(H) Lecture 19 Lecturer: Zuoqin Wang Time: May 19, 2020

VAN KAMPEN’S THEOREM

Last time we learned:

• invariance: If X ∼ Y , then π1(X) ' π1(Y ). 1 • π1(S ) ' Z. • Many applications. n n Recall that in lecture 17, we proved π1(S ) = {e} for n ≥ 2 by covering S with two simply-connected open sets whose intersection is path-connected, and then represent each loop in Sn (up to path homotopy) by a number of loops each of which lies in a single simply-connected open sets. Today we will extend this method to the case where we have more than two open sets and that these open sets are no longer simply-connected.

1. Some group theory

1 1 1 2 We have seen π1(S ) ' Z. As an immediate consequence, we have π1(S ×S ) ' Z . Note that the of S1 × S1 is an abelian group with two generators. Here is a picture showing why the two generators commutes:

Next we consider the following “figure 8” :

1 2 VAN KAMPEN’S THEOREM

Geometrically it is quite obvious that π1(“∞”) is also generated by two elements, namely [α] and [β] as shown in the picture. However, there is no reason that [α][β] = [β][α] in this case, because, intuitively, there is no space to “move” one loop the other. As a result, a product of like [α]3[β]2[α]−2 should be represents a loop that you first go along α for 3 cycles, then go along β for two cycles, and then go along α but in the reverse direction for 2 cycles. It is different from the loop [α][β]2. More generally, any (finite) word consisting of [α], [β], [α]−1 and [β]−1 should represents a loop in the “figure 8” and different words should represent different (=non-homotopic) loops. So the fundamental group π1(“∞”) is no longer the product group Z×Z, instead it should be Z ∗ Z, the free product of two copies of Z, which we will recall the definition and basic properties below. Definition 1.1. Given any set S (here we don’t assume S to be a finite set), the free group with free generating set S is the group −1 hSi = {c1c2 ··· cn | n ≥ 0, ci ∈ S or ci ∈ S} in which the group multiplication is given by

c1c2 ··· cn · cn+1 ··· cn+m = c1c2 ··· cncn+1 ··· cn+m and group inverse is given by −1 −1 −1 −1 −1 (c1c2 ··· cn) = cn cn−1 ··· c2 c1 , where for each c ∈ S, we let (c−1)−1 = c. The group identity element is defined to be the empty word which is denoted by e. Any element in hSi is called a word. Example. If S = {a, b, c}, then the word abbba−1a−1ccb−1 · aa−1bbb = abbba−1a−1ccb−1aa−1bbb = ab3(a−1)2c2b2 ∈ hSi, where we used the fact b−1aa−1b = b−1b = e ∈ hSi.

Remark. If S = {a}, then hSi ' Z since n hSi = {c | n ∈ Z}, with the group multiplication cn · cm = cn+m

A remarkable fact about the free group hSi is the following universal property: Proposition 1.2 (Universal Property). For any group G and for any map f : S → G, there exists a unique group homomorphism ϕ : hSi → G such that f = ϕ ◦ i where i : S,→ hSi is the inclusion map. VAN KAMPEN’S THEOREM 3

As a consequence, if G is any group and S is any set of generators of G (you may take S = G if you want), then there exists a unique surjective group homomorphism ϕ from hSi to G. As a consequence, Proposition 1.3. Any group is isomorphic to a quotient group of a free group hSi. G = hSi/ ker ϕ. The kernel of ϕ is a subgroup of hSi. So the group G is obtained by the free group hSi by setting all elements in ker(ϕ) to be the identity. Of course since ker(ϕ) is a group, it is enough to set generators of ker(ϕ) to be the identity. These equations are called “relations”. So we can express any group via generators and relations (called a presentation of G) G = hS | Ri, where S is a set of generators and R is a set of relations. For example, Example.

(1) Let G = Zn. Then G is generated by one element, which we denote by a. Let hSi = hai ' Z. Then the surjective group homomorphism is given by ϕ : hSi → G, ak 7→ [k], with kernel ker ϕ = {· · · , a−2n, a−n, a0, an, a2n, ···}. It follows n Zn = ha | a = 1i. (2) For G = Z2 = Z ⊕ Z, it is an abelian group with two generators, and we have 2 Z = ha, b | ab = bai. Conversely, given any presentation of a group G = hS|Ri with generators and relations, we can write G as a quotient group G = hSi/N, where N is the smallest normal subgroup of hSi generated by all those elements s so that R is given by s = 1. Next we define the free products of groups. Definition 1.4. Let G, H be groups. As above we can define a word to be a (formal) product s = s1s2 ··· sn where si ∈ G or H and n ≥ 0. The set of all such words form a group

G ∗ H = {s1s2 ··· sn | si ∈ G or H} with the obvious group operations “connect two words” and “reverse a word” as before. This is called the free product of G and H. 4 VAN KAMPEN’S THEOREM

Similarly we can define the free product of a family of groups Gα to be

∗αGα = {s1s2 ··· sn | for each i, there exists α s.t. si ∈ Gα}.

