I am currently reading the paper "Newton-Okounkov bodies, semigroups of integral points, graded algebras and intersection theory" by Kiumars Kaveh, Askold Georgievich Khovanskii (paper). In the page 937 the following paragraph appears:
Let $L$ be a linear subspace in $\mathbb{R}^n$ and $M$ a half-space in $L$ with boundary $\partial M$. A half-space $M \subset L$ is rational if the subspaces $L$ and $\partial M$ can be spanned by integral vectors, i.e., are rational subspaces.
$\hspace{5mm}$ With a rational half-space $M \subset L$ one can associate $\partial M_{\mathbb{Z}} = \partial M \cap\mathbb{Z}^n$ and $L_{\mathbb{Z}} = L \cap \mathbb{Z}^n$. Take the linear map $\pi_M : L \rightarrow \mathbb{R}$ such that $ker(\pi_M) = \partial M$, $\pi_M(L_{\mathbb{Z}}) = Z$ and $\pi_M(M \cap \mathbb{Z}^n) = \mathbb{Z}_{\geq 0}$, the set of all nonnegative integers. The linear map $\pi_M$ induces an isomorphism from $L_{\mathbb{Z}}/\partial M_{\mathbb{Z}}$ to $\mathbb{Z}$.
I have the following question:
Q) Why the linear map $\pi_M$ exists?
Thanks in advance.
$\mathbb{Z}^n$ contains a vector normal to $\partial M$ whose endpoint is in $M$ because $\partial M$ is a rational subspace. Projection onto one of these normal vectors has all but one ($\pi_M(L_{\mathbb{Z}}) = \mathbb{Z}$) of the listed properties because the integers form a ring.
Let $v \in \mathbb{Z}^n$, $v \perp \partial M$, $v \in M$ and $\pi_v$ be projection onto $v$. Then $\pi_v(L_{\mathbb{Z}}) = n\mathbb{Z}$ for some $n \in \mathbb{Z}_{>0}$ (because the image is a torsion-free, rank $1$ $\mathbb{Z}$-module). Take $\pi_M = \frac{1}{n} \pi_v$.
Why does $\mathbb{Z}^n$ contain a normal to $\partial M$?
A constructive method labels the coordinates $x_1, \dots , x_n$ then picks a permutation of these, $x_{\sigma(1)}, \dots, x_{\sigma(n)}$. Then find the intersections of $\partial M$ with the $x_{\sigma(i)}x_{\sigma(i+1)}$ coordinate planes, $i \in [1,n-1]$. These intersections are lines with rational slopes. The perpendiculars to these intersections are also lines with rational slopes. Then, using the LCM, iteratively glue these vectors together to give a normal in $\mathbb{Z}^n$.
For example, suppose the permutation is $x_1, x_3, x_2$, where I will use the friendlier labels $x, z, y$, respectively, and $\partial M$'s intersections gives us coordinate plane perpendiculars through $(x,z) = (2,3)$ on the $xz$-plane and through $(y,z) = (7,2)$ on the $yz$-plane. The LCM of the $z$-components is $6$, so $2 \cdot(2,3) = (4,6)$ and $3 \cdot(7,2) = (21,6)$ assemble into $(4,21,6)$. If $n > 3$, we continue iterating through the permutation, having only one component of our current vector in common with the next coordinate plane perpendicular, so needing only one LCM to glue the new vector to the current vector.