INTRODUCTION TO ALGEBRAIC TOPOLOGY Exercise 7, solutions 1) Let k < j 1 0 j n, where 1 n. We want to show that e j n e k n 1 = e k n e j 1 n 1. Recall that the map e j n : n 1 n is defined by e j nt 0,..., t n 1 ) = t 0,..., t j 1, 0, t j,..., t n 1 ). e j n e k n 1t 0,..., t n 2 ) = e j nt 0,..., t k 1, 0, t k,..., t n 2 ) = t 0,..., t k 1, 0, t k,..., 0, t j 1,..., t n 2 ) e k n e j 1 n 1t 0,..., t n 2 ) = e k nt 0,..., t j 2, 0, t j 1,..., t n 2 ) = t 0,..., t k 1, 0, t k,..., 0, t j 1,..., t n 2 ). e j n e k n 1t 0,..., t n 2 ) = e k n e j 1 n 1t 0,..., t n 2 ), for every t 0,..., t n 2 ) n 2. It follows that e j n e k n 1 = e k n e j 1 n 1. 2) Let X be a topological space, S 1 X) be the free abelian group with basis all paths σ : I X, S 0 X) be the free abelian group with basis X. a) Claim: There is a homomorphism 1 : S 1 X) S 0 X) with 1 σ = σ1) σ0) for every path σ X. Proof. An element in S 1 X) is of the form Σ σ n σ σ, where n σ = 0 except for finitely many σ, the sum Σ σ is over all paths σ : I X. Similarly, an element in S 0 X) is of the form Σ x n x x, where n x = 0 except for finitely many x, the sum Σ x is over all points in X. 1 : S 1 X) S 0 X), Σ σ n σ σ Σ σ n σ σ1) σ0)). 1 Σσ n σ σ + Σ σ m σ σ ) = 1 Σσ n σ + m σ )σ ) = Σ σ n σ + m σ )σ1) σ0)) = Σ σ n σ σ1) σ0)) + Σ σ m σ σ1) σ0)) = 1 Σσ n σ σ ) + 1 Σσ m σ σ ). Date: November 8, 2017. 1
2 INTRODUCTION TO ALGEBRAIC TOPOLOGY 1 is a homomorphism. If σ : I X is a path, then 1 σ = σ1) σ0). b) Let x 0, x 1 X. Claim: The points x 0 x 1 lie in the same path component of X if only if x 1 x 0 im 1. Proof. Assume first that x 0 x 1 lie in the same path component of X. there is a path σ : I X with σ0) = x 0 σ1) = x 1. 1 σ = σ1) σ0) = x 1 x 0, it follows that x 1 x 0 im 1. Assume then that x 1 x 0 im 1. Let s consider an arbitrary element x im 1. The element x is of the form Σ σ n σ σ1) σ0)). In this sum, the sum of the coefficients of σ1) σ0) equals n σ n σ = 0., if we write x = Σ k i=1n i x i, then the sum of the coefficients of the x i in the same path component of X equals 0. Therefore, x 1 x 0 must be in the same path component of X. If not, then the coefficient of x 1 is 1 0. Since x 0 is in a different path component, that would make the sum of the coefficients of points in the same path component as x 1 to be 1 0, a contradiction.) c) Let σ be a path in X. Claim: The path σ ker 1 if only if it is a closed path. Proof. The claim follows from the fact that σ ker 1 σ1) σ0) = 1 σ = 0 σ1) = σ0) σ is a closed path. Let σ 1 σ 2 be closed paths in X. 2σ 1 + 3σ 2 S 1 X) but 2σ 1 + 3σ 2 is a not a closed path. However, 1 2σ 1 + 3σ 2 ) = 2 1 σ 1 + 3 1 σ 2 = 0 + 0 = 0. 3) Let α: 1 X be a one-dimensional singular simplex of a topological space X. β : 1 X by βt 0, t 1 ) = αt 1, t 0 ). Claim: The element α + β is a boundary in S 1 X). Proof. Let ϱ: 2 X be a singular 2-simplex. 2 ϱt 0, t 1 ) = ϱ0, t 0, t 1 ) ϱt 0, 0, t 1 ) + ϱt 0, t 1, 0). ϱ 1 : 2 X, t 0, t 1, t 2 ) αt 0 + t 2, t 1 ).
