = X,G = { [γ] | γ : I → X}, s([γ]) = γ(0), t([γ]) = γ(1), e(x) = [x] (the homotopy class of the constant path at the point x), m([β], [α]) = [α · β], and i(γ) =γ ¯. Deﬁnition 1.3. A path connected space is called simply connected iﬀ its fundamental group π1(X, x0) is trivial.

1 n Theorem 1.4. π1(S ) = Z and S is simply connected for n > 1.

2 Covering Spaces

2.1. A map p : E → B is called locally trivial iff every x0 ∈ B there is a −1 neighborhood U of x0 ∈ B and a homeomorphism φ : U × F → p (U) such that p(φ(x, v)) = x for (x, v) ∈ U × F . The map p is called the projection, E is called the total space, B is called the base, and the topological space F is called the fiber. It is easy to see that if B is connected, then the fiber is unique up to homeomorphism. A locally trivial map is also called a fiber bundle, but sometimes the latter term has a more restricted meaning. A covering space is a locally trivial map with discrete fiber. Usually we denote covering spaces with notation like p : Y → X or p : X˜ → X. Note that the projection p of a covering space is a local homeomorphism. Example 2.2. The archetypal example of a covering space is the exponential map p : R → S1 defined by p(θ) = eiθ. The restriction of p to an open interval is a local homeomorphism but not a covering. Example 2.3. The map p : S1 → S1, p(z) = zn is an n-sheeted covering space. Example 2.4. The various covers of a wedge S1 ∨S1 of circles shown on page 58 of [3] are a good source of examples.

2 2.5. When p : X˜ → X and f : Y → X we call a map f˜ : Y → X˜ a lift of f iﬀ ˜ p ◦ f = f. If p :(X,˜ x˜0) → (X, x0) and f :(Y, y0) → (X, x0) a pointed lift of ˜ ˜ f is a lift f such that f(y0) =x ˜0.

Proposition 2.6 (Path Lifting). Let p :(X,˜ x˜0) → (X, x0) is a covering −1 space, γ : I → X, and y0 ∈ p (γ(0)). Then there is a unique lift γ˜ : I → X˜ of γ with γ˜(0) = y0. Corollary 2.7 (Homotopy Lifting). Let p : X˜ → X is a covering space, {ft : Y → X}t∈I be a homotopy, and g : Y → X˜ of be a lift of f0. Then there ˜ ˜ ˜ is a unique homotopy {ft : I → X˜}t∈I such that f0 = g and ft is a lift of ft. ˜ Proof. The uniqueness of ft(y) follows from 2.6 applied to the path I 7→ X : ˜ t 7→ ft(y) with starting point g(y) ∈ X˜ and this also deﬁnes ft(y) for y ∈ Y and ˜ t ∈ I. We show that (t, y) 7→ ft(y) is continuous at (t0, y0). Bu compactness write write the interval [0, t0] as a union of closed intervals [0, t0] = I1 ∪ · · · ∪ Ik so that ft(y0) ∈ Ui for t ∈ Ii where Ui ⊂ X is such that p is trivial over Ui. By induction on i there are open sets U˜i ⊂ X˜ such that p maps U˜i homeomorphically ˜ onto Ui and ft(y0) ∈ U˜i for t ∈ Ii. By continuity there is a neighborhood V of y0 ∈ Y such that ft(y) ∈ Ui for t ∈ Ii and y ∈ V . By induction on i, the fact that p : U˜i → Ui is a homeomorphism, and uniqueness of path lifting, ˜ ft(y) ∈ U˜i for t ∈ Ii and y ∈ V . Now, since p : U˜i → Ui is a homeomorphism, ˜ the continuity of [0, t0] × V → X˜ :(t, y) 7→ ft(y) at points t ∈ [0, t0] and y ∈ V follows from the continuity of [0, t0] × V → X :(t, y) 7→ ft(y)

Corollary 2.8 (Injectivity). Let p :(X,˜ x˜0) → (X, x0) is a covering space. Then the induced map p∗ : π1(X,˜ x˜0) → π1(X, x0) is injective.

Theorem 2.9 (Lifting Criterion). Assume p :(X,˜ x˜0) → (X, x0) be a covering space and let f :(Y, y0) → (X, x0) be a map. Assume Y is path connected and locally path connected. Then (i) Two pointed lifts of the same map f are equal.

(ii) There is a pointed lift of f if and only if f∗π1(Y, y0) ⊂ p∗π1(X,˜ x˜0).

