I am trying to prove that every zonotope $\mathcal{Z}(b_1, \cdots, b_n) := \{\lambda_i \cdot b_i \mid \lambda_i \in [0,1]\}$ for some generators $b_1, \cdots, b_m \subseteq \mathbb{R}^d$ can be disjointly decomposed into (translated) parallelepipeds (possibly lower dimensional). To be precise, the parallelepipeds are half open and translated, meaning of the form $t + \sum_{i=1}^n [0, a_i)$. This is a proof / exercise of http://math.sfsu.edu/beck/papers/ccd.pdf (see page 170).
Very roughly, what is done there, is to argue by induction (the statement is trivially true for $n=1$, $\mathcal{Z}(b_1) = \{0\}, (0, b_1]$). Specifically $\mathcal{Z}(b_1, \cdots, b_{n-1})$ can be decomposed in parallelepipeds $\Pi_1, \cdots, \Pi_m$ (because we have one generator less). Similarly, we consider the decomposition of $\mathcal{Z}(\pi(b_1), \cdots, \pi(b_{n-1}))$ into $\phi_1, \cdots, \phi_{m'}$, where $\pi(\cdot)$ is the orthogonal projection onto $H := \{x \mid u_n^T \cdot x = 0\}$. Then, it is claimed that the desired union of parallelepipeds are the $\Pi_i$ and $\mathcal{P}_i$, where $\mathcal{P}_i = \pi^{-1}(\phi_i) + (0,u_n]$.
Intuitively, this makes sense to me. To prove it, I try to show that any $x \in \mathcal{Z}$ belongs to (exactly) one such parallelepiped. Specifically, given $x$, we calculate the minimum $\gamma_m \in [0,1]$ such that $x-\gamma_m \in \mathcal{Z}(b_1, \cdots, b_{m-1})$. If $\gamma_m = 0$, then, then, by induction, there must be some $\Pi_i \ni x$. So suppose this minimum $\gamma_m \in ]0,1]$ and define $y = x-\gamma_m b_m$. Now (ideally ;) ) $\pi(y) \in \phi_i$ for a unique $\phi_i$ that arises in the decomposition of $\mathcal{Z}(\pi(b_1), \cdots, \pi(b_{n-1}))$.
I cannot conclude that $x \in \phi_i + (0,u_n]$. I think, the issue is that I do not really know how to properly define the $\Pi_i$ and $\phi_i$ (at least geometrically, for all choices of $\phi_i$, we need to pick the one "closest" in the direction of $u_n$, but I have no idea how to make this rigorous). Any help is appreciated.