For $k\in\mathbb{N}$, let $H_k$ be the free Heyting algebra on $k$ variables $p_1,\ldots,p_k$ and $B_k$ be the free Boolean algebra on the same $k$ variables. Thus, $B_k$ has $2^{2^k}$ elements (corresponding to all possible truth-value tables with $k$ entries), while $H_k$ is infinite as soon as $k\geq 1$.

Since Boolean algebras are, in particular, Heyting algebras, there is a natural morphism $\psi \colon H_k \to B_k$ (taking each $p_i$ to the corresponding Boolean variable).

**Example:** For $k=1$ (and writing $p:=p_1$), the algebra $H_1$ is the (infinite) Rieger-Nishimura lattice (see here), while $B_1$ has four elements namely $\bot,p,\neg p,\top$. The morphism $\psi$ takes the single element $\bot$ to $\bot$, the two $p$ and $\neg\neg p$ to $p$, the single $\neg p$ to $\neg p$, and every other element of $H_1$ to $\top$. So three of its fibers are finite while the last is infinite.

Question:Which elements $u \in B_k$ have finite fiber $\psi^{-1}(u)$, and how can we describe their cardinalities or, better, their elements?