Computing $H_1(X)$ using Hurewicz Can you tell me if the following 3 computations are correct:


*

*$X = S^1$, $\pi_1 (X) = \mathbb{Z}$, $Ab(\mathbb{Z}) = \mathbb{Z}$ $\implies H_1(X) = \mathbb{Z}$

*$X = S^1 \vee S^1$, $\pi_1(X) = \mathbb{Z} \ast \mathbb{Z}$, $Ab(\pi_1(X)) = \pi_1(X) / \pi_1(X) = 0$, $\implies H_1(X) = 0$

*$X = T^2$, $\pi_1(X) = \mathbb{Z} \times \mathbb{Z} = Ab(\pi_1(X))$, $\implies H_1(X) = \mathbb{Z} \times \mathbb{Z}$ 


Thanks for your help!
 A: 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 $H_1(X_1 \vee X_2  \vee \cdots  \vee X_n)=H_1(X_1)\times \cdots \times H_1(X_n)$.
A: 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.