In general, if Gα = hSα|Rαi, then

∗αGα = h∪αSα| ∪α Rαi. Example.

• Z ∗ Z = ha, bi. (So the free product of abelian groups could be non-abelian) •h a | a3 = 1i ∗ hb | b4 = 1i = ha, b | a3 = b4 = 1i. (so the free product of finite groups could be infinite) 2 2 • Z2 ∗ Z2 = ha, b | a = b = 1i = {1, a, b, ab, ba, aba, bab, abab, ···}. (which is the semi-product of the subgroups Z and Z2 generated by ab and a)

One of the most important property for free products ∗αGα of groups Gα is the following universal property: Proposition 1.5 (Universal Property). For any group H and any collection of group homomorphisms ϕα : Gα → H, there exists a unique “lifted” group homomorphism

ϕ : ∗αGα → H which extends each ϕα, namely for gk ∈ Gαk , we have

(*) ϕ(g1 ··· gn) = ϕα1 (g1) ··· ϕαn (gn). Remark. In fact, one can use (*) to define the lifted group homomor- phism ϕ : ∗αGα → H (and check it is well-defined).

Free product with amalgamation. More generally, let G, H be groups, and let ϕ : F → G, ψ : F → H be group homomorphisms. Let N be the smallest normal subgroup of G ∗ H that contains all elements of the form ϕ(a)ψ(a)−1, a ∈ F Definition 1.6. The free product with amalgamation of G and H w.r.t. ϕ and ψ is

G ∗F H := G ∗ H/N Equivalently, it is the free product G ∗ H modulo the relation ϕ(a)ψ(a)−1 = 1, ∀a ∈ F . (Instead of using all elements a ∈ F , it is enough to take all a’s from a set of generator in F .) VAN KAMPEN’S THEOREM 5

2. van Kampen’s Theorem Now we are ready to state the following powerful tool using which we can compute the fundamental group of a via the knowledge of the fundamental groups of some subsets.

Setting: Suppose X = U1 ∪ U2 where U1,U2 are open in X. Suppose U1,U2 and U1 ∩ U2 are path connected, and x0 ∈ U1 ∩ U2. Then the inclusion maps

U1 ∩ U2 ,→ U1 and U1 ∩ U2 ,→ U2 induce group homomorphisms

ϕ : π1(U1 ∩ U2, x0) → π1(U1, x0) and ψ : π1(U1 ∩ U2, x0) → π1(U2, x0). The van Kampen’s theorem1 claim that the fundamental group of X is the amalgamated free product of π1(U1) and π1(U2) with respect to the group homomorphisms ϕ and ψ: Theorem 2.1 (van Kampen’s Theorem).

π1(X, x0) ' π1(U1, x0) ∗π1(U1∩U2,x0) π1(U2, x0) Van Kampen’s theorem has the following generalized version in which a covering of X by arbitrarily many open sets is allowed. Since the proof of the original version is as complicated 2 as the general version, and since the general version is not a consequence of the original version, we shall state the prove the general version.

Theorem 2.2 (Van Kampen’s theorem, generalized version). Suppose {Uα} is an open covering of X such that each Uα is path-connected and there is a common base point x0 sits in all Uα. Let

jα : π1(Uα) → π1(X) be the group homomorphism induced by the inclusion Uα ,→ X. Let

Φ: ∗απ1(Uα) → π1(X) be the lifted group homomorphism as described by the universal property.

(1) Suppose each intersection Uα ∩ Uβ is path connected. Then Φ is surjective. (2) If in addition, each intersection Uα ∩ Uβ ∩ Uγ is path connected, then Ker(Φ) is −1 the normal subgroup N generated by all elements of the form ιαβ(ω)ιβα(ω) , where ιαβ : π1(Uα ∩ Uβ) → π1(Uα) is the group homomorphism induced by the inclusion Uα ∩ Uβ ,→ Uα. As a result, Φ induces a group isomorphism

∗απ1(Uα)/N ' π1(X).

