Homology groups of torus

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...

Solution 1:

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!