Remark 2.10. Since p∗ is injective, we have f∗π1(Y, y0) ⊂ p∗π1(X,˜ x˜0) if and only if there is a homomorphism h : π1(Y, y0) → π1(X,˜ x˜0) such that p∗ ◦h = f∗. ˜ When these equivalent conditions hold we have h = f∗ as follows: If p∗ ◦ h = f∗, ˜ ˜ then p∗ ◦ h = f∗ = p∗ ◦ f∗ so h = f∗ by the injectivity of p∗. Thus, in the two diagrams

(X,˜ x˜0) π1(X,˜ x˜0) s9 p8 f˜ ss h pp ss p pp p ss ppp ∗ sss ppp f f∗ (Y, y0) / (X, x0) π1(Y, y0) / π1(X, x0) we can ﬁnd f˜ so that the diagram on the left commutes if and only if we can ﬁnd h so that the diagram on the right commutes.

3 ˜ Proof of Theorem 2.9. When f is a pointed lift of f and α is a path from y0 to ˜ ˜ y, then f ◦ α is a path from f(y0) to f(y) and f ◦ α is a path from f(y0) to f˜(y), so uniqueness of the lift follows from uniqueness of path lifting. ’Only if’ ˜ ˜ in part (ii) follows from (p ◦ f)∗ = p∗ ◦ f∗. We prove ’if’. ˜ Assume that f∗π1(Y, y0) ⊂ p∗π1(X,˜ x˜0). As just noted, we must define f(y) ˜ by f(y) =γ ˜(1) whereγ ˜ is the lift of γ := f ◦ α withγ ˜(0) =x ˜0 and α is a path ˜ from y0 to y in Y . We must show that f is well defined (i.e. independent of the choice of α) and continuous. ¯ Suppose that β is another path from y0 to y. Then [αβ] ∈ π1(Y, y0) so ¯ [(f ◦ α)(f ◦ β)] ∈ f∗π1(Y, y0) ⊂ p∗π1(X,˜ x˜0) so there is an element of [˜γ] ∈ ¯ ¯ π1(X,˜ x˜0) which projects to [(f ◦α)(f ◦β)], i.e. p◦γ˜ is homotopic to (f ◦α)(f ◦β) with endpoints fixed so the liftγ ˜ of which starts atx ˜0 has the same endpoint ¯ as the lift of (f ◦ α)(f ◦ β) of which starts atx ˜0. But this endpoint isx ˜0 asγ ˜ is ¯ closed. Hence the lift of (f ◦ α)(f ◦ β) of which starts atx ˜0 is closed so the lifts of f ◦ α and f ◦ β which start atx ˜0 have the same endpoint as required. ˜ Now we show that f is continuous. Choose y1 ∈ Y and a neighborhood W˜ ˜ ˜ of f(y1) in X˜. We must construct a neighborhood V of y1 such that f(V ) ⊂ W˜ . As p is a covering we may shrink W˜ so that p maps W˜ homeomorphicly onto a neighborhood W of f(y1). By the continuity of f there is a neighborhood V of y1 such that f(V ) ⊂ U. By local path connectedness there is a path connected neighborhood U of y1 with U ⊂ V . For y ∈ V there is a path β from y1 to y in ˜ V . Then β lies in U so f ◦ β lies in W so the lift of f ◦ β starting at f(y1) lies in W˜ . As Y is path connected there is a path α from y0 to y1. Then αβ is a path from y0 to y and the lift of (f ◦ α)(f ◦ β) starting at f(y0) ends the same ˜ ˜ ˜ point as the lift of f ◦ β starting at f(y1). This lift is f(y) so f(y) lies in W˜ as required.

2.11. Denote by Cov(X, x0) the category whose objects are covering spaces p :(Y, y0) → (X, x0). A morphism in this category from the covering space p :(Y, y0) → (X, x0) to the covering space q :(Z, z0) → (X, x0) is a map φ :(Y, y0) → (Z, z0) such that q ◦ φ = p. For any group G let Sub(G) denote the category whose objects are subgroups H of G and where the morphisms are inclusions K ⊂ H. There is a functor

Cov(X, x0) → Sub(π1(X, x0)) which assigns the subgroup p∗π1(Y, y0) to the object p :(Y, y0) → (X, x0) and assigns the inclusion

p∗π1(Y, y0) ⊂ p∗φ∗π1(Z, z0) = q∗π1(Z, z0) to the morphism φ :(Y, y0) → (Z, z0)

Corollary 2.12. Let p :(Y, y˜0) → (X, x0) and q :(Z, z0) → (X, x0) be pointed covering spaces over the same pointed space (X, x0). Then (i) There is at most one morphism from p to q.

