A categorical third isomorphism theorem Given is a regular category $\mathcal A$ (including finite limits) and an object $A$. A subobject of $A$ is a mono into $A$. 
Let $R =(r_1 : R \to A, r_2 : R\to A)$, $S = (s_1 : S\to A, s_2 : S \to A)$ be (internal) congruences on $A$. Let $r := \langle r_1, r_2\rangle : R \to A^2$ and $s:= \langle s_1, s_2 \rangle : S\to A^2$  be the corresponding monos (given by the universal property of $A^2$). Let $R$ also be a subrelation of $S$, that is there exists a morphism $m : R \to S$ such that $s\circ m = r$.
There is a well-known relation $S/R$ in universal algebra which occurs in the formulation of the third isomorphism theorem which roughly says:
$$\frac{A/R}{S/R} \cong A/S$$
How can we recover this relation and the third isomorphism theorem in category theory?
 A: Interestingly, the relation $S/R$ can be easily seen to be a kernel relation! (the relevant information can be found in "Basic Algebra 2 - Jacobson")
Given a relation $R'$ let $e_{R'} : A \to A /R'$ be the corresponding quotient.
We have $e_S\circ s_1 = e_S \circ s_2$, so $e_S\circ r_1 = e_S \circ r_2$ by composing with $m : R\to S$. Hence there is a unqiue morphism $A/m : A/R\to A/S$ with $A/m\circ e_R = e_S$ by the universal property of $e : A \to A /R$. Let $S/R = (S/R, \tilde r_1 : S/R \to A/R, \tilde r_2 : S/R \to A/R)$ be the kernel relation of $A/m$ (pullback along itself). 
This is known to be an equivalence relation and it's the one we are looking for.
Now let's see how to derive the desired isomorphism:
$$\frac{A/R}{S/R} \cong A/S$$
Since $A/m\circ e_R = e_S$ and $e_S$ is a strong epi, $A/m$ is also a strong epi. Now we have unique factorization $A/m = m\circ e_{S/R}$, where $m$ is a mono. Since $A/m$ is a strong epi $m$ is too, hence $m : \frac{A/R}{S/R} \to A/S$ is an iso. $\checkmark$

Remark: I had a different formulation in mind originally:
Let $\tilde r = \langle r_1, r_2\rangle : S/R \to A/R^2$ be the corresponding mono.
By the universal property of $S/R$ there is a unique morphism $u : S \to S/R$ with $r_1\circ u = e\circ s_1$ and $r_2\circ u = e\circ s_2$. Then: 
$$\tilde r\circ u = \langle r_1\circ u, r_2\circ u\rangle = \langle e\circ s_1, e\circ s_2\rangle = (e\times e)\circ s$$
I don't know yet whether or why $u$ is a regular epi, hence whether is the / an image of $(e\times e)\circ s$, but I suspect that.
