probability - Is a Markov process uniquely determined?

Let

• $E$ be a Polish space and $\mathcal E$ be the Borel $\sigma$-algebra on $E$


• $I\subseteq[0,\infty)$ be closed under addition and $0\in I$

Please consider the following result:

Let $(\kappa_t:t\in I)$ be a Markovian semigroup on $(E,\mathcal E)$ $\Rightarrow$ There is a measurable space $(\Omega,\mathcal A)$ and a Markov process $X$ with distributions $(\operatorname P_x)_{x\in E}$ such that $$\operatorname P_x\left[X_t\in B\right]=\kappa_t(x,B)\;\;\;\text{for all }x\in E,B\in\mathcal E\text{ and }t\in I\;.\tag 1$$ Conversely, given a Markov process $X$ with distributions $(\operatorname P_x)_{x\in E}$ on a measurable space $(\Omega,\mathcal A)$, a Markovian semigroup $(\kappa_t:t\in I)$ is defined by $(1)$.

It turns out that $X$ in the first part of the statement can be constructed as the family of coordinate maps on $(\Omega,\mathcal A)=(E^I,\mathcal E^{\otimes I})$.

I've seen that many authors assume that Markov processes are such coordinate maps. Why can they do that?

The statement above doesn't state, that given $(\Omega,\mathcal A)$ there is one unique Markov process, does it? However, the finite-dimensional distributions of $X$, i.e. $$\operatorname P_x\left[X\in\;\cdot\;\right]\circ\pi_J^{-1}\;\;\;\text{for }J\subseteq I\text{ with }|J|<\infty\;,\tag 2$$ where $\pi_J:E^I\to E^J$ are the canonical projections, are uniquely determined by $(1)$.

Maybe $(\operatorname P_x)_{x\in E}$ (not only the finite-dimensional distributions) are uniquely determined by $(1)$, if $I\subseteq \mathbb N_0$ or $I$ is at least almost countable or when $E$ is almost countable.

So, why does the stated result allows us to think about $X$ as being uniquely determined?