4 (ii) There is a morphism from p to q if and only if p∗π1(Y, y0) ⊂ q∗π1(Z, z0). Proof. This is a special case of Theorem 2.9. Exercise 2.13. Let φ : Y → Z be a morphism from the covering space p : Y → X to the covering space q : Z → X, as in 2.11. Assume that X (and hence also Y and Z) is locally connected. Show that φ : Y → Z is a covering space. Give an example showing that the hypothesis that X is locally connected cannot be dropped.

Solution. Choose z0 ∈ Z and let x0 = q(z0). We must ﬁnd a neighborhood −1 V of z0 and a trivialization of the restriction φ to φ (V ). Trivialize p and q over a neighborhood of x0 and then shrink this neighborhood so it is connected. Restricting to this neighborhood we may assume w.l.o.g. that

Y = X × A, Z = X × B, p(x, a) = q(a, b) = x where A and B are discrete and X is connected. The condition q ◦φ = p implies that φ(x, a) = (x, f(x, a)) where f : X × A → B is continuous. Let z0 = (x0, b0) and C = {a ∈ A : f(x0, a) = b0}. Let V = X × {b0}. Then for (x, a) ∈ X × A we have

φ(x, a) ∈ V ⇐⇒ f(x, a) = b0 ⇐⇒ f(x0, a) = b0 ⇐⇒ a ∈ C where the middle ⇐⇒ holds because f is locally constant and X is connected. Thus φ−1(V ) = X × C so the inclusion X × C → X × A is a local trivialization. 1 1 1 Consider X = {0, 1, 2 , 3 , 4 ,...}, A = N, B = {1, 2, }, and f(x, n) = 1 or 2 according as nx < 1 or xn ≥ 1. Then the fiber φ−1(x, 1) is finite for x 6= 0 and infinite for x = 0 so φ is not trivial over any neighborhood of (0, 1). 2.14. A covering space p : X˜ → X of a path connected space X is called universal iff X˜ is connected and simply connected. A space X is called semi locally simply connected iff every point x0 ∈ X has arbitrarily small neighborhoods U such that the homomorphism π1(U, x0) → π1(X, x0) is trivial. Theorem 2.15. A path connected space and locally path connected space X has a universal cover if and only if it is semi locally simply connected.

Proof. For “only if” assume that p : X˜ → X is universal. Choose x0 ∈ X and a neighborhood V of x. Choose a neighborhood U of x0 in V so that P is trivial over U. Let U˜ ⊂ X˜ map homeomorphicly to U by p andx ˜0 ∈ U˜ be deﬁned by p(˜x0) = x0. The inclusion π1(U, x0) → π1(X, x0) factors as π1(U, x0) → π1(U,˜ x˜0) → π1(X,˜ x˜0) → π1(X, x0) so it is trivial as π1(X,˜ x˜0) is trivial. For “if” choose x0 ∈ X and deﬁne p : X˜ → X by

X˜ := {[γ]: γ : I → X, γ(0) = x0}, p([γ]) = γ(1).

5 For each [γ] ∈ X˜ and each path connected neighborhood U of γ(1) such that π1(U, γ(1)) → π1(X, γ(1)) is trivial deﬁne

U[γ] := {[γη]: η : I → U, η(0) = γ(1)}.

(The set U[γ] is clearly independent of the choice of the representative γ of [γ].) ˜ 0 These sets form a basis for a topology on X. If U[γ] ∩ U[γ0] 6= ∅ then [γ] = [γ ] so U[γ] = U[γ0]. This deﬁnes a homeomorphism

−1 U × {[γ]: γ(1) = x0} → p (U) so p : X˜ → X is a covering. A function λ : I → X˜ is continuous if and only if there is a (continuous) map Λ : I × I → X with λ(s) = [Λ(s, ·)] and Λ(s, 0) = x0 for s ∈ I. Letx ˜0 be the class of the constant path at x0. Then the path λ begins and ends atx ˜0 if and only if there is a map Λ with Λ(0, t) = Λ(1, t) = x0 for t ∈ I. The set {0, 1} × I ∪ I × {0} is a deformation retract of the square I × I. Composing Λ with this deformation gives a homotopy from λ to a constant loop. Hence π1(X,˜ x˜0) is trivial. For details see [3] page 64. Exercise 2.16. Show that if two locally path connected spaces are homotopy equivalent, then so are their universal covers. (Assume that the spaces are semi locally simply connected so that they have universal covers.)

