space-time process of an non-homogeneous markov process is a homogeneous markov process Let $(X_t)_{t\geq 0}$ a non-homogeneous Markov process. I have read several times, that the associated space-time process $(t, X_t)_{t\geq 0}$ is then a homogeneous Markov process.
I tried to come up with a proof, but it seems not very intuitive to me. Grateful for any hints, that push me in the right direction. 
 A: I have not heard the term space-time process before,  given the rest of your question I think I understand though (if not, I apologise beforehand). 
Let $\{Y_t\}$ be the what you call the space-time process. Now consider the probability a transition $\operatorname P \left (Y_{t + d} = (\tau, s') \mid Y_t = (\tau_0, s) \right)$. The transition probability can be non-zero only if $d = \tau - \tau_0$ (otherwise the transition is not possible since the time parameter is "deterministic").
If $d = \tau - \tau_0$ then the probability is given by the transition probability $\operatorname P \left ( X_\tau = s' \mid X_{\tau_0} = s \right )$, independently of $t$.
Hope that answers your question.
A: If $X$ has t.f.p $P_{s,t}$ then $t\to(t,X_t)$ has t.f.p $$ P_{t_1,t_2}(r,x,dr,dx):=\varepsilon_{r+(t_2-t_1)}(dr) P_{r,r+(t_2-t_1)}(x,dx), $$ this for
\begin{align*}
E[f(t,X_t)|\mathcal{G}_s]=\int P_{s,t}(X_s,dy) f(t,dy)
\end{align*}
but on other hand
\begin{align*}
P_{s,t} f(s,X_s)&=\int P_{s,t}(dx,dy) f(x,y)=\int\int f(x,y)P_{s,t}(s,X_s)(dx,dy)\\&=\int\int f(x,y) \varepsilon_{s+(t-r)}(dx) P_{s,s+(t-s)}(X_s,dy)\\
&=\int f(t,y) P_{s,t}(X_s,dy)
\end{align*}
