Let $A,B$ be complete Boolean algebras and $\varphi,\psi:A\rightarrow B$ be 
maps preserving $0,1$, and arbitrary joins and meets. Let $C$ be the equalizer of these two; so $C=\left\{a\in A:\varphi(a)=\psi(a)\right\}$. It can be easily checked that $C$ itself is a complete Boolean algebra. 


**Question:** Let $a\in A$ be given so that for any prime ideal $\mathfrak{r}$ in $B$, we have $$\varphi(a)\notin\mathfrak{r}\quad\Leftrightarrow\quad
\psi(a)\notin\mathfrak{r}.$$
Then for any prime ideal $\mathfrak{p}$ in $A$ with $a\notin\mathfrak{p}$, is $\mathfrak{p}\cap C=\mathfrak{q}\cap C$ implies $a\notin\mathfrak{q}$?


A motivation of the question was to check continuity of a certain functor. 
Let us denote the Stone space of prime (or equivalently, maximal) ideals as $\mathrm{Spec}$, so that $A$ and $B$ can be identified with the clopen algebras of 
$\mathrm{Spec}A$ and $\mathrm{Spec}B$. Then $\mathrm{Spec}C$ becomes a quotient space of $\mathrm{Spec}A$, by defining an equivalence relation $\sim$ on $\mathrm{Spec}A$ as $$\mathfrak{p}\sim\mathfrak{q}\quad
\textrm{if and only if}\quad\mathfrak{p}\cap C=\mathfrak{q}\cap C.$$
By the well-known duality, certainly $\mathrm{Spec}C$ should be the coequalizer of the induced maps $\varphi^{\*},\psi^{\*}:\mathrm{Spec}B\rightarrow\mathrm{Spec}A$ in the category of Stone spaces (with continuous maps as morphisms). Completeness conditions in fact says that this is the coequalizer in the category of *Stonean* spaces. But it is not clear to me about the relation between the "usual" quotient space of $\mathrm{Spec}A$ by the relation $\varphi^{\*}(\mathfrak{r})\sim'\psi^{\*}(\mathfrak{r})$ and the space $\mathrm{Spec}C$. The question asks that if we have a clopen set $a$ in $\mathrm{Spec}A$ which is closed under the relation $\sim'$, then it is also closed under $\sim$.


I'm also not sure about that completeness really matters or not.


Thank you.



About equalizer of Boolean algebras
The condition $\varphi(a)\notin\mathfrak{r}\iff \psi(a)\notin\mathfrak{r}$ is equivalent to $\varphi(a)=\psi(a)$, since an element of a Boolean algebra is determined by the prime ideals that contain it (if $a\neq b$, then there is a prime ideal containing $1+a+b$, which then must contain exactly one of $a$ and $b$). So in fact $a\in C$, so trivially if $\mathfrak{p}\cap C=\mathfrak{q}\cap C$ then $a\not\in\mathfrak{p}$ implies $a\not\in\mathfrak{q}$.


