I computed the homology groups of the torus, can someone tell me if this is correct? The calculation, not the result that is. Thanks!
The cells of $T^2$ are $e^0, e^1_a, e^1_b, e^2$
The chain groups are
$$ C_0(T^2) = \{ k e^0 | k \in \mathbb{Z} \} = \mathbb{Z}$$
$$ C_1(T^2) = \{ k_1 e^1_a + k_2 e^1_b | k_1 , k_2 \in \mathbb{Z} \} = \mathbb{Z} \oplus \mathbb{Z}$$
$$ C_2(T^2) = \mathbb{Z}$$
$$ C_k(T^2) = \{ 0 \} , k > 2$$
Now the homology groups:
$$ H_0(T^2) = \ker \partial_0 / im \partial_1 = \mathbb{Z} / 0 = \mathbb{Z}$$
where $im \partial_1 = 0$ because there is no chain in $C_1(T^2)$ whose boundary is a zero-chain in $C_0(T^2)$. (Is this reasoning correct?)
$$ H_1(T^2) = \ker \partial_1 / im \partial_2 = \mathbb{Z} \oplus \mathbb{Z}$$
where $\ker \partial_1 = \mathbb{Z} \oplus \mathbb{Z} $, i.e. again everything maps to zero because there is no element in $C_1$ whose boundary maps to an element in $C_0$.
$im \partial_2 = 0$ again because there is no element in $C_2$ whose boundary is an element of $C_1$.
I don't want to use Hurewicz to do $H_1$.
$$ H_2(T^2) = \mathbb{Z}$$ using similar arguments as above.
Thanks for your help.
Edit
I posted the computations as an answer below. I got two up votes but I don't know by who so I'm not yet sure I can trust them…
Best Answer
Ok, so here are the boundary homomorphisms:
$$ \dots \rightarrow 0 \xrightarrow{d_3} \mathbb{Z} \xrightarrow{d_2} \mathbb{Z}^2 \xrightarrow{d_1} \mathbb{Z} \xrightarrow{d_0} 0$$
$d_0 = 0$
$d_1 = 0$ because the attaching map ($f = const.$) as there is one $0$-cell.
$d_2 = c_1 e_1^1 + c_2 e_2^1= 0$ because $f = ab a^{-1}b^{-1}$ so the coefficients are $c_i = +1 - 1$ respectively.
The homology groups as claimed above in the question follow from these maps.
Is this right? Many thanks for your help!