I want to explicitly understand the $\mathbb{Z}$-module I constructed as $M = \mathbb{Z}^4/\mathrm{im}(A)$, where $A\colon \mathbb{Z}^6 \to \mathbb{Z}^4$ is represented by the matrix $$ A = \begin{pmatrix} 1 & -1 & 0 & 1 & 0 & 0 \\ 1 & 0 & -1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & -1 & 1 \end{pmatrix} .$$
In order to do so, I have computed the Smith Normal Form of A as $$PAQ = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ,$$ where $$ P = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 1 & -1 & 1 \end{pmatrix} \text{ and } Q = \begin{pmatrix} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & -1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \end{pmatrix} $$
As I understand it, this tells me that the image of $A$ is isomorphic to $\mathbb{Z}^3 \oplus \{0\}$, and so $M \cong \mathbb{Z}$. However, knowing this in the abstract isn’t enough for my purposes, I need to explicitly know the map $\mathbb{Z}^4 \to M \to \mathbb{Z}$, where the first arrow is the projection $v \mapsto v + \mathrm{im}(A)$ and the second is an isomorphism. I suspect that this map can be constructed somehow via the matrix $P$, but I must admit I don’t understand the Smith Normal Form well enough to see how.
Ended up being able to work out the answer myself:
Since $Q$ is a change of basis matrix of $\mathbb{Z}^6$, we have $\mathrm{im}(A) = \mathrm{im}(AQ)$, which is the column space of $$ AQ = \begin{pmatrix} 1 & -1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix}, $$ i.e. $\left\{\begin{pmatrix} a-b+c & a & b & c \end{pmatrix}^T\ \middle|\ a,b,c \in \mathbb{Z}\right\}$.
Set $f_1, \dots, f_4$ to be the columns of $$ P^{-1} = \begin{pmatrix} 1 &-1 & 1 &-1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}. $$ Then we see that $\mathbb{Z}^4 / \mathrm{im}(A) = \frac{\mathbb{Z} f_1 \oplus \mathbb{Z} f_2 \oplus \mathbb{Z} f_3 \oplus \mathbb{Z} f_4}{\mathbb{Z} f_1 \oplus \mathbb{Z} f_2 \oplus \mathbb{Z} f_3} \cong \mathbb{Z} f_4$. The canonical dual basis $\{\phi^i\}$ of the $f_i$ is given by (left-multiplication with) the rows of $P$. In particular, the map $\phi^4 \colon \mathbb{Z}^4 \to \mathbb{Z}$ given by $\begin{pmatrix} a & b & c & d \end{pmatrix}^T \mapsto (-a+b-c+d)$ induces the desired isomorphism $\mathcal{M} \to \mathbb{Z}$.