I struggled to understand the first 3 lines of this paragraph (see linked image), can someone elaborate? Not sure what is the inverse image in $\mathbb{Z}^2$ and how to understand the corresponding sublattice $L_H$ $\subset \mathbb{Z}^2$ in case for example $a$, $b$ are not both $0$ and $$ H = \bigl\{ (x, y)\in (\mathbb{Z}/p\mathbb{Z})^2: ax+by \equiv c\pmod p \bigr\} $$
(I'm only interested in the case $n=2$ and $j=1$, using the paragraph's notations). If someone can give an example to explain how this correspondence works and then to treat the "general case", i.e., to find $L_H$ for the $H$ I wrote above, it would be very appreciated.
This paragraph is from page 8 of this article: https://www.math.ias.edu/~akshay/research/andreas.pdf
I happened to have a few relevant pictures on my laptop in a folder dedicated to a suitable course in algebra, so I will share.
As commented by @CyclotomicField, this is all about the preimages of the natural projection (modulo $p$) $\pi:\Bbb{Z}^2\to(\Bbb{Z}/p\Bbb{Z})^2$. This is a homomorphism of groups, so the preimage of a subgroup is necessarily a subgroup. As a finitely generated subgroup of a free abelian group, it is then necessarily a free abelian group itself. Hence it is also lattice.
What's going on here is an instance of the correspondence theorem giving a bijection between subgroups $H$ of $(\Bbb{Z}/p\Bbb{Z})^2$ and the collection of such subgroups $M$ of $\Bbb{Z}^2$ that contain $p\Bbb{Z}^2$. The correspondence is via $M=\pi^{-1}(H)$. Sometimes the term lattice is reserved for certain subgroups $\Lambda$ of $\Bbb{R}^n$ (here obviously $n=2$ only). Namely the discrete, full rank, free abelian groups. But your source seems to also call cosets of such subgroups $\lambda+\vec{x}\subseteq\Bbb{Z}^2$ also lattices. This is fine (geometrically), but may explain the confusion. More precisely, the choice $c=0$ corresponds to the subgroup $\Lambda$ itself, and non-zero choices to its cosets.
The first picture is about the case $p=2$, $a=b=1$. Here $$H=\{(\overline{x},\overline{y})\in\Bbb{Z}_2^2\mid \overline{x}+\overline{y}=\overline{0}\}.$$ The preimage $\pi^{-1}(H)$ then consists of the lattice points $(x,y)$ such that $x+y$ is even. Those are marked red. The coset corresponding to $c=1$ (in other words, odd $x+y$) is marked blue.
The second picture is about $p=5$ and the case where $H=\langle (\overline{2},\overline{1})\rangle$. In other words, $H$ consists of the solutions of $x-2y=0$ inside $\Bbb{Z}_5^2$. This time $\pi^{-1}(H)$ (so $c=0$) is marked with blue dots, the coset $c=1$ with red dots, the coset $c=2$ green, the coset $c=3$ grey, and the coset $c=4$ yellow.
I apologize to viewers suffering from a form of color-blindness that makes the image less useful. I'm open to suggestions for other ways of visually distinguishing the cosets.