Let $X_t : \Omega \to E$ be a cadlag process with Polish state space $E$, $T$ a stopping time w.r.t. the canonical filtration $\mathcal{F}_t$ of $X$ and $X^T_t$ the stopped process. Then it should hold that $\mathcal{F}_T = \sigma(X^T_t | t \geq 0)$. Note, that this is an equality of $\sigma$-algebras on the unknown sample space $\Omega$!
From Dellacherie-Meyer IV.100 this equality follows only for the special case of $\Omega = D([0,\infty), E)$ being the space of cadlag maps. In general, $\Omega$ is unknown, but provides at least all the necessary samples $\omega$ such that $t \mapsto X_t(\omega)$ is cadlag for almost all $\omega$ (resp. has a cadlag version). How can I relate the equality of the sigma-algebras to the unknown $\Omega$ on which my original $X_t$ is defined?
Of course I can factor for each $t$ the random variable $X_{[0,t]} : \Omega \to E^{[0,t]}$ into a measurable projection $\pi_t: \Omega \to \Omega / X_{[0,t]}$ onto equivalence classes induced by $X_{[0,t]}$ (i.e. the saturated sets) with the quotient $\sigma$-algebra and a well-defined random variable $Y_{[0,t]} : \Omega / X_{[0,t]} \to E^{[0,t]}$. In this way, I can identify $\Omega / X_{[0,t]}$ with a subset of $D([0,t], E)$. Now, I could somehow extend $Y_{[0,t]}$ to $D([0,t], E)$ and get there the equality by Dellacherie-Meyer. Is it then just that simply, that I can push forward both $\sigma$-algebras to $\Omega / X_{[0,t]}$ (i.e. consider the trace $\sigma$-algebras on $\Omega / X_{[0,t]}$) and then pull-back to $\Omega$ by $\pi_t$? So, if I can do that for each deterministic stopping time $t$, can I then also try to proceed in an analogue way with a stopping time $T$?