Theorem of Van Kampen, Covering Spaces, Examples and Complements

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Theorem of Van Kampen, Covering Spaces, Examples and Complements Notes on Algebraic Topology 1.5 Theorem of Van Kampen The theorem of Van Kampen(28) allows the computation of the fundamental group of a space in terms of the fundamental groups of the open subsets of a suitable cover. Let X be an arcwise connected topological space, (Λ, ) an ordered set, U : λ Λ an ≤ { λ ∈ } open cover of X such that: (a) all Uλ are arcwise connected; (b) λ µ if and only if U U ; ≤ λ ⊆ µ (c) the family U : λ Λ is stable under finite intersections; { λ ∈ } (d) there exists x0 λ Λ Uλ. ∈ ∈ Let us denote by ι :U X and ι : U U the canonical inclusions (where λ λ −→ λ,µ λ −→ µ λ µ), and write for short π (U ) instead of π (U ,x ). We then have morphisms ≤ 1 λ 1 λ 0 ι : π (U ) π (X) and ι : π (U ) π (U )(whereλ µ), which simply λ# 1 λ −→ 1 λ,µ# 1 λ −→ 1 µ ≤ associate to the class of a loop the class of the same loop viewed in the larger space. In particular, one has an inductive system π (U ),ι : λ, µ Λ (see Appendix A.1). { 1 λ λ,µ ∈ } Let M Λsuch that X = λ M Uλ. ⊂ ∈ Proposition 1.5.1. π (X)is generated by ι (π (U )) : λ M . 1 { λ# 1 λ ∈ } 1 Proof. Let γ : I X be a loop at x , δ>0 the Lebesgue number relative to the cover γ− (Uλ):λ M −→ 0 { ∈ } of I,0=t0 <t1 < <tk 1 <tk =1withtj tj 1 <δand λj M (with j =1,...,k)suchthat ··· − − − ∈ γ([tj 1,tj ]) Uλ .Sinceγ(tj ) Uλ Uλ (arcwise connected), let σj (for j =1,...,k 1) be an arc − ⊂ j ∈ j ∩ j+1 − into Uλ Uλ from x to γ(tj ). Setting γj = γ (reparametrized by sending tj 1 to 0 and tj to j j+1 0 [tj 1,tj ] − ∩ 1 | − 1 1 1) it clearly holds [γ]=[γ1] [γk]=[γ1 σ1− ] [σ1 γ2 σ2− ] [σk 2 γk 1 σk− 1][σk 1 γk]. ····· · · · · ····· − · − · − − · An obvious consequence of Proposition 1.5.1 is: Corollary 1.5.2. If the open subsets U (λ M) are simply connected, such is also X. λ ∈ n Example. S is simply connected for n 2. Namely, let N = en+1 be the North pole, S = N the n ≥ n − South pole, and set U = S N and V = S S :notingthatU V is arcwise connected and that \{ } \{ } ∩ both U and V are simply connected, just apply Corollary 1.5.2. Note that this argument does not apply 1 for S (in that case U V is not arcwise connected). ∩ In general one has the following result (see Appendix A.1 for the notion of “inductive limit” of an inductive system). Theorem 1.5.3. (Van Kampen) In the category Groups it holds π1(X)=limπ1(Uλ). λ Λ −→∈ (28)The result has been proved independently also by Karl Seifert in the 30s of last century; in fact, it is often referred to as “Seifert - Van Kampen theorem”. Corrado Marastoni 22 Notes on Algebraic Topology Proof. Consider a group L and a family of morphisms ψλ : π1(Uλ) L such that ψλ = ψµ ιλ,µ# for λ µ, −→ ◦ ≤ and let us see if there exists a unique morphism ψ : π1(X) L such that ψλ = ψ ιλ# for any λ Λ. So let −→ ◦ ∈ [γ] π1(X): by Proposition 1.5.1 we may write [γ]=ιλ #([γ1]) ιλ #([γk]) with [γj ] π1(Uλ ). If ψ ∈ 1 ····· k ∈ j exists, it must be necessarily unique because ψ([γ]) = ψ(ιλ #[γ1]) ψ(ιλ #[γk]) = ψλ ([γ1]) ψλ ([γk]) 1 ··· k 1 ··· k (the products are in L). We are left with showing that this is actually a well-posed definition for ψ,i.e.thatif [c ]=ι ([σ ]) ι ([σ ]) then also ψ ([σ ]) ψ ([σ ]) = e (where e is the identity element of L). x0 λ1 # 1 λk # k λ1 1 λk k ····· j 1 j ··· Let σ = σ1 σk be in X (hence, if t [ −k , k ]itholdsσ(t)=σj (kt (j 1))), and let h : I I X be a ····· ∈ − − 1× −→ homotopy rel ∂I between σ and c .Letε>0 the Lebesgue number relative to the cover h− (U ):λ Λ x0 λ √2 i 1 i j 1{ j ∈ } of I I,andletr N be such that kr <ε.Hence,settingRi,j = k−r , kr k−r , kr I I (for × r ∈ ×i j ⊂ × i, j =1,...,k ), there exists λi,j Λsuchthath(Ri,j ) Uλi,j .Letvi,j = kr , kr (hence Ri,j is the r ∈ ⊂ square with side k− and opposed vertices vi 1,j 1 and vi,j ), Uµ(i,j) the intersection of the (one, two or − − t+(i 1) j four) Uλ such that vi,j Rl,m,andγi,j apathinU from x to h(vi,j ). Let αi,j (t)=h − , l,m ∈ µ(i,j) 0 kr kr i