3 Group Actions

3.1. A left action (resp. right action) of a group G on a set X is a homomorphism (resp. anti-homomorphism) G → Perm(X) where Perm(X) denotes the group of permutations of X, i.e. the group of all bijections g : X → X. For a left action it is customary to write the evaluation map as G × X → X :(g, x) 7→ gx whereas for a right action we write G × X → X :(g, x) 7→ xg. This notation makes the homomorphism property look like the associative law. An action is called effective iff the homomorphism G → Perm(X) is injective; for an effective action we view G as a subgroup of Perm(X). An action is called free iff no element of G other than the identity has a fixed point, i.e. iff gx = x =⇒ g = id. When X is a topological space and G is a topological group (meaning that G is a topological space and the group operations are continuous) it is required that the evaluation map be continuous; it then follows the action takes values in the subgroup homeomorphism subgroup Homeo(X) ⊂ Perm(X) of homeo- morphisms from X onto itself. In this section the group G will always have the discrete topology, i.e. every subset is open. 3.2. An action of G on X is called transitive iff for every pair of points x, y ∈ X there exists g ∈ G with gx = y (or xg = y for a right action). For each x ∈ X the subgroup of all g ∈ G which satisfy gx = x is called the stabilizer group (some authors say isotropy group) of x and is denoted Gx. When the action is transitive the map G/Gx → X : gGx 7→ gx

6 is well defined and bijective. Here G/H denotes the set of left cosets gH of the subgroup H ⊂ G: the group G acts on the set G/H on the left. The map G/Gx → X intertwines the two actions. As a right action determines a left action and vice versa (precompose with the anti-automorphism g 7→ g−1), all this holds for right actions as well. We denote the set of right cosets of H by H\G. 3.3. Let p : Y → X be a covering space and assume that Y and X are path connected. For x0 ∈ X the fundamental group π1(X, x0) acts on the fiber −1 p (x0) on the right by the rule y0[γ] = y1 iff the liftγ ˜ of γ withγ ˜(0) = y0 satisfiesγ ˜(1) = y1. As Y is path connected this action is transitive. The −1 stabilizer group of the point y0 ∈ p (x0) is precisely p∗π1(Y, y0) so there is a bijection −1 p∗π1(Y, y0)\π1(X, x0) → π1 (x0) −1 between the set of right cosets of p∗π1(Y, y0) in π1(X, x0) and the fiber π1 (x0). We may summarize this as: the cardinality of the fiber is the index of the subgroup.

3.4. An automorphism of a covering space p : Y → X is called a deck trans- formation; we denote the group of all deck transformations of the cover by Aut(p). Thus an element of Aut(p) is a homeomorphism f : Y → Y such that p ◦ f = f. 3.5. A group G acts on the set Sub(G) of it subgroups via the adjoint action,

Sub(G) → Sub(G) : H 7→ gHg−1 for g ∈ G. The stabilizer of a subgroup H under this action is called the normalizer of H in G and denoted

N(H,G) := {g ∈ G : gHg−1 = H}.

It is the largest subgroup of G containing H as a normal subgroup; the subgroup H is a normal subgroup of G if and only if N(H,G) = G.

Remark 3.6. If G acts on X, H is a subgroup of G, and

XH := {x ∈ X : Gx = H}. denotes the set of points in X having stabilizer group H, then

N(H,G) = {g ∈ G : g(XH ) = XH } and the formula

(G/N) × XH → XH :(gH, x) 7→ gx, N := N(H,G) gives a well deﬁned action of G/N on XH .

7 Theorem 3.7. Let p : Y → X be covering space with X and Y path connected −1 and locally path connected and assume x0 ∈ X and y0 ∈ p (x0). Abbreviate N := N(p∗π1(Y, y0), π1(X, x0)). Then the groups Aut(p) and N/p∗π1(Y, y0) are anti isomorphic. Corollary 3.8. The following are equivalent:

−1 (i) The group Aut(p) acts transitively on the ﬁber p (x0).

(ii) The subgroup p∗π1(Y, y0) is a normal subgroup of π1(X, x0).

(iii) For every element [γ] ∈ π1(X, x0) either every lift of γ is closed or no lift is closed. When these equivalent conditions hold the covering is called normal (regular by some authors).

−1 Remark 3.9. For a normal covering p : Y → X and y0 ∈ p (x0), the groups Aut(p) and π1(X, x0)/p∗π1(Y, y0) are anti isomorphic. In particular, for the universal cover p the groups Aut(p) and π1(X, x0) are anti isomorphic. 3.10. Let a discrete group G act on the right on a topological space X and X/G denote the orbit space, i.e.

