Show that \(H_1(X, A)\) is not isomorphic to \(\tilde{H}_1(X/A)\) if \(X = [0, 1]\) and \(A\) is the sequence \(1, {^1_2}, {^1_3}, \cdots\) together with its limit 0. [See Example 1.25.]
Solution
Let \[ X=[0,1], \qquad A=\{0,1,1/2,1/3,\dots\}. \]
1. The group \(H_1(X,A)\) is a countably generated free abelian group.
Since \(X\) is contractible and \(A\) is totally disconnected, the long exact sequence of the pair contains \[ 0=H_1(X)\longrightarrow H_1(X,A) \longrightarrow H_0(A)\xrightarrow{\epsilon}H_0(X)\longrightarrow0. \] Each point of \(A\) is a path component, so \[ H_0(A)\cong\bigoplus_{a\in A}\mathbb Z[a], \qquad H_0(X)\cong\mathbb Z, \] and \(\epsilon\) is the augmentation map. Hence \[ H_1(X,A)\cong\ker\epsilon. \] Taking \(0\) as a distinguished point gives the free basis \[ [a]-[0],\qquad a\in A-\{0\}. \] Thus \(H_1(X,A)\) is free abelian of countable rank, in particular it is countable as a set.
2. The quotient \(X/A\) is the shrinking wedge of countably many circles (the Hawaiian earring).
For each \(n\ge1\), the interval \[ I_n=[1/(n+1),1/n] \] has both endpoints in \(A\). After all of \(A\) is collapsed to one point, \(I_n/\partial I_n\) becomes a circle. The diameters of these intervals tend to zero at the collapsed limit point, so the quotient topology is exactly the standard shrinking-wedge topology of the Hawaiian earring.
3. The group \(H_1(X/A)\) is uncountable.
Let \(C_n\) denote the \(n\)th circle of the shrinking wedge. For each binary sequence \[ \varepsilon=(\varepsilon_1,\varepsilon_2,\dots)\in\{0,1\}^{\mathbb N}, \] construct a loop \(\gamma_\varepsilon\) by traversing \(C_n\) once if \(\varepsilon_n=1\) and staying at the wedge point if \(\varepsilon_n=0\), using pairwise consecutive time intervals whose lengths tend to zero and accumulate only at the endpoint of the parameter interval. Because the circles \(C_n\) shrink to the wedge point, this infinite concatenation is continuous.
For each \(n\) there is a continuous retraction \[ r_n:X/A\longrightarrow C_n \] that collapses all other circles to the wedge point. Hence the induced homomorphism \[ (r_n)_*:H_1(X/A)\longrightarrow H_1(C_n)\cong\mathbb Z \] sends \([\gamma_\varepsilon]\) to \(\varepsilon_n\). Therefore distinct binary sequences give distinct homology classes. Since \(\{0,1\}^{\mathbb N}\) is uncountable, so is \(H_1(X/A)\).
4. Consequently\[ \boxed{H_1(X,A)\not\cong\widetilde H_1(X/A).} \]
The quotient is path connected, so \(\widetilde H_1(X/A)=H_1(X/A)\). By 1 the relative group is countable, while by 3 the quotient-space homology group is uncountable. Hence they cannot be isomorphic.