Calculation of Klein bottle Integral Simplicial Homology the $H_{1}$ case I am new to algebraic topology. 
Martin Crossley in his Essential Topology shows how to calculate $$H_{1}(K)$$ if $K$ is the Klein bottle triangulation using simplicial complex, in example 9.17 on page 164. 
It is shown that $$H_{1}(K)= \mathbb{Z} \oplus \mathbb{Z}/2$$
But I cannot follow this result and how we calculate it. 
I see that based on the definition we have $$H_{1}(K)= Ker \delta_{1}/ Img \delta_{2}$$
Could you fill in the gap for me here. How we get from this definition to the result above?
In particular, could you list a number of elements in $H_{1}(K)$ so that I get a sense of what are the elements in the quotient group?
 A: I can choose the simplicial complex of a square to be :
$C_0 = \mathbb Z \langle a_1, a_2, a_3, a_4 \rangle$
$C_1 = \mathbb Z \langle x_1, x_2, x_3, x_4, x_5\rangle$
$C_2 = \mathbb Z \langle \Delta_1,\Delta_2 \rangle$
according to this drawing :

If I glue $x_1$ to $x_4$ and $x_2$ to $x_5$ (hence also $a_1$ to $a_2$ to $a_3$ to $a_4$), this induces a simplicial complex 
$C_0 = \mathbb Z \langle a_1 \rangle$
$C_1 = \mathbb Z \langle x_1, x_2, x_3\rangle$
$C_2 = \mathbb Z \langle \Delta_1,\Delta_2 \rangle$
for the Klein bottle $K$.
Now the boundaries are :
$\partial_0 a_1 = 0$
$\partial_1 x_1 = 0$
$\partial_1 x_2 = 0$
$\partial_1 x_3 = 0$
$\partial_2 \Delta_1 = x_1 + x_2 + x_3$
$\partial_2 \Delta_2 = - x_4 + x_5 - x_3 = -x_1 + x_2 - x_3$
We can now calculate explicitly $H_1(K)$ :
$$
H_1(K) = \frac{ker \partial_1}{im \partial_2}
\\ = \frac{\mathbb Z \langle x_1, x_2, x_3\rangle}{\mathbb Z \langle x_1 + x_2 + x_3, -x_1 + x_2 - x_3\rangle}
\\ = \mathbb Z \langle x_1, x_2, x_3| x_1 + x_2 + x_3, -x_1 + x_2 - x_3\rangle
\\ = \mathbb Z \langle x_1, x_2, -x_1 - x_2| 2x_2 = 0\rangle
\\ \cong \mathbb Z \oplus \mathbb Z_2
$$
