Let $(X\_t)\_{t \in \mathbb{N}}$ be a discrete-time Markov process on the finite state space $\mathcal{X}$, with transition matrix $T$. Suppose $f: \mathcal{X} \to \mathcal{Y}$ is a (deterministic) function such that $(f(X\_t))\_{t \in \mathbb{N}}$ is a Markov process on the state space $\mathcal{Y}$. What can we say about $f$ in relation to the transition matrix $T$ of the original chain? 


Ideally, I would like an exact characterization of when the derived process is Markov. If $\mathcal{X} = \mathcal{Y} \times \mathcal{Z}$ and $f$ is projection onto the first component, then I think it suffices for the transition probability $T((y\_1,z\_1),(y\_2,z\_2))$ to be constant in $z\_1$ **or** constant in $z\_2$. There is also the case when $f$ is injective. Are there any other cases?



When is a coarsening of a Markov chain Markov?

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Sure there are. Say, $(X\_t)\_{t\in\mathbb{N}}$ is the same as a chain $(Y\_t)\_{t\in\mathbb{N}}$ except that it has two copies $a\_0,a\_1$ of a certain state $a$ of $(Y\_t)\_t$. Every time $(Y\_t)\_t$ has a transition to $a$, $(X\_t)\_t$ flips a coin to decide which of $a\_0$ or $a\_1$ to go. For transitions out of $a$, $(X\_t)\_t$ simply ignores the distinction between $a\_0$ and $a\_1$. The function $f$ maps $a\_0$ and $a\_1$ to $a$ and every other state to itself.


The general criterion on $f$ to ensure that $(f(X\_t))\_t$ is a Markov chain is that for every $x,x'\in\mathcal{X}$ with $f(x)=f(x')$ and every $y\in\mathcal{Y}$, the transition probabilities should satisfy $T(x,f^{-1}(y))=T(x',f^{-1}(y))$. 


A related question: *under what conditions the projected process is a $k$-step Markov chain?*