1The theorem is also known as the Seifert-van Kampen theorem 2Let me emphasis here: the proofs are as complicated as each other, not as simple as each other! 6 VAN KAMPEN’S THEOREM

Proof. Step 1. Φ is surjective.

Given any loop γ : [0, 1] → X with base point x0, by Lebesgue number lemma, there exists a partition 0 = s0 < s1 < ··· < sm = 1 such that each γ([si, si+1]) is contained in a single Uα. Repeating the proof of Proposition 2.5 in Lecture 17 we get ¯ ¯ γ ∼ (γ1 ∗λ1)∗(λ1 ∗γ2 ∗λ2)∗· · ·∗(λm−1 ∗γm), p ¯ where each λi ∗γ2 ∗λi+1 is a loop in one Uα. So [γ] ∈ Image(Φ), i.e. Φ is surjective. [Note: The above argument also shows that Φ is not injective, since if we denote the Uα containing f([si, si+1]) by Ui, then the loop λi∗γi+1∗λi+1 lies in both Ui and Ui+1, and thus can be viewed either as a loop in Ui or as a loop in Ui+1, which will give us different representation in ∗απ1(Uα) as long as Ui 6= Ui+1.] Step 2. Reduction of the problem.

Let [γ] ∈ π1(X, x0). If γ ∼ γ1 ∗γ2 ∗· · ·∗γn, where each γi is a loop in some Uα with base point x0, then we write formally [γ] = [γ1]∗[γ2]∗· · ·∗[γn] and call it a factorization of [γ]. [One can regard a factorization of [γ] as a word which lies in the free product ∗απ1(Uα).] Note that the surjectivity of Φ implies that each class [γ] ∈ π1(X, x0) admits a factorization. As we have explained above, one class could have many different factorizations. We say two factorizations of [γ] are equivalent if they are related by a finite number of the following two moves and their inverses:

• If [γi], [γi+1] are in the same π1(Uα, x0), then replace [γi]∗[γi+1] by [γi ∗γi+1]. • If γi is a loop in Uα ∩ Uβ, then identify [γi] ∈ π1(Uα, x0) with [γi] ∈ π1(Uβ, x0).

Note that the first move does not change the element of ∗απ1(Uα) defined by the factorization, while the second move does not change the image of the element defined by the factorization in the quotient group ∗απ1(Uα)/N because N is normal. As a result, equivalent factorizations given the same element in ∗απ1(Uα)/N. We will prove Reduced problem: Any two factorizations of [γ] are always equivalent.

This is enough for our purpose, since it implies that the map ∗απ1(Uα)/N → π1(X, x0) induced by Φ is injective. Step 3. Decomposition of [0, 1] × [0, 1]. Now suppose we have two factorizations of [γ], namely, 0 0 [γ1]∗[γ2]∗· · · [γM ] = [γ] = [γ1]∗· · ·∗[γN ]. By definition, we have 0 0 γ1 ∗· · ·∗γM ∼ γ ∼ γ1 ∗· · ·∗γN . VAN KAMPEN’S THEOREM 7

0 0 Let F : [0, 1] × [0, 1] → X be a path homotopy connecting γ1 ∗· · ·∗γM and γ1 ∗· · ·∗γN . By Lebesgue number lemma, we can decompose [0, 1] × [0, 1] into finitely many small i i rectangles Rij = [tj, tj+1] × [si, si+1], with 0 = s0 < s1 < ··· < sK < sK+1 = 1, s.t.

• F (Rij) is contained in a single Uα, which we will denoted by Uij. 0 • F (t, 0)|[t0 ,t0 ] = γk(t) and F (t, 1)|[tK ,tK ] = γ (t). nk nk+1 mk mk+1 k • Each point in [0, 1] × [0, 1] lies in at most three rectangles Rij. The decomposition3 might looks like the following:

For simplicity, we label these “bricks” be R1,R2, ··· ,RL, say, from left to right, from bottom to top. Step 4. Equivalent factorizations from bottom to top.

Note: F maps the left edge and the right edge to x0. So if λ is a path in [0, 1]×[0, 1] from the left edge to the right edge, then F |λ is a loop with base point x0.