t+(j 1) (path from h(vi 1,j )toh(vi,j )) and βi,j (t)=h r , r− (path from h(vi,j 1)toh(vi,j )): note that − k k − α r 1 α r 1 =[σ ] (for m =1 ,...,k)andthat α r (t)=β (t)=β r (t) x (for (m 1)k − +1,0 mk − ,0 m i,k 0,j k ,j 0 − ····· r ≡ t I and i, j =1,...,k ). From the equality [αi,j 1 βi,j ]=[βi 1,j αi,j ] one gets (by inserting the paths ∈ − · − · 1 1 γi,j and their inverses to base at x ) the relations (γi 1,j 1 αi,j 1) γ− (γi,j 1 βi,j ) γ− = 0 − − − i,j 1 − i,j 1 1 · · − · · · (γi 1,j 1 βi 1,j ) γ− (γi 1,j αi,j ) γ− in the group π1(Uλ ). Applying ψλ and setting − − − i 1,j − i,j i,j i,j · · − · 1 · · 1 ai,j = ψλ (γi 1,j αi,j ) γ− and bi,j = ψµ(i,j) (γi,j 1 βi,j ) γ− ,onethenhastheequality i,j − · · i,j − · · i,j ai,j 1 bi,j = bi 1,j ai,j in L.Knowingthata1,kr akr ,kr = e and that b0,j = bkr ,j = e (for any j = − r − ··· 1,...,k ), one has e = a1,kr akr ,kr =(b0,kr a1,kr )a2,kr akr ,kr = a1,kr 1(b1,kr a2,kr ) akr ,kr = ··· ··· − ··· = a1,kr 1 akr ,kr 1; by repeating the procedure one obtains a1,0 akr ,0 = e, as required. ··· − ··· − ··· Now the problem is to understand what lim π1(Uλ)seemslike. λ Λ −→∈ In general, the free product λ Λ Gλ of a family of groups Gλ : λ Λ is the group ∈ { ∈ } formed by finite “words” a ∗a constructed with “letters” a G (j =1,...,k; k 1), 1 ··· k j ∈ λj ≥ where any letter is different from the identity element in the respective group and where two adjacent letters must belong to different groups (one often says “reduced letters”); also the “empty word” is considered to be an element. The operation is given by the natural juxtaposition (a a ) (b b )=a a b b where, in the case a and 1 ··· k · 1 ··· h 1 ··· k 1 ··· h k b1 belong to a same group, the expression “akb1” should be replaced by their product in that group (and possibly removed if akb1 is the identity, causing then the same procedure for ak 1b2, and so on); the identity element is clearly the empty word. − Example. The free product of any number of copies of Z is called free group, in the sense that there is one generator for each copy of Z and the elements of the group are words formed by powers of these generators. For example, Z Z is formed by the words r1s1 rksk where all rj ’s and sj ’s are integer (the ∗ ··· rj ’s are meant to belong to the first copy of Z,andthesj ’s to the second); or also, in abstract notation, by ak1 bh1 akr bhr where a and b denote the two generators and the exponents are integers. ··· Note that for any µ Λthere is a natural monomorphism Gµ λ Λ Gλ; in fact one sees ∈ −→ ∈ that the free product λ Λ Gλ is the inductive limit in Groups of∗ the system Gλ : λ Λ ∈ { ∈ } with trivial preorder, i.e.∗ without considering morphisms(29). More generally, if morphisms f : G G are given for some pairs (λ, µ)withf f = f whenever defined, λ,µ λ −→ µ λ,µ ◦ µ,ν λ,ν then to obtain the inductive limit of the system G ,f : λ, µ Λ one must quotient { λ λ,µ ∈ } out the previous free product λ Λ Gλ by its normal subgroup N generated by all the elements of type f (a)f (a)∗1 ∈for a G whenever the morphisms f and f are λ,µ λ,ν − ∈ λ λ,µ λ,ν (29) Namely, given a family of morphisms ψλ : Gλ L, the definition (necessary, hence unique) −→ ψ(a1 ak)=ψλ (a1) ψλ (ak) is a morphism. ··· 1 ··· k Corrado Marastoni 23 Notes on Algebraic Topology defined(30): this procedure is said to be an amalgamation of the free product with respect to the given morphisms fλ,µ. When applying the above notions to the framework of Van Kampen, the groups are π1(Uλ) and the morphisms are the maps ι : π (U ) π (U ), which send the class of a loop λ,µ# 1 λ −→ 1 µ in the small open subset Uλ to the class of same loop viewed in the larger open subset Uµ: by Proposition 1.5.1 and the subsequent discussion it is then enough to consider the free product of the groups π1(Uλ)’s of a selected family of open subsets Uλ with indices λ M which cover X and are not contained in each other, and then to amalgamate only ∈ with respect to the double intersections U U for λ, µ M.(31) λ ∩ µ ∈ Example.
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