How can we show that $\left(X^{(1)}_t\right)_{t\in[0,\:\tau_1)}$ and $\left(X^{(2)}_t\right)_{t\in[0,\:\tau_2)}$ are independent here?

Question

Let

• $(\Omega_i,\mathcal A_i)$ be a measurable space;
• $(E_i,\mathcal E_i)$ be a measurable space with $E_1\cap E_2=\emptyset$ and $(E,\mathcal E):=(E_1\cup E_2,E_1\vee E_2)$;
• $\Delta_i\not\in E_i$ and $(E_i^\ast,\mathcal E_i^\ast)$ denote the one-point extension of $(E_i,\mathcal E_i)$ by $\Delta_i$
• $\left(X^{(i)}_t\right)_{t\ge0}$ be an $(E_i^\ast,\mathcal E_i^\ast)$-valued process on $(\Omega_i,\mathcal A_i)$ and $$\tau_i:=\inf\left\{t\ge0:X^{(i)}_t=\Delta_i\right\};$$
• $T_1$ be a Markov kernel with source $(\Omega_1,\mathcal A_1)$ and target $(E_1,\mathcal E_2)$
• $p_i$ denote the projection from $\Omega:=\Omega_1\times\Omega_2$ onto the $i$th coordinate and $$\tilde X^{(i)}:=X^{(i)}\circ p_i$$
• $\operatorname P^{(i)}_{x_i}$ be a probability measure on $(\Omega_i,\mathcal A_i)$ with $$\operatorname P_{x_i}^{(i)}\left[X^{(i)}_0=x_i\right]=1$$ for $x_i\in E_i$
• $\pi_i(x_i,\;\cdot\;):=\operatorname P_{x_i}^{(i)}$ for $x_i\in E_i$
• $[\Delta_i]\in\Omega_i$ with $$X^{(i)}[\Delta_i]\equiv\Delta_i$$
• $\tilde\tau_i:=\tau\circ p_i$

Now define $$\operatorname P_{x_1}:=\operatorname P^{(1)}_{x_1}\otimes(T_1\pi_2)$$ for $x_1\in E_1$ and $$\operatorname P_{x_2}:=\delta_{\left[\Delta_1\right]}\otimes\operatorname P^{(2)}_{x_2}$$ for $x_2\in E_2$.

I would like to show that if $f:E\to\mathbb R$ is bounded and $\mathcal E$-measurable, then $$\int_0^{\tilde\tau_1}f\left(\tilde X^{(1)}_t\right)\:{\rm d}t\tag1$$ and $$\int_0^{\tilde\tau_2}f\left(\tilde X^{(2)}_t\right)\:{\rm d}t\tag2$$ are $\operatorname P_{x_1}$-independent for all $x_1\in E_1$.

Further assumptions on $T_1$ are obviously necessary. I would like to assume that $T_1(\;\cdot\;,B_2)$ is $\mathcal F_{\tau_1-}$-measurable for all $B_2\in\mathcal E_2$. This should be enough.

However, I don't know how to proceed. Maybe we can show the stronger claim that $\left(X^{(1)}_t\right)_{t\in[0,\:\tau_1)}$ and $\left(X^{(2)}_t\right)_{t\in[0,\:\tau_2)}$ are $\operatorname P_{x_1}$-independent. But I struggle to show the latter ... It might be useful to note that $$\operatorname E_{x_1}\left[g\left(\tilde X^{(2)}_0\right)\mid\mathcal F_{\tilde\tau_1-}\right]=T_1\left.f\right|_{E_2}\circ p_1\tag3$$ for all bounded $\mathcal E$-measurable $g:E\to\mathbb R$. In particular, if $$T_1(\omega_1,\;\cdot\;)=\mu_1\left(X^{(1)}_{\tau_1-}(\omega_1),\;\cdot\;\right)\;\;\;\text{for all }\omega_1\in\Omega_1$$ (assuming $\mathcal E_i$ is the Borel $\sigma$-algebra of a topology on $E_i$ and $X^{(1)}$ is left-regular with respect to that topology), then $$\operatorname E_{x_1}\left[g(\tilde X^{(2)}_0)\mid\tilde X^{(1)}_{\tilde\tau_1-}\right]=(\mu_1\left.f\right|_{E_2})\left(\tilde X^{(1)}_{\tilde\tau_1-}\right)\tag4.$$ So, my interpretation is that $\tilde X^{(2)}$ and $\tilde X^{(1)}$ are conditionally independent given $\tilde X^{(1)}_{\tilde\tau_1-}$ (or maybe better $\mathcal F_{\tilde\tau_1-}$)?

Note that it is easy to show that $\tilde X^{(2)}$ and $\tilde X^{(1)}$ are actually $\operatorname P_{x_1}$-independent for all $x_1\in E_1$, when $T_1$ does not depend on the first argument (i.e. is simply a measure).