markov property of Brownian motions

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In Schilling's "Brownian Motion" chapter 6, p60, I found the followin theorem:

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but I have several questions:

(1) Does "$\cdot$" in (6.3) mean "$\omega$" ? and we don't write it explicitly by convention? just like we write $\mathbb EX$ instead of $\mathbb EX(\omega)$ ?

(2)I can't understand the proof of (6.5) because I can't figure out what $\Psi$ should be in (6.3). In (6.3) $\Psi$ has 2 arguments but in (6.5) $\Psi$ only has 1 argument.

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(1) In (6.3), $\Psi(X(\cdot),\cdot)$ is a name for the map $\omega\mapsto \Psi(X(\omega),\omega)$.

(2) Schilling and Partzsch make an awkward notational choice here. Let's instead use $\Phi$ in (6.5). Then $\Phi(B_{\cdot+s}(\omega))=\Phi(B_{\cdot+s}(\omega)-B_s(\omega)+B_s(\omega))=\Phi(W(\omega)+B_s(\omega))=\Psi(X(\omega),\omega)$, where $\Psi(x,\omega):=\Phi(W(\omega)+x)$, and the other ingredients are as in the last sentence of the proof.