# 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.

• 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. – Paul Frost Nov 25 '19 at 1:12
• Thanks so much. – James Nov 25 '19 at 5:51

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.

• $X/A$ is in fact homeomorphic to the Hawaiin earring. – Paul Frost Nov 23 '19 at 12:50
• @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 ! – Maxime Ramzi Nov 23 '19 at 12:52
• Thanks again. I have seen the example of Hawaiin earring in Hatcher. Now, It's clear. Thanks again. – James Nov 23 '19 at 13:26

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.

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