INTRODUCTION TO ALGEBRAIC TOPOLOGY 3 2 ϱ 1 t 0, t 1 ) = ϱ 1 0, t 0, t 1 ) ϱ 1 t 0, 0, t 1 ) + ϱ 1 t 0, t 1, 0) = α0 + t 1, t 0 ) αt 0 + t 1, 0) + αt 0 + 0, t 1 ) = αt 1, t 0 ) α1, 0) + αt 0, t 1 ) = αt 0, t 1 ) + βt 0, t 1 ) α1, 0). ϱ 2 : 2 X, t 0, t 1, t 2 ) α1, 0). 2 ϱ 2 t 0, t 1 ) = ϱ 2 0, t 0, t 1 ) ϱ 2 t 0, 0, t 1 ) + ϱ 2 t 0, t 1, 0) = α1, 0) α1, 0) + α1, 0) = α1, 0). for every t 0, t 1 ) 1, 2 ϱ 1 t 0, t 1 ) + 2 ϱ 2 t 0, t 1 ) = αt 0, t 1 ) + βt 0, t 1 ) α1, 0) + α1, 0) = αt 0, t 1 ) + βt 0, t 1 ). 2 ϱ 1 + ϱ 2 ) = 2 ϱ 1 + 2 ϱ 2 = α + β. Let s redo the proof by using different notation: Proof. Let e 0 : 1 2, t 0, t 1 ) 0, t 0, t 1 ), e 1 : 1 2, t 0, t 1 ) t 0, 0, t 1 ), e 2 : 1 2, t 0, t 1 ) t 0, t 1, 0). Let ϱ: 2 X be a singular 2-simplex. 2 ϱ = ϱ e 0 ϱ e 1 + ϱ e 2. ϱ 1 : 2 X, t 0, t 1, t 2 ) αt 0 + t 2, t 1 ) f : 2 1, t 0, t 1, t 2 ) t 0 + t 2, t 1 ) so that ϱ 1 = α f. Let For every t 0, t 1 ) 1, c: 1 1, t 0, t 1 ) 1, 0). 2 ϱ 1 = ϱ 1 e 0 ϱ 1 e 1 + ϱ 1 e 2 = α f e 0 α f e 1 + α f e 2. α f e 0 )t 0, t 1 ) = αt 1, t 0 ) = βt 0, t 1 ),
4 INTRODUCTION TO ALGEBRAIC TOPOLOGY α f e 1 )t 0, t 1 ) = αt 0 + t 1, 0) = α1, 0) = α c)t 0, t 1 ), α f e 2 )t 0, t 1 ) = αt 0, t 1 ). ϱ 2 : 2 X, t 0, t 1, t 2 ) α1, 0). 2 ϱ 2 = ϱ 2 e 0 ϱ 2 e 1 + ϱ 2 e 2, where ϱ 2 e i )t 0, t 1 ) = α1, 0) = α c)t 0, t 1 ), for all t 0, t 1 ) 1 for all i {0, 1, 2}. 2 ϱ 1 + ϱ 2 ) = 2 ϱ 1 + 2 ϱ 2 = α f e 0 α f e 1 + α f e 2 + ϱ 2 e 0 ϱ 2 e 1 + ϱ 2 e 2 = β α c + α + α c α c + α c = β + α = α + β. 4) Let σ τ be one-dimensional singular simplices of S 1, i.e., continuous maps 1 S 1, given by τt 0, t 1 ) = sin2t 1 π), cos2t 1 π)) σt 0, t 1 ) = sin2t 0 π), cos2t 0 π)). a) Claim: τ σ are cycles. Proof. Since 0 = {1}, it follows that 1 σ1) = σ0, 1) σ1, 0) = sin 0, cos 0) sin 2π, cos 2π) = 0, 1) 0, 1) = 0. Consequently, σ Z 1 S 1 ). Similarly, 1 τ1) = τ0, 1) τ1, 0) = sin 2π, cos 2π) sin 0, cos 0) = 0, 1) 0, 1) = 0. Consequently, τ Z 1 S 1 ). b) Claim: τ + σ is a boundary. Proof. Since σt 0, t 1 ) = τt 1, t 0 ), for every t 0, t 1 ) 1, the claim follows immediately from Exercise 3.
INTRODUCTION TO ALGEBRAIC TOPOLOGY 5 5) Consider the following sequence of groups group homomorphisms: 0 3 Z Z 2 Z 1 0, where the groups are C 1 = Z, C 2 = Z Z C n = 0, for n 1, 2, the homomorphism 2 is defined by the rule 2 m, n) = 3m + 3n. Here 0 denotes the trivial group having just one element.) a) Calculate the groups im n ker n, n 1. The kernels are as follows: i) ker 1 = Z, ii) ker 2 = {m, n) Z Z 3m + 3n = 0} = {n, n) n Z}, iii) ker 3 = 0, iv) ker n = 0, for n 4. The images are as follows: i) im 1 = 0, ii) im 2 = {3m + 3n m, n Z} = {3m m, n Z} = 3Z, iii) im 3 = 0, iv) im n = 0, for n 4. b) Calculate the groups ker n /im n+1, for n 1. The groups are: i) ker 1 /im 2 = Z/3Z = Z 3, ii) ker 2 /im 3 = {n, n) n Z} = Z, iii) ker n /im n+1 = 0/0 = 0, for n 3.