X/G := {xG : x ∈ X}, xG := {xg : g ∈ G}.

Then the projection X → X/G (with the quotient topology) is a covering space if and only if every point x ∈ X has a neighborhood U such that (Ug1)∩(Ug2) = ∅ for g1 6= g2. When X → X/G is a covering, the group G is (anti isomorphic to) the group of deck transformations so the covering is normal. Conversely, if p : X → Y is a normal covering and G = Aut(p) then Y =∼ X/Gop. (Note. The group Aut(p) is a subgroup of the homeomorphism group and acts on the left. To get a right action we replace G by Gop where the opposite group Gop of G has the same underlying set as G and the product of a and b in Gop is product of b and a in G. Thus the map G → Gop : g 7→ g−1 is an isomorphism.)

Theorem 3.11 (Classiﬁcation Theorem). Let (X, x0) be a path connected pointed space which admits a universal cover. Then the functor described in 2.11 is an equivalence of categories as explained in the proof below. Proof. This means that

(i) For every subgroup H ⊂ π1(X, x0) there is a covering p :(Y, y0) → (X, x0) with H = p∗π1(Y, y0).

(ii) If p :(Y, y0) → (X, x0) and q :(Z, z0) → (X, x0) are connected covering spaces of (X, x0) satisfying p∗π1(Y, y0) ⊂ q∗π1(Z, z0), then there exists a unique map φ :(Y, y0) → (Z, z0) with q ◦ φ = p.

(iii) If p∗π1(Y, y0) = q∗π1(Z, z0), then φ is a homeomorphism and hence an isomorphism of pointed coverings.

8 Part (ii) is a restatement of Corollary 2.12 and part (iii) follows immediately from the uniqueness. For part (i) we construct the inverse functor. Let u : X˜ → −1 X be a universal cover. Abbreviate G := π1(X, x0) and choosex ˜0 ∈ u (x0). The group G acts on X˜ on the right by path lifting starting atx ˜0. The projection u induces a homeomorphism from the orbit space X/G˜ to X. A subgroup H ⊂ G determines a covering ∼ pH : X/H˜ → X/G˜ = X. An inclusion K ⊂ H ⊂ G of subgroups determines a commutative diagram

fK.H X/K˜ / X/H˜ JJ t JJ tt JJ tt pK JJ tt pH J% ty t X/G˜ = X

Example 3.12. The universal cover of a wedge X = S1 ∨ S1 of circles is an inﬁnite tree with four edges at each vertex. Various other coverings of X are shown in [3] on page 58. Example 3.13. A graph is a one dimensional cell complex. A contractible graph is called a tree. A connected graph X contains a maximal tree A and for any maximal tree A in X we have that X/A is a wedge of circles. Now (X,A) is a cellular pair and A is contractible so the projection X → X/A is a homotopy equivalence. Thus the fundamental group π1(X) is free for any graph X. This implies that a subgroup of a free group is free as follows. Let F be a free group and X be a wedge of circles with one circle for every generator of X so that F = π1(X). For any subgroup G of F there is a connected cover p : Y → X with p∗π1(Y ) = G. As p∗ is injective it follows that G is free. 3.14. Now assume that X is connected and locally path connected and that p : (X,˜ x˜0) → (X, x0) is a universal cover. Then X = X/G˜ where G = π1(X, x0) = Aut(p). Given any discrete space F and any left action ρ : G → Perm(F ) deﬁne −1 X˜ ×ρ F := {[˜x, v] :x ˜ ∈ X,˜ v ∈ F }, [˜x, v] := {(˜xg, ρ(g) v): g ∈ G}.

It is not hard to prove that X˜ ×ρ F → X : [˜x, v] 7→ p(˜x) is a covering space, connected if and only if the action on F is transitive, and that this operation determines a bijective correspondence between isomorphism classes of left actions and isomorphism classes of coverings. (See [3] page 68.) Remark 3.15. This construction is a special case of a more general construction in the theory of fiber bundles as follows. A principal fiber bundle is a fiber bundle π : P → B such that the total space P is equipped with a free transitive right action action of a topological group G whose orbits are the fibers of π. Given any left action ρ : G → Homeo(F ) on a topological space F the associated fiber bundle P ×ρ F → B :[u, ξ] → π(u) is defined by −1 P ×ρ F := {[u, ξ]: u ∈ P, ξ ∈ F }, [u, ξ] := {(ug, ρ(g) ξ): g ∈ G}.

