Q:

Chinese remainder theorem in a regular category

I want to prove a Chinese remainder theorem for regular categories (if this is possible).  I'm stuck with the following question, which is really easy to answer positively in the set-theoretical case:

Given congruences $R,S$ such that $SR = \top_A$ (the biggest relation
  on $A$) on an object $A$ and respective quotients $e_1 : A \to A/R$,
  $e_2 : A\to A/S$, is the morphism $$e := (e_1, e_2) : A\to A/R\times A/S$$ a regular epi? 

The condition "$SR = \top_A$" is equivalent to saying that $(R_1\circ p_2, S_2\circ p_1)$ is a regular epi, where the upper left corner in the following diagram is a pullback:

But I don't see how to get anything useful from this. 

A:

$\require{AMScd}$
Consider the diagram
\begin{CD}A  @>{e_2}>> A/S \\
@V{e_1}VV  @VV{\tau_2}V \\
A/R  @>>{\tau_1}> 1
\end{CD}
and denote $\tau_A$ the unique arrow $A\to 1$, which is the diagonal of this square.  Then by Proposition 2.2 in the paper "Some remarks on Mal'tsev and Goursat categories", the induced arrow to the pullback (or in this case, product) $A/R\times A/S$ is a regular epimorphism if and only if $\tau_1^{o} \tau_2=e_1e_2^{o}$. Now since $R=e_1^{o}e_1$, $S=e_2^{o}e_2$ and $\top_A=\tau_A^{o}\tau_A$, we have
$$e_1^{o}e_1e_2^{o}e_2=\tau_A^{o}\tau_A=e_1^{o}\tau_1^{o}\tau_2e_2;$$
composing with $e_1$ on the left and $e_2$ on the right gives us
$$e_1e_1^{o}e_1e_2^{o}e_2e_2^{o}=e_1e_1^{o}\tau_1^{o}\tau_2e_2e_2^{o}.$$
Now $e_1$ and $e_2$ are regular epimorphisms, so we have $e_1e_1^{o}=id_A=e_2e_2^{o}$, and thus $e_1e_2^{o}=\tau_1^{o}\tau_2$.
Note : the proof above is more or less the same as the one of Theorem 5.2 in the paper I mentioned.