Now for each r, we let λr be the path from left to right separating the first r rectangles R1,R2, ··· ,Rr form the remaining rectangles. For example, λ0 is the bottom edge while λL is the top edge. For each vertex v of Rr with F (v) 6= x0, then by construction, F (v) belongs to the intersection of at most three open sets Uα’s. Since any non-empty intersection of three Uα’s is path connected, there is a path µv from x0 to F (v) which lies in the intersection of those (two or three) Ui’s that corresponds to the Rj’s containing v. As in the proof of surjectivity we can insert into these µv’s and

µ¯v’s into F |λr to decompose the loop into a number of loops based at x0, so that each loop lies in a single Uα. By this way we obtain a factorization of [F |λr ]. Note:

• Different choices of Ur will change the factorization of [F |λr ] to an equivalent factorization.

• Furthermore, the factorization of [F |λr ] and the factorization of [F |λr+1 ] are equivalent, since we may push the path λr across Rr+1 to the path λr+1, which

relates F |λr to F |λr+1 by a homotopy within Ur. 0 0 Step 5. The factorizations [γ1]∗[γ2]∗· · · [γM ] and [γ1]∗[γ2]∗· · · [γN ] are equivalent.

3This is a “brick” decomposition. One can also use a hexagonal decomposition instead, to guarantee that each point lies in at most three elements in the decomposition. 8 VAN KAMPEN’S THEOREM

Finally we show that the factorization [F |γ0 ] is equivalent to [γ1]∗[γ2]∗· · · [γM ]. To do so, for each vertex v in the bottom edge, we choose the path λv to lie not only in the two Uk’s corresponding to the two Rk’s containing v, but also in the Uα for the γi. [Note: If v is is the common end point of two consecutive γi’s, then F (v) = x0 and thus there is no need to choose gv.] 0 0 By the same way the factorization [F |γL ] is equivalent to [γ1]∗[γ2]∗· · · [γN ]. This solves the reduced problem and thus finishes the proof of theorem. 

Remark. The conditions “Uα ∩ Uβ and Uα ∩ Uβ ∩ Uγ are path connected” can’t be removed. (1) One can’t use van Kampen’s theorem to compute the fundamental group of S1. 1 1 In fact if we let U1 = S \{1} and U2 = S \{−1}, then π1(U1) = π1(U2) = {e}, so the group homomorphism

1 Φ: π1(U1) ∗ π1(U2) → π1(S )

can’t be surjective. The reason is U1 ∩ U2 is not path connected. (2) To see why we can’t remove the condition “each Uα ∩Uβ ∩Uγ is path connected”, let’s consider the following graph:

Let U1 = X \{a},U2 = X \{b} and U3 = X \{c}. Then

π1(Ui) ' Z, 1 ≤ i ≤ 3.

By applying van Kampen’s theorem to the cover {U1,U3} of X, we get

π1(X) ' Z ∗ Z.

But if we try to apply van Kampen’s theorem to the cover {U1,U2,U3} (which does not satisfy the condition of van Kampen’s theorem since the intersection is not path-connected), we would get a wrong answer π1(X) ' Z ∗ Z ∗ Z. [It is not obvious that Z ∗ Z ∗ Z 6' Z ∗ Z. One way to see this is to prove that they have different abelianization.] VAN KAMPEN’S THEOREM 9

3. Applications of van Kampen’s theorem In what follows, we use van Kampen’s theorem to compute the fundamental groups. In most cases, it is enough to use the original version, i.e. cover X by two open sets. Example.

(1) If U1,U2 are simply connected, and U1 ∩ U2 is path connected, then

π1(U1 ∪ U2) '{e}. n In particular, π1(S ) = {e} (See Lecture 17). (2) We first introduce a conception.

Definition 3.1. Let (Xα, xα) be a family of pointed topological spaces. The wedge sum of these pointed spaces is the “one-point-union” space defined by _ a Xα = Xα/ ∼, α

where ∼ is the equivalence relation that identify all base points xα’s. For example, the wedge sum of two S1 is the “figure-8” graph in page 1. By using van Kampen’s theorem, it is easy to see 2 2 π1(S ∨ S ) '{e} ∗{e} {e}'{e}, 1 2 π1(S ∨ S ) ' Z ∗{e} {e}' Z, 1 1 π1(S ∨ S ) ' Z ∗{e} Z ' Z ∗ Z = ha, bi. we can also compute the fundamental group of the wedge sum of k copies of S1, using the general version of van Kampen’s theorem: 1 1 aaaaaπ1(S| ∨ ·{z · · ∨ S}) ' Z| ∗ ·{z · · ∗ Z} /{e}' Z| ∗ ·{z · · ∗ Z} ' ha1, a2, ··· , aki. k k k

More generally, suppose (Xα, xα) are pointed topological spaces such that each xα is a deformation retract of some simply connected open set Uα in Xα. Then we can choose Uα to be the wedge sum of Xα with all Uβ’s with β 6= α. Ap- plying the general version of van Kampen’s theorem, we get a contractible neighborhood in Xi. Then the same argument above would give us _ π1( Xα) ' ∗απ1(Xα). α (3) The fundamental group of a connected graph. Let’s explain by an example.

