Section 6.13

Reduced homology

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Problem 6.13.1

With ϵ\epsilon, H0(C)H_0(C), Hˉ0(C)\bar{H}_0(C) as defined below, show that the short exact sequence

0→ker⁡(ϵ)→C0→ϵZ→0, 0 \to \ker (\epsilon ) \to C_0 \xrightarrow {\epsilon } \mathbb {Z} \to 0,

leads to a short exact sequence

0→Hˉ0(C)→H0(C)→Z→0, 0 \to \bar{H}_0(C) \to H_0(C) \to \mathbb {Z} \to 0,

which splits to give H0(C)≃Hˉ0(C)⊕ZH_0(C) \simeq \bar{H}_0(C) \oplus \mathbb {Z}.

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Remark.
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Given a chain complex of free abelian groups (Ci,∂i)(C_i, \partial_i), i≥0i \geq 0, extend it by introducing C−1=ZC_{-1} = \mathbb {Z}. Define ∂0=ϵ\partial_0 = \epsilon by first selecting a basis of generators of C0C_0 and setting ϵ(g)=1\epsilon (g) = 1 for each generator gg; then ker⁡(ϵ)\ker (\epsilon ) is generated as a free group on differences of generators. When ∂1\partial_1 has the property that, for a generator gg of C1C_1, ∂1(g)\partial_1(g) is a difference of generators, we have im(∂1)⊂ker⁡(ϵ)\mathrm{im}(\partial_1) \subset \ker (\epsilon ), so the extended chain complex is still a chain complex; ∂0=ϵ\partial_0 = \epsilon is called the augmentation.

Definition 6.13.1: the reduced homology Hˉk(C)\bar{H}_k(C) is the homology of the augmented chain complex for k≥0k \geq 0. This agrees with Hk(C)H_k(C) in dimensions greater than 00, but Hˉ0(C)=ker⁡(ϵ)/im(∂1)\bar{H}_0(C) = \ker (\epsilon )/\mathrm{im}(\partial_1), whereas H0(C)=C0/im(∂1)H_0(C) = C_0/\mathrm{im}(\partial_1).

For the singular complex S(X)S(X) of a topological space XX, the augmentation ϵ:S0(X)→Z\epsilon : S_0(X) \to \mathbb {Z} is defined by ϵ(σ)=1\epsilon (\sigma ) = 1 for a singular 00-simplex σ\sigma, and the reduced homology Hˉk(X)\bar{H}_k(X) is the homology of this augmented singular complex.

Problem 6.13.2
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(a)

Verify that for S(X)S(X) we have ϵ∂1=0\epsilon \partial_1 = 0 (notation as in Exercise 6.13.1).

(b)

Show that H0(X)≃Hˉ0(X)⊕ZH_0(X) \simeq \bar{H}_0(X) \oplus \mathbb {Z}.

(c)

Show that a topological space is path connected iff Hˉ0(X)≃0\bar{H}_0(X) \simeq 0.

Problem 6.13.3

Show that the reduced homology of the sphere can be expressed as

Hˉk(Sn)={Zk=n,0k≠n. \bar{H}_k(S^n) = \begin{cases} \mathbb {Z} & k = n, \\ 0 & k \neq n. \end{cases}
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Problem 6.13.4

Show that the homology of the suspension ΣX\Sigma X of a path-connected space can be expressed as Hk+1(ΣX)≃Hˉk+1(ΣX)=Hˉk(X)H_{k+1}(\Sigma X) \simeq \bar{H}_{k+1}(\Sigma X) = \bar{H}_k(X) for k≥0k \geq 0.

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Problem 6.13.5

By using the augmented complex in place of the original complex and using S−1(X,A)=S−1(X)/S−1(A)=Z/Z=0S_{-1}(X, A) = S_{-1}(X)/S_{-1}(A) = \mathbb {Z}/\mathbb {Z} = 0, show that there is a long exact sequence in reduced homology which ends in

H1(X,A)→Hˉ0(A)→Hˉ0(X)→H0(X,A)→0. H_1(X, A) \to \bar{H}_0(A) \to \bar{H}_0(X) \to H_0(X, A) \to 0.
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Problem 6.13.6

Assume that X=A∪BX = A \cup B, A∩B≠∅A \cap B \neq \emptyset, and the hypotheses giving the usual Mayer--Vietoris exact sequence hold. The Mayer--Vietoris exact sequence can be modified to use reduced homology as long as A∩B≠∅A \cap B \neq \emptyset, augmenting the usual singular chains by S−1(A∩B)=ZS_{-1}(A \cap B) = \mathbb {Z}, S−1(A)⊕S−1(B)=Z⊕ZS_{-1}(A) \oplus S_{-1}(B) = \mathbb {Z} \oplus \mathbb {Z}, S−1(A)+S−1(B)=ZS_{-1}(A) + S_{-1}(B) = \mathbb {Z}, S−1(X)=ZS_{-1}(X) = \mathbb {Z}.

Show that there is a long exact sequence using reduced homology which ends in

H1(X)→δHˉ0(A∩B)→i0Hˉ0(A)⊕Hˉ0(B)→j0Hˉ0(X)→0. H_1(X) \xrightarrow {\delta } \bar{H}_0(A \cap B) \xrightarrow {i_0} \bar{H}_0(A) \oplus \bar{H}_0(B) \xrightarrow {j_0} \bar{H}_0(X) \to 0.
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Problem 6.13.7

Use the reduced Mayer--Vietoris sequence of Exercise 6.13.6 to show that, if AA and BB are path connected, A∩B≠∅A \cap B \neq \emptyset, A∪B=XA \cup B = X, and the hypotheses giving the Mayer--Vietoris exact sequence apply, then XX is path connected.

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