Hatcher Exercise 2.1.26. 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.
 A: 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. 
A: This is not at all trivial.
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 $0 = H_1(X)  \to H_1(X,A) \to \tilde{H}_0(A) \to \tilde{H}_0(X) = 0$ shows that $H_1(X,A)$ is isomorphic to the free abelian group $\tilde{H}_0(A)$ (which has infinitely many generators).
The computation of $H_1(H)$ is difficult. See for example https://web.math.rochester.edu/people/faculty/doug/otherpapers/eda-kawamura2.pdf.
You will see that it is not free abelian.
