Section 6.8

Applications of singular homology

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Problem 6.8.1

Use Theorem 6.6.2 (a homotopy equivalence of pairs induces an isomorphism in homology) and the dimension property to show that

Hk(Dn)={Zk=0,0k>0. H_k(D^n) = \begin{cases} \mathbb {Z} & k = 0, \\ 0 & k > 0. \end{cases}
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Problem 6.8.2

Show that

Hk(S0)={Z⊕Zk=0,0k>0. H_k(S^0) = \begin{cases} \mathbb {Z} \oplus \mathbb {Z} & k = 0, \\ 0 & k > 0. \end{cases}
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Problem 6.8.3

Use the long exact sequence of the pair (D1,S0)(D^1, S^0) to show that

Hk(D1,S0)={Zk=1,0k≠1. H_k(D^1, S^0) = \begin{cases} \mathbb {Z} & k = 1, \\ 0 & k \neq 1. \end{cases}
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Problem 6.8.4

Give an explicit singular simplex σ:Δ1→D1\sigma : \Delta_1 \to D^1 which represents the generator of H1(D1,S0)H_1(D^1, S^0) (as computed in Exercise 6.8.3).

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Remark.
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It should map via ∂\partial to a generator of the kernel of i∗0:H0(S0)→H0(D1)i_*^0 : H_0(S^0) \to H_0(D^1).

Problem 6.8.5

Show that there is a homeomorphism h:D1→S+1={x∈S1:x2≥0}h : D^1 \to S^1_+ = \left\{ x \in S^1 : x_2 \geq 0\right\} and use this to compute Hk(S+1,S0)H_k(S^1_+, S^0) for all k≥0k \geq 0.

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Problem 6.8.6

Suppose B=S+1⊂S1B = S^1_+ \subset S^1 is the upper semicircle x2≥0x_2 \geq 0 and A=S−1A = S^1_- is the lower semicircle.

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(a)

Use Exercise 6.7.7 to show that there is an isomorphism Hk(S+1,S0)→Hk(S1,S−1)H_k(S^1_+, S^0) \to H_k(S^1, S^1_-). Indicate how to find the sets U,U′U, U' used in that exercise.

(b)

Use Exercise 6.7.8 to show that there is an isomorphism Hk(S+1,S0)→Hk(S1,S−1)H_k(S^1_+, S^0) \to H_k(S^1, S^1_-). Indicate how to find the required bicollar neighborhood.

Problem 6.8.7

Use Exercises 6.8.5 and 6.8.6 to compute Hk(S1,S−1)H_k(S^1, S^1_-).

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Problem 6.8.8

Give an explicit singular simplex σ:Δ1→S1\sigma : \Delta_1 \to S^1 which represents a generator of H1(S1,S−1)H_1(S^1, S^1_-) (as computed in Exercise 6.8.7).

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Remark.
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The essential ingredient of a generator is that it corresponds to a generator of H1(S+1,S0)H_1(S^1_+, S^0), and this generator must map to a generator of ker⁡(i∗0)⊂H0(S0)\ker (i_*^0) \subset H_0(S^0). From earlier work (Exercise 6.8.4), ker⁡(i∗0)\ker (i_*^0) is generated by ([1]−[−1])([1] - [-1]). There is a simple trigonometric formula that works for σ\sigma.

Problem 6.8.9

Use Exercises 6.8.7 and 6.8.8 to compute Hk(S1)H_k(S^1) by using the long exact sequence of the pair (S1,S−1)(S^1, S^1_-).

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Problem 6.8.10

Give an explicit singular chain which is a sum of two singular simplices, one of which is the answer to Exercise 6.8.8 and the other is a singular simplex in S−1S^1_-, which represents a generator of H1(S1)H_1(S^1). Explain how this relates to the isomorphism between H1(S1)H_1(S^1) and π1(S1,1)\pi_1(S^1, 1).

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Problem 6.8.11

Suppose p≥1p \geq 1. Show that Hk+1(Dp+1,Sp)≃Hk(Sp)H_{k+1}(D^{p+1}, S^p) \simeq H_k(S^p), k>0k > 0. Show that H1(Dp+1,Sp)=0H_1(D^{p+1}, S^p) = 0.

