I've got a question from example 2.37 from Hatcher's Algebraic Topology.
In that example there's the chain complex coming from the cell decomposition of a non-orientable surface of genus $g$, $N_g$:
$0\to\mathbb{Z}\xrightarrow{d_2}\mathbb{Z}^g\xrightarrow{d_1}\mathbb{Z}\to 0$
where $d_1=0$ and $d_2(1)=(2,\dots, 2)$. Since $d_1=0$, $H_1(N_g)=\mathbb{Z}^g/\mathrm{Im}(d_2)$. I would say that the quotient is $\mathbb{Z}_2\oplus\cdots\oplus\mathbb{Z}_2$, but Hatcher says
If we change the basis for $\mathbb{Z}^g$ by replacing the last standard basis element $(0, \dots , 0, 1)$ by $(1, \dots , 1)$, we see that $H_1(N_g) ≈ \mathbb{Z}^{g−1}⊕\mathbb{Z}_2$.
It is true that $d_2(1)=0\cdot(1, \dots , 0, 0)+\cdots +0\cdot (0, \dots , 1, 0)+2\cdot (1, \dots , 1)$, so I guess that's where we get $\mathbb{Z}^{g−1}⊕\mathbb{Z}_2$ from.
But why was my first intuition wrong?
The image of $d_2$ is a copy of $\Bbb Z$ inside $\Bbb Z^g$. So the quotient of $\Bbb Z^g$ by $d_2(\Bbb Z)$ will be an Abelian group of rank $g-1$. For $g\ge2$ that will be a group containing a copy of $\Bbb Z^{g-1}$ so certainly infinite, and not torsion (unlike $\Bbb Z_2^g$).