⇒ By exactness, ker d 2 = im d 1 = B, ker d 4 = im d 3 = 0 so d 1 is surjective and d 4 is injective.

USTC, School of Mathematical Sciences Winter semester 2018/19 Algebraic topology by Prof. Mao Sheng Hint to exercise sheet 7 MA04311 Tutor: Lihao Huang, Han Wu Posted by Dr. Muxi Li Hint 1. Let the exact sequence be A → B ^d

→ C ^d

→ D ^d

→ E. ^d

⇒ By exactness, ker d ₂ = im d 1 = B, ker d 4 = im d 3 = 0 so d 1 is surjective and d ₄ is injective.

⇐ By exactness and surjection of d ₁ , have im d ₁ = ker d ₂ = B, which implies d 2 = 0. By exactness and injection of d 4 , have im d 3 = ker d 4 = 0, which im- plies d ₃ = 0. So 0 = im d ₂ = ker d ₃ = C.

The long exact sequence · · · → H _n+1 (X, A) → H _n (A) → H ^d

_n (X) → H _n (X, A) →

· · · implies that all d _n are isomorphisms (injection and surjection) iff H _n (X, A) = 0, ∀n.

Hint 2. (a)By exactness of H ₀ (A) → H ⁱ ₀ (X) → H 0 (X, A) → 0, the H 0 (X, A) = 0 iff i is surjective. H 0 (−) is free Z-module generated by path-component of the given space. Thus, the surjection of i means that A meets each path- component of X.

(b) By exactness of H ₁ (A) → H ₁ (X) → H ₁ (X, A) → H ^∂ ₀ (A) → H ⁱ ₀ (X) and exercise 1 above, H ₁ (X, A) = 0 iff H 1 (A) → H 1 (X) is surjective and i is injective. The same analyses as (a),the injective of i means that each path-component of X contains at most one path-component of A.

Hint 3. (a) Let the number of points of A be m, then

H n (A) = (

Z ^⊕m if n = 0 0 if n > 0.

It is easy to calculate that

H _n (S ² ) = (

Z if n = 0 or n = 2

0 otherwise H _n (S ¹ ×S ¹ ) =



 

 

Z if n = 0 or n = 2 Z ^⊕2 if n = 1

0 otherwise By the exact sequence H ₂ (A)(= 0) → H 2 (X) → H ^≈ ₂ (X, A) → H 1 (A)(= 0) → H ₁ (X) → H ₁ (X, A) → H ₀ (A)(= Z ^⊕m )  H 0 (X)(= Z) → H 0 (X, A)(=

0) → 0, we have

H _n (S ² , A) =



 

 

Z ^⊕m−1 if n = 1

Z if n = 2

0 otherwise

H _n (S ¹ × S ¹ ) =



 

 

Z ^⊕m+1 if n = 1

Z if n = 2

0 otherwise

(b) Method 1, using the 4 complex structure of X pictured in textbook P102 second picture, having one vertex, four edges, six 2-simplices, we can

1

calculate the simplicial homology group of X

H _n (X) = H _n ⁴ (X) =



 

 

Z if n = 0 or n = 2 Z ^⊕4 if n = 1

0 otherwise And from example 2,2 in textbook P106,

H _n (S ¹ ) = H _n ⁴ (S ¹ ) = (

Z if n = 0 or n = 1 0 otherwise

We have a long exact sequence · · · → H ₂ (S ¹ )(= 0) → H 2 (X)(= Z) ,→

H 2 (X, S ¹ ) → H ^∂ ₁ (S ¹ )(= Z) → H ^d ₁ (X)(= Z ^⊕4 ) → H 1 (X, S ¹ ) → H ⁰ ₀ (S ¹ )(=

Z) → H ^≈ ₀ (X)(= Z) → H 0 (X, S ¹ )(= 0) → 0.

For S ¹ = A as in the picture, the image of the generator of H ₁ (A) in H ₁ (X) is 0 because this is the boundary of three adjunctive 2-simplices. So we have

H n (X, A) =



 

 

Z ^⊕4 if n = 1 Z ^⊕2 if n = 2 0 otherwise

For B as in the picture, the image of the generator of H ₁ (A) in H 1 (X) is one generator of H ₁ (X), So d is an injection and ∂ = 0.

H _n (X, B) =



 

 

Z ^⊕3 if n = 1 Z if n = 2 0 otherwise

Method 2, using the fact that H _n (X, S ¹ ) = ˜ H _n (X/S ¹ ). For S ¹ = A as in the picture, X/A = T ¹ ∨ T ¹ . By Mayer-Vietoris sequences or Corollary 2.25 in textbook P126, we have

H n (X, A) = ˜ H n (X/A) =



 

 

Z ^⊕4 if n = 1 Z ^⊕2 if n = 2 0 otherwise

For B as in the picture, X/B is homotopy to T ¹ ∨ S ¹ . Similarly, we have

H _n (X, B) = ˜ H _n (X/B) =



 

 

Z ^⊕3 if n = 1 Z if n = 2 0 otherwise

Hint 4. By the exact sequence H ₁ (R)(= 0) → H 1 (R, Q) → H ^∂ ₀ (Q)(= L

Q

Z)

 i

H 0 (R)(= Z), we have H 1 (R, Q) = L

Q

Z.

2

Hint 5. Let this space be X, and let A = I ×[0, 2/3]∩X, B = I ×[1/3, 1]∩X.

Using Mayer-Vietoris sequences, we have · · · → H _n (A ∩ B) → H n (A) ⊕ H n (B) → H n (X) → · · · . For A, B are contractible, we have ˜ H n (A) = H ˜ _n (B) = 0. And A ∩ B is homotopy to Q ∩ I. So we have

H n (X) =



 



 



Z if n = 0

L

Q

∩I

Z if n = 1 0 otherwise

Hint 6. For the first part, consider X ,→ CX  CX/X ∼ = SX. There is a long exact sequence · · · → ˜ H _n+1 (SX) ^∂ →

H ˜ _n (X) → ˜ H _n (CX) → H ˜ _n (SX) → · · · . For ˜ ^∂

H _n (CX) = 0 ∀n, we have ˜ H _n (X) = ˜ H _n+1 (SX) ∀n.

For the second part, let S ^r X be the union of r + 1 cones CX with their basis identified. For r ≥ 2, consider S ^r−1 X ,→ S ^{i r} X  S ^r X/S ^r−1 X ∼ = SX. There is a long exact sequence · · · → ˜ H n+1 (SX) ^∂ →

H ˜ n (S ^r−1 X) → ˜ H n (S ^r X) → H ˜ n (SX) → · · · . There exists a section of i, which means every ∂ ^∂

_n = 0 and H ˜ n (S ^r X) = ˜ H n (S ^r−1 X) ⊕ ˜ H n (SX). Inductively, we have

H ˜ n (S ^r X) =

( ˜ H _n (SX) ^⊕r = ˜ H _n−1 (X) ^⊕r if r > 0

0 if r = 0

Hint 7. Let σ : ∆ ⁿ → X. It induces ˜ s : C∆ ⁿ → CX. Let linear map

∆ ⁿ⁺¹ ∼ = C∆ ⁿ given by mapping the last vertex of ∆ ⁿ⁺¹ to the vertex of the cone. Define s(σ) : ∆ ⁿ⁺¹ ∼ = C∆ ⁿ → CX → CX/X = SX. It is ^{˜ s} easy to check that ∂ ◦ s = s ◦ ∂. This chain map induces an isomorphisms H ˜ n (X) = ˜ H n+1 (SX), which is the same as that in exercise 6.

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