Let $X$ be a Markov proces with state space $(E,\mathcal{E})$with initial distribution $\nu$ and transition function $P_{t}$, so $$E_{\nu}(f(X_{t+s})\mid\mathcal{F}_{s})=P_{t}f(X_{s})$$ Suppose that $(E',\mathcal{E}')$ is a measurable space and let $\phi:E\rightarrow E'$ be measurable and onto. Let $Q_{t}$ be a collection of kernels such that $$P_{t}(f\circ\phi)=(Q_{t}f)\circ\phi$$
I would now like to prove that $\phi(X)$, is a Markov process with respect to the state space $(E',\mathcal{E}')$, initial measure $\nu'$, where $\nu'(B')=\nu(\phi^{-1}(B'))$ and transition function $Q_{t}$.
So basically I would like to prove that $$E_{\nu'}(f(\phi(X_{t+s}))\mid\mathcal{F}_{s})=Q_{t}f(\phi(X_{s}))$$ Here is my try as $\phi$ is onto and f is assumed to be a bounded function $$E_{\nu'}(f(\phi(X_{t+s}))\mid\mathcal{F}_{s})=E_{\nu}(\phi^{-1}(f(\phi(X_{t+s})))\mid\mathcal{F}_{s})$$ $$=P_{t}\phi^{-1}(f(\phi(X_{s})))=(Q_{t}\phi^{-1})(f(\phi(X_{s})))$$
From this point on I do not know how to proceed. Could anyone help me further?
No, basically you would like to prove that, if $Y_t=\phi(X_t)$ and $\mathcal{F}^Y_t=\sigma(Y_s;s\leqslant t)$ for every $t$, then $$ E_{\nu'}(f(Y_{t+s})\mid\mathcal{F}^Y_{s})=Q_{t}f(Y_{s}), $$ not what you wrote. You could then show that this is equivalent to $$ E_{\nu}(f(\phi(X_{t+s}))\mid\mathcal{F}_{s})=Q_{t}f(\phi(X_{s})), $$ the definition of $\nu'$ allowing you to replace $\nu'$ by $\nu$, and the hypothesis that $\phi$ is onto allowing you to replace each $\mathcal{F}^Y_{s}$ by $\mathcal{F}_{s}$. You might want to proceed from there.