Let G0 = a maximal tree in G (which is simply-connected and contains all vertices G = of G), as marked by red in the picture. Label the remaining edges by e1, e2, e3, ··· 10 VAN KAMPEN’S THEOREM

To compute the fundamental group of G, we start with G0. Since G0 is a tree which is contractible,

π1(G0) '{e}.

Now we add the edge e1 and compute the fundamental group of G1 = G0 ∪ e1. The decomposition of G1 into open subsets is shown as the following picture:

1 Since U1 ∩ V1 is contractible, while V1 is homotopy equivalent to S , we get

π1(G1) ' π1(G0) ∗{e} Z = Z.

Then we repeat the same procedure by adding e2 to get

π1(G2) = π1(G1 ∪ e2) ' π1(G1) ∗{e} Z ' Z ∗ Z and finally

π1(G) ' Z ∗ Z ∗ Z. In general, given any connected graph G, • there always exists a maximal tree G0 (which is simply-connected and contains all vertices of G). [Proof: Induction for finite tree. Use Zorn lemma for general case.] • If we let {eα} be the set of all edges in G that are not in G0, Then

π1(G) ' ∗eα Z. For the case of finite graph G, if we denote |V (G)| = the number of vertices in G and |E(G)| = the number of edges in G,

then the number of edges in G0 is |V (G)| − 1. It follows

π1(G) ' Z| ∗ ·{z · · ∗ Z}, k where k = |E(G)| − |V (G)| + 1. VAN KAMPEN’S THEOREM 11

2 (4) Next we compute the fundamental group of Σ1 = T . Although we already know the fundamental group by using the product formula,

2 1 1 2 π1(T ) = π1(S × S ) ' Z , we want to re-compute its fundamental group via van Kampen, because the computation shed a light on how to compute the fundamental group of Σg (the closed oriented surface with g holes), the latter can not be computed via the product formula. We first write T2 to a union of 2 f U1 = T \ D and U2 = D, where D is a small disc and Df is a small disc which contains D. We first compute 2 the fundamental group π1(T \ D). According to the following picture,

we have

π1(U1) ' π1(“∞”) ' Z ∗ Z = ha, bi.

On the other hand, obvious U2 is contractible, and U1 ∩ U2 is a “thin annulus” which is homotopic equivalent to a circle, so

1 π1(U2) ' π1(“pt”) = {e} and π1(U1 ∩ U2) ' π(S ) ' Z.

Unfortunately the above information is still not enough 2 to determine π1(T ), since we still need the map

ϕ : π1(U1 ∩ U2) → π1(U1) This group homotopy is induced by the inclusion map. According to the picture, the generator of π1(U1 ∩ U2) is a circle which can be deformed inside U1 to the boundary loop aba−1b−1. In other words, ϕ(1) = aba−1b−1 So we conclude

2 π1(T ) ' (Z ∗ Z) ∗Z {e} = ha, b |aba−1b−1 = 1i = ha, b | ab = bai ' Z ⊕ Z. 12 VAN KAMPEN’S THEOREM

2 2 (5) The fundamental group of Σ2 = T #T . The decomposition of Σ2 is given as below:

2 Note that in computing π1(T ), we have already computed the fundamental group of U1 and U2. So we get

π1(U1) ' ha1, b1i, π1(U2) ' ha2, b2i, π1(U1 ∩ U2) ' Z. Moreover, we also explained just now that −1 −1 −1 −1 ϕ(1) = a1b1a1 b1 , ψ(1) = b2a2b2 a2 . It follows from van Kampen’s theorem that 2 2 −1 −1 −1 −1 π1(T #T ) ' ha1, b1, a2, b2 | |a1b1a1 b1 {za2b2a2 b2 } = 1i ϕ(1)ψ(1)−1 Note that this is no longer abelian. In general by induction one can compute the fundamental group of 2 2 Σg = T| # ···{z #T}, g

and the result is 2 2 −1 −1 −1 −1 π1(|T # ···{z #T}) ' ha1, b1, ··· , ag, bg | a1b1a1 b1 ··· agbgag bg = 1i g