Let $X_t\in S$ be a Markov Chain. Does there exist a function $f\colon S\to S$ such that $X_{t+1}=f(X_t)$?
One way I came up with representing $X_t$ in terms of $X_t$ is $$ X_{t+1} = \begin{cases} 0 & \text{, with probability $p_{00}$ if $X_t = 0$} \\ 0 & \text{, with probability $p_{01}$ if $X_t = 1$} \\ 1 & \text{, with probability $p_{10}$ if $X_t = 0$} \\ 1 & \text{, with probability $p_{11}$ if $X_t = 1$} \\ \end{cases} $$ but this this is not a function since $f(0)$ maps both to $0$ and $1$.