I met some confusion when I was working on the following exercise: $X_t$ is a non-time homogeneous Markov process with state space $E$, prove $Y_t:=(X_t,t)$ is a time homogeneous Markov process with state space $E\times [0,\infty)$. The solution is considering the probability transition kernel of $Y_t$, \begin{equation} \begin{split} Q_{t_1,t_2}[(u,r), A\times B]&=\mathbb{P}(X_{t_2}\in A, t_2\in B|X_{t_1}=u, t_1=r)\\ &=\mathbb{P}(X_{r+(t_2-t_1)}\in A|X_{r}=u)\cdot\mathbb{P}(t_2\in B|t_1=r)\\ &=\mathbb{P}(X_{r+(t_2-t_1)}\in A|X_{r}=u)\cdot\mathbb{P}((t_2-t_1)\in B|0=r)\\ \end{split} \end{equation} So it's only depend on $t_2-t_1$.
My confusion is whether the initial position of $Y_t$ is determined, in other words, if $r\neq t_1$, is such conditional probability well-defined? The time homogeneous Markov property means for any $u>0$ \begin{equation} \mathbb{P}(Y_{t_2+u}\in A|Y_{t_1+u}=x)=\mathbb{P}(Y_{t_2}\in A|Y_{t_1}=x) \end{equation} And now if the initial position of $Y_t$ is determined w.r.t. $t$, $Y_{t_1}$ and $Y_{t_1+u}$ are impossible to start from the same point. For example, $Y_3$ can only start from $(X_3,3)$, $Y_5$ can only start from $(X_5,5)$, $X_3$ can equal $X_5$, but $3\neq 5$. In this case, how to say the time homogeneous Markov property?
In short, is the initial position of $Y_t$ arbitrarily chosen in space? I completely cannot understand the meaning of the transition of this process. And the what's the meaning of $\mathbb{P}(t_2\in B|t_1=r)=\mathbb{P}(t_2-t_1\in B|0=r)$ in the solution.
Let $Q_{s, t}$, $s\leq t$, be the stochastic kernels from $(E,\mathcal{E})$ to itself such that $$\mathbb{E}[f(X_t)|\mathcal{F}_s]=\mathbb{E}[f(X_t)|X_s]=\int f(x)Q_{s, t}(X_s,dx)$$ for all measurable bounded function $f$ on $(E,\mathcal{E})$. Since $Q_{s, t}=Q_{s,s+(t-s)}$, by introducing a purely deterministic process $u_t=t$, and looking at $Y_t=(X_t,u_t)=(X_t,t)$, we can create a time homogeneous process on $E\times[0,\infty)$. More explicitly, consider the stochastic kernel $\tilde{Q}_{s,t}g(x,u)=\delta_{u+t-s}\otimes Q_{s, t}g$ for any bounded measurable function $g$ on $E\times[0,\infty)$. Then \begin{align} \mathbb{E}[g(X_{t+h},t+h)|(X_t,t)]&=\int g(x,u)\, \delta_{t+h} \otimes Q_{t,t+h}(X_t,u)\\ &=\int\Big(\int_E g(x,u)Q_{t,t+h}(X_t,dx)\Big)\delta_{t+h}(du)\\ &=\int_Eg(x,t+h)Q_{t,t+h}(X_t,dx) \end{align}
The check that $\tilde{Q}$ is the correct kernel on $E\times[0,\infty)$, notice that for $g(x,u)=\mathbb{1}_A(x)\mathbb{1}_B(u)$, $A\in\mathcal{E}$ and $B\in\mathcal{B}$, then \begin{align} \tilde{\mathbb{E}}[g(X_t,t)|(X_s,s)=(x,u)]&=\tilde{Q}_{s, t}g(x,u)\\ &=\int \mathbb{1}_B(w)\int\mathbb{1}_A(y)Q_s(x,dy)\,\delta_{u+t-s}(dw)\\ &=\mathbb{1}_B(u+t-s)Q_{s, t}(A)=\mathbb{1}_B(u+(t-s))Q_{u, t}(A) \end{align} on account that $s=u$. In particular, for $B=\mathbb{R}$, we recover the distribution of $X$.