What's going wrong with my cohomology computation? Use the following $\Delta$ complex for the klein bottle: 
we may obtain the chain complex:
$$\mathbb{Z}^2 \rightarrow \mathbb{Z}^3 \rightarrow \mathbb{Z} \rightarrow 0$$
My goal is to find the cohomology of this chain complex with $\mathbb{Z}$ coefficients: so I want to apply $Hom(-, \mathbb{Z})$ to these groups and find the homology groups for the corresponding (co)chain complex. There may be more powerful tools for doing this but I just started cohomology so I need to do it using just the definitions.
Now it is clear that $Hom(\mathbb{Z}^n, \mathbb{Z}) = \mathbb{Z}^n$ (identifying $(a_1, .. a_n)$ with the function that sends $(0,..1,..0)$ to $a_i$ where $1$ is in the $i$th slot)
So our cochain complex is $$0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}^3 \rightarrow \mathbb{Z}^2$$
I think I'm making mistakes in computing the coboundary maps $\delta^n$.
Given $(a_1, ..., a_n)$ denote $\overline{(a_1, ..a_n)}$ to be the associated element of $Hom(\mathbb{Z}^n, \mathbb{Z})$.
So $\partial_1 : \mathbb{Z}^3 \rightarrow \mathbb{Z}$ is the $0$ map. Notice then that given any $\overline{a} \in \mathbb{Z}$, $\delta^1 (\overline{a}) = \overline{a}(\partial_1) \in Hom(\mathbb{Z}^3, \mathbb{Z})$. Since $\partial_1 = 0$, so is $\delta^1(\overline{a}) = \overline{a}\partial_1$. So this means that $\delta^1$ is the $0$ map as well, right? Is there any mistake in this type of argument?
And similarly, computing $\delta^2 : \mathbb{Z}^3 \rightarrow \mathbb{Z}^2$. $\delta(\overline{(1,0,0)}) =  \overline{(1,0,0)}(\partial_2) \in Hom(\mathbb{Z}^2, \mathbb{Z})$. Notice that $\overline{(1,0,0)}\partial_2(1,0) = \overline{(1,0,0)}((1,1,-1)) = 1$ , and similar $\overline{(1,0,0)}\partial_3(0,1) = 1$ as well, so $\delta^2(\overline{(1,0,0)}) = (1,1)$. Similarly we have that $\delta^2(\overline{(0,1,0)}) = (1,-1)$ and $\delta^2(\overline{(0,0,1)}) = (-1,1)$.
This means that ultimately $\delta^2(x,y,z) = (x + y - z, x-y+z)$ describes $\delta^2$ which feels right, sorta.
So now we now our coboundary map and we may compute the cohomology groups. Let $C$ refer to the original chain complex.
Then $H^1(C; \mathbb{Z}) = ker(\delta^1) / 0 = \mathbb{Z}$ since $\delta^1 = 0$.
Next $H^2 (C; \mathbb{Z}) = ker(\delta^2)/im(\delta^1)$
And here's where the problem is; since $\delta^2(x,y,z) = (x+y-z, x-y+z)$, this means that $ker(\delta^2) = 0$, which means that $H^2(C; \mathbb{Z}) = 0$. But this can't be right, right?
Cos then we would have a sequence $$0 \rightarrow \mathbb{Z} \rightarrow 0$$ which makes no sense.
My suspicion is that I'm confusing direct sums and products or something like that. I'm still slightly confused about that stuff. Or perhaps I'm just going completely in the wrong direction and have no idea what I'm doing.
Please let me know what I'm doing wrong.
 A: Your cochain complex is correct and generated by the dual basis elements via $$0 \xrightarrow{\delta_{-1}} \langle v^*\rangle \xrightarrow{\delta_0} \langle a^*,b^*,c^*\rangle \xrightarrow{\delta_1} \langle U^*, L^*\rangle \xrightarrow{\delta_2} 0$$
$\delta_0$ is the $0$-map since $\delta_0v^*$ evaluated on the generators yields $0$, i.e. $$\delta_0v^*(a) = v^*(\partial_1a) = v^*(\partial_1b) = v^*(\partial_1c) = v-v = 0 $$
$\delta_2$ is obviously the $0$-map.
Evaluating $\delta_1$ on the generators $U,L$ gives $$\delta_1a^*(U) = a^*(\partial_2U) = a^*(a+b-c) = 1$$
$$\delta_1a^*(L) = a^*(\partial_2L) = a^*(c+a-b) = 1$$
$$\delta_1b^*(U) = b^*(\partial_2U) = b^*(a+b-c) = 1 $$
$$\delta_1b^*(L) = b^*(\partial_2L) = b^*(c+a-b) = -1 $$
$$\delta_1c^*(U) = c^*(\partial_2U) = c^*(a+b-c) = -1 $$
$$\delta_1c^*(L) = c^*(\partial_2L) = c^*(c+a-b) = 1 $$
Thus, $\delta_1$ is fully determined by $$\delta_1 = \begin{pmatrix} 1 & 1 & -1 \\ 1 & -1 &1 \end{pmatrix}$$
$$H^0(K;\mathbb{Z}) = \ker \delta_0/{\operatorname{im} \delta_{-1}} = \mathbb{Z}/0 =  \mathbb{Z}$$
The kernel of $\delta_1$ is generated by $\langle b^*+c^*\rangle$ and since $\operatorname{im} \delta_0 = 0$, we get $$H^1(K,\mathbb{Z}) = \langle b^*+c^*\rangle \cong \mathbb{Z}$$
Now $\operatorname{im} \delta_1$ is generated by $\langle U^*+L^*, U^*-L^*\rangle$ and since $\ker \delta_{2} = \mathbb{Z}^2$ we get
$$H^2(K,\mathbb{Z}) = {\langle U^*,L^*\rangle}\big/{\langle U^*+L^*, U^*-L^*\rangle} = \langle U^*, U^*-L^*\rangle\big/{\langle 2U^*, U^*-L^*\rangle} = \langle U^*\rangle\big/\langle 2U^*\rangle \cong \mathbb{Z}/2\mathbb{Z}$$
A: You can write the map $\delta^2$ as a matrix:
$$
\begin{pmatrix}
1 & 1 & -1 \\
1 & -1 & 1
\end{pmatrix}.
$$
If you subtract the first row from the second (an elementary row operation), this becomes
$$
\begin{pmatrix}
1 & 1 & -1 \\
0 & -2 & 2
\end{pmatrix}.
$$
This second matrix will have the same kernel and image as the first. Its kernel consists of elements of the form $(0, n, n)$ for $n \in \mathbb{Z}$. That is isomorphic to $\mathbb{Z}$, and that gives you $H^1$. To compute $H^2$, you have to compute $C^3/\textrm{im}\, \delta^2 = \mathbb{Z}^2/\textrm{im}\, \delta^2$. I claim that the image is given by all elements of the form $(a,b)$ where $a$ can be any integer but $b$ must be even. From that, you can get $H^2$.
