Exact all-preparations coarse Markov criterion

Let $\mathcal{H}$ be the subspace of functions on $E$ that are constant on each block. For $f(a)={\bar{f}}_i$ when $a\in C_i$, $(Qf)(a)={\sum }_{j\mathrm{\ne }i}q(a,C_j)({\bar{f}}_j-{\bar{f}}_i)$ for $a\in C_i$. If the block-rate sums are constant within each $C_i$, this value is independent of the chosen $a\in C_i$. Thus $Q$ maps $\mathcal{H}$ into itself, and its restriction is the matrix $\bar{Q}$ in the Statement. Every power $Q^n$ preserves $\mathcal{H}$, so the matrix series $P_t=e^{tQ}$ does too; for any block $C_j$, $P_t{1}_{C_j}$ is constant on each block and defines ${\bar{P}}_t=e^{t\bar{Q}}$.

Conditional on the full microscopic history through time $t$, the Markov property gives $\mathrm{Pr}({\bar{X}}_{t+s}=j\mid {\mathcal{F}}_t^X)=(P_s{1}_{C_j})(X_t)=({\bar{P}}_s{)}_{{\bar{X}}_t,j}$. The right side depends only on the current block. Taking conditional expectation with respect to the coarser history proves that $\bar{X}$ is a time-homogeneous Markov chain for every microscopic initial law. Moreover, ${\bar{\pi }}_i{\bar{q}}_{ij}={\sum }_{a\in C_i}{\sum }_{b\in C_j}{\pi }_a{w}_{ab}={\sum }_{a\in C_i}{\sum }_{b\in C_j}{\pi }_b{w}_{ba}={\bar{\pi }}_j{\bar{q}}_{ji}$, so its stationary block weights satisfy detailed balance.

Conversely, assume one time-homogeneous block generator applies for every microscopic initial law. Start once from a point mass at $a\in C_i$ and once at $a^{\mathrm{\prime }}\in C_i$. For $j\mathrm{\ne }i$, the finite matrix expansion $e^{hQ}=I+hQ+O(h^2)$ gives ${\mathrm{Pr}}_a({\bar{X}}_h=j)=hq(a,C_j)+O(h^2)$ and ${\mathrm{Pr}}_{a^{\mathrm{\prime }}}({\bar{X}}_h=j)=hq(a^{\mathrm{\prime }},C_j)+O(h^2)$. Both probabilities must use the same coarse transition entry from block $i$ to block $j$. Divide by $h>0$ and let $h\downarrow 0$; then $q(a,C_j)=q(a^{\mathrm{\prime }},C_j)$. This holds for each block pair, proving necessity and the equivalence.