Past exam of the options I course of the University of Oxford / 2010 / E / E2 / b / ii

If $x\in A=g^{-1}(B)$ and $yRx$, then $g(y)=g(x)\in B$, so $y\in A$. Thus $A$ contains every equivalence class it meets: it is a saturated subset. Always $A\subseteq p^{-1}(p(A))$; conversely, a point of the latter set is equivalent to some $x\in A$ and hence belongs to $A$. Therefore