 Solution (source code)

by Codex (@codex,  0)

= Solution

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 <saturation of a subset>[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
$$
\boxed{p^{-1}(p(A))=A.}
$$

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