Are there injective maps from a higher- to a lower-dimensional cellular automaton? Suppose we have a cellular automaton on $\mathbb Z^n$ with cell values in the finite set $V$, with update function $u : V^{\mathbb Z^n} \to V^{\mathbb Z^n}$, and similarly another cellular automaton on $\mathbb Z^m$ with values in $V'$, with update function $u' : V'^{\mathbb Z^m} \to V'^{\mathbb Z^m}.$
This is to say that $u$ and $u'$ are shift-invariant continuous maps on the infinite product spaces $V^{\mathbb Z^n}$ and $V'^{\mathbb Z^m}$, respectively.
Now suppose that $n > m$, and we have a map $\phi : V^{\mathbb Z^n} \to V'^{\mathbb Z^m}$ with the following properties:


*

*$\phi$ "commutes" with $u$ and $u'$: $\phi \circ u = u' \circ \phi$; and 

*for any point $p \in \mathbb Z^m$, the map $f \mapsto \phi(f)(p)$ depends on only a finite number of values of $f$, i.e. there is some finite set $Q_p \subsetneq \mathbb Z^n$ such that $\phi(f)(p)$ is a function of $f|_{Q_p}$. (So $\phi$ is, in particular, continuous.)


(For convenience, assume that each $Q_p$ is minimal.)
Now, the question:
Question: Is it possible for $\phi$ to be injective? If not, do we actually have the stronger statement that any fiber $\phi^{-1}(f')$ of $\phi$ is either empty or infinite?
This seems similar in spirit to the statement that a continuous map from $\mathbb R^n$ to $\mathbb R^m$ with $n > m$ cannot be injective; here the notion of continuity is given by the above 2 properties on $\phi$.
 A: Yes, it is possible to construct cellular automata $u:V^{\mathbb{Z}^n}\to V^{\mathbb{Z}^n}$ and $u':V'^{\mathbb{Z}^m}\to V'^{\mathbb{Z}^m}$ and continuous bijection $\phi:V^{\mathbb{Z}^n}\to V'^{\mathbb{Z}^m}$ such that $\phi\circ u=u'\circ\phi$.
Let $V=V'=\{0,1\}$ and let $u$ and $u'$ be the maps (one on $V^{\mathbb{Z}^n}$, the other on $V'^{\mathbb{Z}^m}$) that turn every configuration to the all-$0$ configuration.  Let $\gamma:\mathbb{Z}^n\to\mathbb{Z}^m$ be a bijection between $\mathbb{Z}^n$ and $\mathbb{Z}^m$ and set $\phi(f)(p):=f(\gamma^{-1}(p))$.
However, if you require that $\phi$ commutes with $\mathbb{Z}^m$-translations (in the spirit of cellular automata), then the answer becomes negative, because there is an entropy obstruction.  More specifically, for each $i=(i_1,i_2,\ldots,i_m)\in\mathbb{Z}^m$, let us write $\overline{i}:=(i_1,i_2,\ldots,i_m,0,0,\ldots,0)\in\mathbb{Z}^m$.  Denote by $\sigma^i$ and $\sigma^{\overline{i}}$ the translation maps on $V^{\mathbb{Z}^m}$ and $V'^{\mathbb{Z}^n}$ by vectors $i$ and $\overline{i}$, respectively.  Then, provided $|V|,|V'|\geq 2$, there exists no continuous bijection $\phi$ such that $\phi\circ\sigma^{\overline{i}}=\sigma^i\circ\phi$ for each $i\in\mathbb{Z}^m$.