9 Exercise 3.16. The projective plane is the quotient P 2 = S2/G where G = 2 2 {±idS2 }. Find all coverings of P × P . Exercise 3.17. The commutator subgroup of a group G is the subgroup [G, G] of G generated by all commutators [a, b] := aba−ab−1 with a, b ∈ G. It is a normal subgroup; the quotient group G/[G, G] is called the Abelianization of G. Show that the Abelianiazation is characterized by the following universal mapping property: For every homomorphism φ : G → A where A is an Abelian group, there is a unique homomorphism ψ : G/[G, G] → A such that ψ ◦ π = φ where π : G → G/[G, G] is the projection.

2 2 m Exercise 3.18. Deﬁne gm,n : R → R by gm,n(x, y) = (x + m, (−1) y + n) 2 and let G = {gm,n : m, n ∈ Z}. The quotient R /G is called the Klein bottle. Determine all double coverings (i.e. the cardinality of the ﬁber is two) of the Klein bottle. Hint: Compute the Abelianization of G.

4 Van Kampen’s Theorem

Deﬁnition 4.1. Let {Gα}α∈Λ be an indexed collection of groups. The free product of this collection is theF set of all equivalence classes of words (ﬁnite sequences) in the disjoint union α∈Λ Gα where the equivalence relation is generated by the following operations: (1) adding or deleting an identity element so that

w1 ··· wkeαwk+1 ··· wn ∼ w1 ··· wkwk+1 ··· wn

where eα is the identity element of Gα; and (2) multiplying adjacent elements if they belong in the same group so that

w1 ··· wkwk+1 ··· wn ∼ w1 ··· (wkwk+1) ··· wn

if wk, wk+1 ∈ Gα.

The free product G of the collection {Gα}α∈Λ is often denoted by Y∗ G = Gα α∈Λ

(Hatcher [3] uses the notation G = ∗αGα) and when Λ = {1, 2, . . . , n} is ﬁnite the free product is often denoted by

G = G1 ∗ G2 ∗ · · · ∗ Gn. The free product G is a group: the group operation is induced by catenation and the identity element is the equivalence class of the empty word. When each Gα is isomorphic to Z, the free product is the free group with one generator for each index α ∈ Λ.

10 4.2. Below we will consider an indexed collection of groups {Gα}α∈Λ and a doubly indexed collection

{iαβ : Gαβ → Gα}α,β∈Λ

of group homomorphisms such that Gαα = Gα, iαα = idGα , and Gαβ = Gβα. To such a system we associate a subgroup N of the free product G; it is the −1 normal subgroup generated by all elements of form iαβ(ω)iβα(ω) . We call the normal subgroup N of G the Van Kampen subgroup of the system. (When all the Gαβ for α 6= β are the same the quotient G/N is commonly called the amalgamated product.)

Theorem 4.3 (Van Kampen). Assume X is a space, x0 ∈ X, and [ X = Aα α∈Λ where each Aα is open and path connected and contains the base point x0. Let G be the free product of the indexed collection {π1(Aα, x0)}α∈Λ of groups and N be the Van Kampen subgroup determined by the doubly indexed collection

iαβ : π1(Aα ∩ Aβ, x0) → π1(Aα, x0) of homomorphisms induced from the inclusions (Aα ∩Aβ, x0) → (Aα, x0). Then

(i) If the double intersections Aα ∩ Aβ (α, β ∈ Λ) are path connected, then the natural homomorphism G → π1(X, x0) is surjective.

(ii) If the triple intersections Aα ∩ Aβ ∩ Aγ (α, β, γ ∈ Λ) are path connected. then the kernel of this homomorphism is N so that π1(X, x0) is isomorphic to G/N.

Proof. Let f :(I, ∂I) → (X, x0). Choose a partition 0 = a0 < a1 < ··· < an = 1 so ﬁne that for each k there is an index αk ∈ Λ with f([ak−1, ak]) ⊂ Aαk . Let fk = f|[ak−1, ak]. For k = 1, . . . , n − 1 choose a path gk from x0 to f(ak) lying in Aαk ∩ Aαk+1 and take g0 and gn to be the constant loop at x0. Then

[f] = [g0f1g¯1][g1f2g¯2] ··· [gn−2fn−1g¯n−1][gn−1fngn]

where [gk−1fkgk] is the image in π1(X, x0) of an element of π1(Aαk , x0). This proves (i).

