Hatcher Exercise 2.1.26.

Question

Given that $ X =[0, 1]$ and $A$ = {$\frac{1}{n} :n$ is any natural number} $\cup$ {$0$}. I want to show that $H_1(X, A)$ is not isomorphic to $H_1(X/A)$. Could you please help me with this problem?

My argument: The quotient space $X/A$ is basically the wedge of countably many circles, and hence its first homology group is the infinite direct sum of $Z$'s. On the other hand, the $X$ is contractible, and therefore, $H_1(X, A)$ is trivial. Is my argument good enough/correct? Any help will be appreciated. Thanks so much.

Answer 1

Your argument is not correct, for two reasons: first of all, $X/A$ is not exactly a wedge if countably many circles (this would not take into account the topology, the fact that $1/n\to 0$), and secondly $X$ is contractible does not imply $H_1(X,A) = 0$ : it implies $H_1(X) = 0$; but note that you have an exact sequence $H_1(X)\to H_1(X,A)\to H_0(A)\to H_0(X)$ so since $H_0(A)$ is very big and $H_0(X), H_1(X)$ are quite small, $H_1(X,A)$ has to be quite big (I'll let you give the precise statements for this)

What you want to do is actually compute $H_1(X,A)$ via this exact sequence (this is not too hard), and then identify $X/A$ more carefully. It should be closer to the Hawaiian earrings than a wedge of circles.

Answer 2

The precise computation of $H_1(X/A)$ is not at all trivial. Luckily, we don't need to know much about $H_1(X/A)$ to complete this exercise.

In fact, the space $X/A$ is homeomorphic to the Hawaiian earring $H = \bigcup_{n=1}^\infty C_n$, where $C_n = \{ z \in \mathbb C \mid \lvert z - \frac{1}{n} \rvert = \frac{1}{n} \}$ is the plane circle with center $\frac{1}{n} \in \mathbb R$ and radius $\frac{1}{n}$. In Hatcher's Example 1.25 $H$ is denoted as "The Shrinking Wedge of Circles". To see this, define $f : [0,1] \to H, f(t) = \frac{1}{n} + \frac{1}{n}e^{(2n(n+1)t -1)\pi i}$ for $t \in [\frac{1}{n+1},\frac{1}{n}]$ and $f(0) = 0$. This is easily seen to be a continuous surjection; the interval $[\frac{1}{n+1},\frac{1}{n}]$ is wrapped counterclockwise once around $C_n$ starting at the "cluster point" $0$. We have $f(A) = \{0 \}$, thus $f$ induces a continuous closed surjection $f' : X/A \to H$. It is easy to see that $f'$ is injective, thus it is a homeomorphism.

The exact sequence $0 = H_1(X) \to H_1(X,A) \to \widetilde{H}_0(A) \to \widetilde{H}_0(X) = 0$ shows that $H_1(X,A)$ is isomorphic to the free abelian group $\widetilde{H}_0(A)$ (which has countably many generators). That is, $\bigoplus_{n\in\Bbb N}\Bbb Z$ is a model for $\widetilde{H_0}(A)$ (and also $H_0(A)$) and thus for $H_1(X,A)$.

It is shown in Hatcher's book that the first homology is the Abelianisation of the first fundamental group. It is also shown by Hatcher (alternatively make reference to the end of this answer) that there is a surjective homomorphism $F:\pi_1(H)\to\prod_{n\in\Bbb N}\Bbb Z$. By the universal property of Abelianisation (quotient property) there is a unique homomorphism $G:H_1(H)\to\prod_{n\in\Bbb N}\Bbb Z$ such that $Gh=F$, where $h:\pi_1(X)\to H_1(H)$ is the canonical map. The surjectivity of $F$ implies surjectivity of $G$, and (relying on the axiom of choice, if you care about that kind of thing) it follows $H_1(H)$ is uncountable.

However, $\bigoplus_{n\in\Bbb N}\Bbb Z$ is countable; it is the countable union of the countable subgroups $\bigoplus_{n=1}^k\Bbb Z$, $k=1,2,\cdots$. Since $H_1(H)\cong H_1(X,A)$ would imply $H_1(H)\cong H_1(X,A)\cong\bigoplus_{n\in\Bbb N}\Bbb Z$ and we know this is impossible (left hand side is uncountable, right hand side is countable) it follows $H_1(H)$ cannot be isomorphic to $H_1(X,A)$, completing the exercise.

Note we did not need any algebraic observations here, this is purely set-theoretic; for instance, $\Bbb Z$ and $\Bbb Z^2$ have the same cardinality so you need some algebraic observations to include they are not isomorphic as groups.

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The computation of $H_1(H)$ is difficult. See this, for instance. For an overkill method to conclude $H_1(H)\not\cong H_1(X,A)$ using this computation, you can note that $H_1(H)$ is not free Abelian whereas $H_1(X,A)$ is.

Acknowledgement. The proof that $H_1(H)$ is uncountable, and therefore not isomorphic to $H_1(X,A)$, is due to FShrike.