Cohomology of $\mathbb{R}P^2$ Note: this is from some homework exercises which my lecturer has left us which $\textbf{don't}$ count towards our grade in any way.

We have $$0\xleftarrow{0}\mathbb{Z}^2 \xleftarrow{
\begin{pmatrix} 
1&-1&1\\
1&-1&-1
\end{pmatrix}}\mathbb{Z}^3\xleftarrow{
\begin{pmatrix}
1&-1\\
1&-1\\
0&0\end{pmatrix}}\mathbb{Z}\leftarrow 0$$
$\ker \delta_1$ is rank 1 and so is isomorphic to $\mathbb{Z}$.
$\ker \delta_2$ is rank 1, so isomorphic to $\mathbb{Z}$. 
$\ker \delta_3=\mathbb{Z}^2$, as $\delta_3$ is the $0$ map.
$\text{im}\delta_0=0$, as $\delta_0$ is the $0$ map.
What I can't understand is how to get the image of $\delta_1$.
Having looked up solutions(page 5) online, it looks like for $\text{im}\delta_2$ they've noticed that by summing $\psi(U)$ and $\psi(L)$ you get an even number ($2\phi(c)$), and that therefore $\text{im}\delta_2$ is the kernel of the map $\mathbb{Z}^2\to \mathbb{Z}_2$. This, I understand - except for why we chose to sum $\psi(U)$ and $\psi(L)$ in the first place.
Problem is, if I try to apply the same method to $\delta_1$, then I get
$$\{\psi(a)=\phi(w)-\phi(v),\psi(b)=\phi(w)-\phi(v),\psi(c)=\phi(v)-\phi(v)\}$$
Summing $\psi(a),\psi(b),\psi(c)$ I get $2(\phi(w)-\phi(v))$. I don't see how this means that $\text{im}\delta_1=\ker\delta_2$ (which is what I'm supposed to get).
 A: Instead of looking up the answer and back engineering, you can just use good old linear algebra.
The image of a linear map $\mathbb{Z}^m \mapsto \mathbb{Z^n}$ defined by an $m \times n$ matrix is equal to the column space of that matrix, which in the case of $\delta_2$ means the column space of 
$$\begin{pmatrix} 
1&-1&1\\
1&-1&-1
\end{pmatrix}
$$
By applying column operations, you can simply this matrix without altering its column space and therefore without altering im$\delta_2$. First add column (1) to column (2), then subtract column (1) from column (3), then multiply column (3) by $-1$, to get: 
$$\begin{pmatrix}
1 & 0 & 0 \\
1 & 0 & 2
\end{pmatrix}
\quad\text{and then delete column (2) to get}\quad
\begin{pmatrix}
1  & 0 \\
1  & 2
\end{pmatrix}
$$
Since that matrix has determinant 2, its column space has index 2 in $\mathbb{Z}^2$, so the quotient is a group of order 2, isomorphic to $\mathbb{Z}_2$.
By the way, you wrote "im$\delta_2$ is the kernel of the map $\mathbb{Z}^2 \mapsto \mathbb{Z}_2$". Question is, which map $\mathbb{Z}^2 \mapsto \mathbb{Z}_2$? There are four such maps: the image of each of the two basis elements of $\mathbb{Z}^2$ can be chosen independently from the two elements of $\mathbb{Z}_2$.
In this example, the normal form of the matrix shows that neither of the basis vectors $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ or $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ is in the column space and so each must map to the nontrivial element of the quotient $\mathbb{Z}_2$, 