To prove (ii) choose an element [f1]α1 [f2]α2 ··· [fm]αm ∈ G with [fj]αj ∈ 2 πj(Aαj , x0). Let F : I → X be a homotopy from the loop f = f1f2 ··· fm to the constant loop at x0, i.e. F (0, t) = f(t) and F (s, 0) = F (s, 1) = F (1, t) = x0.

We must show that [f1]α1 [f2]α2 ··· [fm]αm ∈ N. 2 Write I as a union of closed rectangles R so small that f(R) ⊂ Aα for some α ∈ Λ. Arrange that the interiors of these rectangles are disjoint and that any vertex v of one of the rectangles lies in at most two other rectangles. (The rectangles are jiggled a bit to achieve this.) By suitably subdividing and reindexing

11 we may assume that each fi is (up to reparameterization) the restriction of F to a union of consecutive intervals in 0 × I each of which is the intersection of 0 × I with one of these rectangles R. Let V be the set of all vertices of these rectangles. For each vertex v ∈ V choose a path gv from x0 to F (v) lying in Aα(R1) ∩ Aα(R2) ∩ Aα(R3) where R1,R2,R3 are the rectangles which contain v. Call two vertices u, v ∈ V adjacent if the line segment [u, v] from u to v lies in an edge of one of the rectangles. For each pair u, v of adjacent vertices let eu,v = F |[u, v]. Then the loop

fu,v := gueu,vg¯v lies in Aα ∩ Aβ where R1 and R2 are the two rectangles adjacent to [u, v] and f(R1) ⊂ Aα and f(R2) ⊂ Aβ. Denote the corresponding homotopy classes as

[fu,v]α ∈ π1(Aα, x0), [fu,v]β ∈ π1(Aβ, x0).

These are distinct elements of the free product G but project to the same element of G/N. Consider an element h of the free product G of form

h = [fv0,v1 ]α1 [fv1,v2 ]α1 ··· [fvn−1,vn ]αn .

Replacing a subscript αk by the another subscript βk as above changes the element h ∈ G but leaves its image in G/N unchanged. If two consecutive factors [fvk−1,vk ]αk [fvk,vk+1 ]αk+1 have the same subscript, i.e. if αk = αk+1 =: α, then vk−1, vk, vk+1 are three of the four vertices of a rectangle R and replacing 0 vk by the fourth vertex vk leaves h unchanged (as an element of G). Thus each 0 of these two operations (replacing αk by βk and replacing vk by vk) leaves the image of h in G/N unchanged. But by means of these operations we can move the constant class in G/N to the class of f in G/N. (Represent f as a product h where the vj lie in (0 × I) ∪ (I × 1).) Thus a preimage of [f] in G lies in N as claimed.

Example 4.4. By Van Kampen, π1(X ∨ Y ) = π1(X) ∗ π1(Y ). In particular, the fundamental group of a wedge of circles is a free group with one generator for each circle. Example 4.5. Any group G is the fundamental group of of a two dimensional cell complex X as follows. Represent G as a quotient G = F/H where F is the free group on the generators {xi}i∈I and H is the normal subgroup generated by (1) the words {rα}α∈Λ. Assume that the 1-skeleton X of X is is a wedge of circles (1) W 1 X := i∈I Si with one circle for each generator xi. Form X by attaching 2 1 1 each a 2-cell eα for each relation rα via a map φα : S → X representing (1) 2 rα ∈ F = π1(X ). For each α ∈ Λ choose zα ∈ eα. Let U be a contractible neighborhood of the wedge point not containing any zα and

(1) A = X \{zα : α ∈ Λ},B = U ∪ (X \ X ).

12 Then A and B are open sets, A deformation retracts onto X(1), B is contractible, and A ∩ B is homotopy equivalent to a wedge of circles with one circle for each relation rα. Since π1(B) is trivial the free product π1(A) ∗ π1(B) is π1(A) and the Van Kampen subgroup N is the normal subgroup generated by {rα}α∈Λ. Remark 4.6. Shelah [8] proves that the fundamental group of a compact metric space is either ﬁnitely generated or uncountable. 4.7. A link is the image K = f(S) of an embedding f : S → R3 of a ﬁnite disjoint union of circles. A link with one component (i.e. S = S1) is called a 3 knot. The fundamental group π1(R \ K) is called the group of the link or knot. One generally assumes that the embedding is smooth or piecewise linear (the precise terminology is tame) to avoid pathologies. There is an algorithm which computes the group of a knot or more precisely produces a presentation of the group known as the Wirtinger representation. It is described on page 55 of [3]. Exercise 4.8. Use the Wirtinger representation to prove that the Abelianiza- tion of the group of a knot is Z. Example 4.9. The complement R3 \ S1 of the unknot deformation retracts 1 2 3 1 onto a space homeomorphic to S ∨ S so π1(R \ S ) = Z. Exercise 4.10. Show that if S3 \ K ' X then R3 \ K ' S2 ∨ X. Hence 3 3 π1(S \ K) = π1(R \ K). Example 4.11. The complement S3 \ L of two linked circles

