# Hawaiian Earring

Let $X=[0,1]$ and $A=\{0\}\cup\{\frac{1}{n}|n\in\mathbb Z\}$. Note that $(X,A)$ is not a good pair. Show that $H_1(X,A)$ is not isomorphic to $H_1(X/A)$.

I have a sequence of homology groups: $\cdots\rightarrow H_1(A)\rightarrow H_1(X)\rightarrow H_1(X,A)\rightarrow H_0(A)\rightarrow H_0(X)\rightarrow H_0(X,A)\rightarrow0$

$X$ is path-connected, so $H_0(X)=\mathbb Z$. $A$ is nonempty and I have managed to prove that $H_0(X,A)=0$.

EDIT: As mentioned: $H_1(X)=0$ as $X$ is contractible.

I also have the following in my notes that I can't seem to justify:

1. $H_1(X,A)$ is countable, but I don't know how to reach that conclusion.
2. The map $H_1(X,A)\rightarrow H_0(A)$ is an inclusion map.
3. $H_0(A)$ is $\mathbb Z<\text{uncountably many things}>$

And how do any of these help me with what I want to show?

• do you mean $H_*(X)=0$, not just $H_0$? And do mean reduced homology or unreduced, because the reduced zeroth hom is $0$ but the unreduced is $\Bbb Z$. – Stefan Hamcke May 6 '13 at 16:21
• Sorry it's a typo. I think $H_0(X)=\mathbb Z$ since it is path connected. I'm referring to unreduced homology. I will make the edit. – Haikal Yeo May 6 '13 at 16:52
• Another remark: The induced homomorphism $H_0(A)\to H_0(X)$ is surjective iff $A$ meets each path-component of $X$ and injective iff $A$ meets each path-component of $X$ at most once. So the morphism is surjective, hence the last arrow is zero, and therefore $H_0(X,A)$ must be zero. – Stefan Hamcke May 6 '13 at 16:58
• Do you mean that you want to show $H_1(A)$ is uncountable? $H_1(X)=0$ as $X$ is contractible. – Kris Williams May 6 '13 at 17:02
• Yes, the Splitting Lemma is what I'm talking about :-) You can easily construct a section for the projection $p$ onto $\Bbb Z$. Since $Z$ is cyclic, you can just choose an element $b$ in the preimage of $1$ define $s(1)=b$. This $s$ is then a right-inverse for $p$ – Stefan Hamcke May 6 '13 at 17:19

$H_0(A) = \mathbb{Z}\langle \text{path components of$A$}\rangle$, and since $A$ itself is countable, so must be the set of its path components (in fact every point is a path component). So in fact 3 should have countable and not uncountable. Now a countable direct sum of a countable group is countable.
$H_1(X) = 0$, and since the sequence is exact, the next map must be an inclusion.
Hence $H_1(X,A)$ is countable.
To show that $H_1(X/A)$ has at least uncountable many $\mathbb{Z}$ independent elements, use the same argument as in example 1.25 of Hatcher. So $H_1(X/A)$ is at least $\mathbb{Z}\langle \text{uncountably many things}\rangle$. This gives you a contradiction with $(X,A)$ being a good pair.