# Is the projection of a vector-valued Markov process still a Markov process?

Let $(X_t\colon \Omega\rightarrow E\times F)_{t\in T}$ be a Markov process, with $E,F$ topological spaces (or metric spaces, or whatever makes sense). I am wondering whether the induced process $(Y_t)_{t\in T}$ with $Y_t=\mathrm{pr_E}\circ X_t$ will still be a Markov process. I tried to prove this directly using the Markov property, but failed to do so. More specifically, if $\mathcal{G}_s=\sigma(Y_t~|~t\leq s)$ and $\mathcal{F}_s=\sigma(X_t~|~t\leq s)$ and $f$ is bounded and measurable on $E$:

\begin{align*} \mathbb{E}[f(Y_t)~|~\mathcal{G}_s]&=\mathbb{E}[\mathbb{E}[f(Y_t)~|~\mathcal{G}_s]~|~\mathcal{F_s}] \\ &=\mathbb{E}[\mathbb{E}[f(Y_t)~|~\mathcal{F}_s]~|~\mathcal{G}_s]\\ &=\mathbb{E}[\mathbb{E}[f(Y_t)~|~\sigma(X_s)]~|~\mathcal{G}_s]\\ &=\mathbb{E}[f(Y_t)~|~\sigma(Y_t)] \end{align*}

but the last equality does not seem to hold (this is related to this question). Am I trying to prove something that is not true at all?

A projection of a Markov process is not Markov in general. For example, a simple random walk on non-negative half-axes in $\mathbb Z^2$ is a Markov process, but its projection to any coordinate is not.