From my reader on interacting particle systems:
Consider a Markov process $\{ X_t : t \geq 0\}$ with three states $\{1,2,3\}$ and "coagulate" (merge) the last two, i.e., define $Y_t = 1$ if $X_t = 1$ and $Y_t = 2$ if $X_t \in \{2,3\}$. In general, $Y_t$ will not be Markov. More generally if $X_t$ is a Markov process and $f$ is a non-bijective map, then in general $f(X_t)$ is not Markov. E.g. components of a multicomponent Markov process are in general not Markov, unless the components are e.g. independent. Notice that it can be Markov, e.g., $|W_t$ is a Markov process, despite the fact that $f(x) = |x|$ is of course not bijective.
Now my question is how to show this. If you can give an example of how the first system with $Y_t$ is non-Markovian that would already be helpful.
The fact that it's non-Markovian follows from the fact that you lose information on what the last state actually was. So as an example, imagine a simple markov chain on 3 vertices connected as:
1--2--3.
Then $P(X_2=1|X_1=3)=0$ but $P(X_2=1|X_1=2)>0$. So $P(Y_2=1|Y_2=2)$ can either be 0 or nonzero, i.e. you can't even evaluate it!
Sometimes however it would be Markovian. As an example, let $Y_t=1$ regardless of $X_t$. Then $Y_t$ is Markovian, despite not being a bijection.