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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.

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    $\begingroup$ Hatcher wants to show that in general $H_*(X,A)$ and $\tilde{H}_*(X/A)$ are not isomorphic. In his example it is not easy to prove. A simpler example is $X$ = topologists's sine curve and $A$ = line segment at which $(x,\sin(1/x))$ clusters. $\endgroup$ – Paul Frost Nov 25 '19 at 1:12
  • $\begingroup$ Thanks so much. $\endgroup$ – James Nov 25 '19 at 5:51
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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.

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    $\begingroup$ $X/A$ is in fact homeomorphic to the Hawaiin earring. $\endgroup$ – Paul Frost Nov 23 '19 at 12:50
  • $\begingroup$ @PaulFrost : I thought so - I hadn't checked it in detail (and didn't want to) so I didn't want to claim it so affirmatively. Good to know my intuition isn't completely stupid ! $\endgroup$ – Maxime Ramzi Nov 23 '19 at 12:52
  • $\begingroup$ Thanks again. I have seen the example of Hawaiin earring in Hatcher. Now, It's clear. Thanks again. $\endgroup$ – James Nov 23 '19 at 13:26
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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.

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  • $\begingroup$ Thank you so much. This answer is really beautiful. Thanks again. $\endgroup$ – James Nov 23 '19 at 23:05

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