Klein bottle and $\text{Ker } \delta_{1}/ \text{Img } \delta_{2}$ 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. 
Here is the triangulation outline for $K$ using labels for set of vertices:
$$acba$$
$$ghig$$
$$defd$$
$$abca$$
The first row is different from torus because we have $$cb$$ instead of $$bc$$
I see that based on the definition we have $$H_{1}(K)= \text{Ker } \delta_{1}/ \text{Img } \delta_{2}$$
It is said that "only $2([a,b], [b,c], [c,a])$ is in Img $\delta_{2}$"
Does this mean that Img $\delta_{2}$ is the trivial group for the Klein bottle?
If not can you list a few of the elements in Img $\delta_{2}$ 
Also if possible can you list a few of the elements in Ker $\delta_{1}$?
I need to see how we calculate the $$\text{Ker } \delta_{1}/ \text{Img } \delta_{2}$$ step by step.
 A: Without attempting to copy the notation, I will try to give the idea of what is happening here. The representation of the Klein bottle is a square. It has four sides and an interior. The opposite sides are identified, one pair with and the other against the orientation. The image of $\delta_2$ is the boundary of the interior of the square, and this is the sum of all four sides. Taking the identifications into account, one pair cancels and the other pair of opposite sides is doubled. This gives the image you mentioned. 
As for the kernel of $\delta_1$ this is simply the two pairs of identified sides so you have 
$$\ker \delta_1=\mathbb{Z}\oplus\mathbb{Z}$$
And $$\operatorname{im} \delta_2=0\oplus 2\mathbb{Z}$$ giving $$H_1(K)=\mathbb{Z}\oplus\mathbb{Z}_2$$ Hope this helps.
A: For a triangulation with $n$-simplices $e^n_0,e^n_1, e^n_2...$ the group $\text{Im}\, \partial_n$ is generated by the set $\partial_n e^n_0, \partial_n e^n_1, \ldots$ so if we have look at a $2$-simplex $[a,b,c]$ with linear ordering $a<b<c$ we explicitly have  $$\partial_2([a,b,c]) = [b,c]-[a,c]+[b,a] \in \text{Im}\, \partial_2 $$
As $\text{Im} \, \partial_2 \subset \text{Ker} \, \partial_1$ we already have an explicit element of $\text{Ker} \, \partial_1$. But if we want in $\gamma \in \text{Ker}\, \partial_1 / \text{Im}\, \partial_2$, take paths that terminate their origin but are not the boundary of any face.