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Remark.
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Exercises 6.8.1--6.8.10 establish a chain of isomorphisms Hk+1(S1)≃Hk+1(S1,S−1)≃Hk+1(S+1,S0)≃Hk+1(D1,S0)H_{k+1}(S^1) \simeq H_{k+1}(S^1, S^1_-) \simeq H_{k+1}(S^1_+, S^0) \simeq H_{k+1}(D^1, S^0). When k>0k > 0, there is also an isomorphism Hk+1(D1,S0)≃Hk(S0)H_{k+1}(D^1, S^0) \simeq H_k(S^0). When k=0k = 0, the argument is slightly different since S0S^0 is not path connected, and instead one uses the isomorphism H1(D1,S0)≃ker⁡(i∗0)H_1(D^1, S^0) \simeq \ker (i_*^0). This exercise and the ones that follow generalize these arguments to compute the homology of SpS^p and (Dp+1,Sp)(D^{p+1}, S^p) for p≥1p \geq 1 via an inductive argument. Since Sp,Dp+1S^p, D^{p+1} are path connected, H0(Sp)≃H0(Dp+1)≃ZH_0(S^p) \simeq H_0(D^{p+1}) \simeq \mathbb {Z}, and the portion of the exact sequence H0(Sp)→H0(Dp+1)→H0(Dp+1,Sp)→0H_0(S^p) \to H_0(D^{p+1}) \to H_0(D^{p+1}, S^p) \to 0 shows that H0(Dp+1,Sp)=0H_0(D^{p+1}, S^p) = 0, since the first map is an isomorphism.

Problem 6.8.12

Suppose p≥1p \geq 1. Show that Hk(Dp+1,Sp)≃Hk(S+p+1,Sp)H_k(D^{p+1}, S^p) \simeq H_k(S^{p+1}_+, S^p), k≥0k \geq 0. Here S+p+1S^{p+1}_+ denotes the part of the sphere where the last coordinate is ≥0\geq 0.

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Problem 6.8.13

Suppose p≥1p \geq 1. Show that Hk(S+p+1,Sp)≃Hk(Sp+1,S−p+1)H_k(S^{p+1}_+, S^p) \simeq H_k(S^{p+1}, S^{p+1}_-), k≥0k \geq 0.

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Problem 6.8.14

Suppose p≥1p \geq 1. Show that Hk(Sp+1)≃Hk(Sp+1,S−p+1)H_k(S^{p+1}) \simeq H_k(S^{p+1}, S^{p+1}_-), k≥1k \geq 1.

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Problem 6.8.15

Suppose p≥1p \geq 1. Combine Exercises 6.8.11--6.8.14 to show that Hk+1(Sp+1)≃Hk(Sp)H_{k+1}(S^{p+1}) \simeq H_k(S^p) for k≥1k \geq 1, and H1(Sp+1)=0H_1(S^{p+1}) = 0.

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Problem 6.8.16

Use an inductive argument (based on Exercises 6.8.11--6.8.15) to show that

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(a)
Hk(Sn)={Z⊕Zn=k=0,Zn=k>0 or n>0,k=0,0otherwise; H_k(S^n) = \begin{cases} \mathbb {Z} \oplus \mathbb {Z} & n = k = 0, \\ \mathbb {Z} & n = k > 0 \text{ or } n > 0, k = 0, \\ 0 & \text{otherwise}; \end{cases}
(b)
Hk(Dn,Sn−1)={Zk=n≥0,0otherwise. H_k(D^n, S^{n-1}) = \begin{cases} \mathbb {Z} & k = n \geq 0, \\ 0 & \text{otherwise}. \end{cases}
Problem 6.8.17

If A⊂XA \subset X with inclusion map ii, a map r:X→Ar : X \to A is called a retraction if ri=1Ari = 1_A. Show that, if r:X→Ar : X \to A is a retraction, then r∗:Hk(X)→Hk(A)r_* : H_k(X) \to H_k(A) is surjective for each kk.

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Problem 6.8.18

Show that, if there is a kk with Hk(X)=0H_k(X) = 0 and Hk(A)≠0H_k(A) \neq 0, then there is no retraction (as defined in Exercise 6.8.17) of XX onto AA.

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Problem 6.8.19

Show that DnD^n does not retract onto Sn−1S^{n-1}, n≥1n \geq 1.

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Problem 6.8.20

Show that if f:Dn→Dnf : D^n \to D^n is continuous, then there is a fixed point x∈Dnx \in D^n with f(x)=xf(x) = x.

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Remark.
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See the proof of Theorem 3.5.4.

Problem 6.8.21
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(a)

If MM denotes the Mobius band with boundary ∂M\partial M, show that H1(∂M)H_1(\partial M) and H1(M)H_1(M) are each isomorphic to Z\mathbb {Z}, but the induced map i∗i_* is given by multiplication by 22.

(b)

Use (a) to show that a Mobius band does not retract onto its boundary even though they have isomorphic homology.