This is inspired from the definition of consistency of states in Brassard and Robichaud https://arxiv.org/abs/1710.01380 . We have a family $(H_A)_{A \in {\cal S}}$ of arbitrary sets indexed by a join-semilattice ${\cal S}$. Physically, we can think of $H_A$ as the set of states of a system $A$ and we can think of $A \leq B$ as saying that $A$ is a subsystem of $B$. The reason why we have a join-semilattice is because, if we have two systems $A$ and $B$, the system $A$ and $B$ together constitute a larger system $AB$.
Physically, if $A \leq B$, the state $N^B \in H_B$ must uniquely determines the state $N^{A} \in H_A$. Mathematically, this is expressed by the existence of a mapping $\pi^{B}_{A}: H_B \mapsto H_A$ and a constraint $N^{A} = \pi^{B}_{A}(N^{B})$, for every pair $A, B \in {\cal S}$ where $A \leq B$. The overall state of all the systems together must be a solution to the following system of equations on the variables $N^{A}$, $A \in {\cal S}$:
$N^{A} = \pi^{B}_{A}(N^{B}) \text{ for every } A, B \text{ where } A \leq B. \tag{1}$
Naturally, we want that, for every system $A$, the projection $\pi_{A}$ from the set of solutions of system $(1)$ to $H_{A}$ is a surjection. Otherwise, it means that the set $H_{A}$ contains states that are physically impossible. If the mappings $\pi^{B}_{A}$ were totally arbitrary, for some choice of mappings, there might not be any solution to these equations, which is far from the surjectivity of the $\pi_{A}$, which we want.
Brassard and Robichaud require the following properties.
- For every $A \in {\cal S}$, the set $H_{A}$ is not empty.
- For every two $A, B \in {\cal S}$, where $A \leq B$, the map $\pi^{B}_{A}$ is a surjection.
- For every three $A, B, C \in {\cal S}$, where $A \leq B \leq C$, we have $\pi^{C}_{A} = \pi^{B}_{A} \circ \pi^{C}_{B}$.
- Also, for every $A \in {\cal S}$, $\pi^{A}_{A} = I$.
Brassard and Robichaud do not require that the maps $\pi_{A}$ are surjections. Therefore, unless proven otherwise, they might not be surjections. They require the existence of one solution to system $(1)$ and that ${\cal S}$ is finite, but, in this question, let's ignore that and only consider these four properties. In particular, the case ${\cal S}$ infinite makes the question more interesting.
An analogous question would be that we have an arbitrary infinite sequence $x_0, x_1, \ldots$ of real numbers $x_i \in \mathbb{R}$ and, for every pair of integers $i,j$ with $i \leq j$, we have a mapping $\pi^{j}_{i}: \mathbb{R} \mapsto \mathbb{R}$. The system $(1)$ is analogous to
$x_i = \pi^{j}_{i}(x_j), \text{ for every } i, j \text{ with } i \leq j. \tag{2}$
Whether there is a solution depends on the mappings. In the special case where the mappings are each the identity, then we have the solution $a = x_0 = x_1 = \ldots$ to system $(2)$.
Back to system $(1)$, Brassard and Robichaud have required that the sets $H_{A}$ and maps $\pi^{B}_{A}$ respect the four properties that are mentioned above. I would like to know if these properties are sufficient to garantee a solution to system $(1)$ and that the projections $\pi_{A}$ from the set of solutions to $H_{A}$ are surjections. I have a strong intuition that these properties are enough, but I cannot find a nice way to formalize that. So much that I doubt my intuition. Perhaps some extra conditions on the sets $H_{A}$ are needed. I would love to see a proof for ${\cal S}$ infinite.