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Can you tell me if the following 3 computations are correct:

  1. $X = S^1$, $\pi_1 (X) = \mathbb{Z}$, $Ab(\mathbb{Z}) = \mathbb{Z}$ $\implies H_1(X) = \mathbb{Z}$
  2. $X = S^1 \vee S^1$, $\pi_1(X) = \mathbb{Z} \ast \mathbb{Z}$, $Ab(\pi_1(X)) = \pi_1(X) / \pi_1(X) = 0$, $\implies H_1(X) = 0$
  3. $X = T^2$, $\pi_1(X) = \mathbb{Z} \times \mathbb{Z} = Ab(\pi_1(X))$, $\implies H_1(X) = \mathbb{Z} \times \mathbb{Z}$

Thanks for your help!

Rudy the Reindeer
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    What makes you think the abelianisation of $\mathbb{Z} * \mathbb{Z}$ is trivial? – Zhen Lin Jul 31 '11 at 16:44
  • I think that the commutator subgroup of the free group over two generators $F_2$ is $F_2$ again. But maybe that's wrong? – Rudy the Reindeer Jul 31 '11 at 16:47
  • @Matt It must be wrong, since $\mathbf{Z} * \mathbf{Z}$ maps onto any abelian group which is generated by two elements, and there are plenty of nontrivial examples of those. – Dylan Moreland Jul 31 '11 at 16:50
  • @Matt In the "Topology II" book (Fuchs ed.) it is stated that it is Poicare theorem (that you using) –  Aug 09 '11 at 19:08
  • Hey, thanks! I think I'm using Hurewicz theorem, maybe it's called Poincare for $n=1$ : ) – Rudy the Reindeer Aug 09 '11 at 19:33

2 Answers2

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Given a group $G$, the abelianization is not $G/G$, it is $G/[G,G]$. If you are given your group by a presentation of generators and relations, a presentation of $G/[G,G]$ has the same generators and relations as $G$, plus the additional relations that $ab=ba$ for each pair of generators $a,b\in G$.

$\mathbb Z * \mathbb Z$ has two generators $a,b$, with no relations between them, and so the abelianization will have two generators $a,b$, and only the relation $ab=ba$. This is $\mathbb{Z}\times \mathbb{Z}$.

More generally, $\operatorname{Ab}(G*H)\cong \operatorname{Ab}(\operatorname{Ab}(G)*\operatorname{Ab}(H))\cong \operatorname{Ab}(G)\times \operatorname{Ab}(H)$. In terms of generators and relations, we can view the isomorphisms as first adding relations that make $G$ and $H$ commutative, and then adding relations which make the elements of $G$ commute with the elements of $H$. Adding the relations in the opposite order also gives $\operatorname{Ab}(G*H)=\operatorname{Ab}(G\times H)$.

Now, if we use induction, we see that $\operatorname{Ab}(G_1 * G_2 * \cdots * G_n)\cong \operatorname{Ab}(G_1)\times \cdots \times \operatorname{Ab}(G_n)$. This, combined with Hurewicz, shows that $H_1(X_1 \vee X_2 \vee \cdots \vee X_n)=H_1(X_1)\times \cdots \times H_1(X_n)$.

Aaron
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The abelianization of $\underbrace{\mathbb{Z}*\ldots *\mathbb{Z}}_n$ is $\underbrace{\mathbb{Z}\times\ldots \times\mathbb{Z}}_n$

The other examples of yours are ok.