L = S1 × 0 ∪ 0 × S1 ⊂ S3 ⊂ C2 is homeomorphic to T 2 × R via the map µ ¶ et+iα e−t+iβ S1 × S1 × R → L :(eiα, eiβ, t) 7→ √ , √ e2t + e−2t e2t + e−2t

3 2 2 2 2 so (under a suitable embedding) π1(R \ L) = π1(S ∨ T ) = π1(T ) = Z . A suitable stereographic projection sends L to the disjoint union of the circle x2 + y2 = 1, z = 0 with the z-axis. Example 4.12. The complement S3 \ L0 of two unlinked circles deformation 1 2 1 2 3 0 retracts X ' S ∨ S ∨ S ∨ S so π1(R \ L ) is a free group on two generators. Example 4.13. Let p, q ∈ Z. The torus knot of type (p, q) is the knot √ K := {(u, v) ∈ S1 × S1 : up = vq} ⊂ S3( 2) √ √ where S3( 2) := {(u, v) ∈ C2 : |u|2 + |v|2 = 2} denotes the sphere of radius 2. Exercise 4.14. The one point compactiﬁcation of R3 is (homeomorphic to) S3. Show that if K ⊂ R3 is compact and S3 \ K ' Y , then R3 \ K ' S2 ∨ Y .

13 Exercise 4.15. Let Y result from the cube I3 by identifying opposite faces with a 90◦ right hand twist. Use Van Kampen’s Theorem to show that the fundamental group of Y is the eight element quaternion group

G = {±1, ±i, ±j, ±k}.

Then give another proof by showing that the universal cover is S3 → S3/G =∼ Y . Hint: S3 =∼ ∂[−1, 1]4 and each face of [−1, 1]4 is a cube. Exercise 4.16. Conley [1] page 26 studies a ﬂow entering a cylinder X = D2 ×I at the top D2 ×1, exiting at the bottom D2 ×0, with the orbits running vertically downward on the side S1×I ⊂ ∂X, and with a knotted orbit segment A inside the cylinder running from the point p = (0, 0, 1) ∈ D2 × 1 in the top of the cylinder to a point q = (0, 0, 0) ∈ D2 × 0 in bottom. He argues that there must be a nonempty invariant set inside the cylinder since otherwise the complement of the knotted orbit in the cylinder would deformation retract onto the punctured disk at the bottom. Prove that (D2 × 0) \ q is not a deformation retract of X \ A. In this context the condition that A is knotted means that π1(X \ A) 3 is not Z. To see that this can occur, show that π1(X \ A) = π1(R \ K) where K is constructed from A by adjoining the polygonal arc with successive vertices p0 = 0, p1 = (0, 0, 2), p2 = (0, 2, 2), p3 = (0, 2, −2) , p4 = (0, 0, −2), and p5 = q.

References

[1] Charles Conley: Isolated Inveariant Sets and the Morse Index, CBMS Con- ferences in Math 38, AMS 1976. [2] William Fulton: Algebraic Topology, A First Course Springer GTM 153, 1995. [3] Allen Hatcher: Algebraic Topology, Cambridge U. Press, 2002.

[4] John G. Hocking & Gail S. Young: Topology, Addisow Wesley, 1961. [5] Morris W. Hirsch: Diﬀerential Topology, Springer GTM 33, 1976.

[6] William S. Massey: A Basic Course in Algebraic Topology, Springer GTM 127, 1991. [7] James R. Munkres: Elements of Algebraic Topology, Addison Wesley, 1984. [8] S. Shelah: Can the fundamental group of a space be the rationals? Proc. Amer. Math. Soc. 103 (1988) 627–632. [9] Stephen Smale: A Vietoris mapping theorem for homotopy. Proc. Amer. Math. Soc. 8 (1957), 604–610

[10] J. W. Vick: Homology Theory (2nd edition), Graduate Tests in Math 145, Springer 1994.

14 [11] J.H.C. Whitehead: Combinatorial homotopy II, Bull. A.M.S 55 (1949) 453-496.

[12] George W. Whitehead: Elements of Homotopy Theory Graduate Tests in Math 61, Springer 1978.