If a morphism $X\xrightarrow{f}Y$ has a kernel pair, denote the kernel pair by $K[f]\rightrightarrows X$. The equivalence of the conditions
- $I\xrightarrow{m}Y$ is a monomorphism
- The kernel pair $K[m]\rightrightarrows I$ exists and its projections are equal
- The kernel pair $K[m]\rightrightarrows I$ exists and its projections are isomorphisms.
implies that the answer to your question hinges on the behavior of the kernel pair of $I\xrightarrow{m}Y$. In order to state the relevant theorem, I will need the fact that kernel pairs are functorial, which I summarize below.
Recall that $K[f]\rightrightarrows X\xrightarrow{f}Y$ is by definition the pullback square of $X\xrightarrow{f} Y$ along itself. Recall also that the pullback square of two morphisms $X_1\xrightarrow{f}Y$ and $X_2\xrightarrow{g}Y$ along one another is the same thing as their binary product in $\mathcal C/Y$, the category whose objects are morphisms with codomain $Y$ and whose morphisms from $X_1\to Y$ to $X_2\to Y$ are morphisms $X_1\to X_2$ that participate in commutative triangles/factorizations $X_1\to X_2\to Y$ of $X_1\to Y$ through $X_2\to Y$.
Thus, a kernel pair $K[f]\rightrightarrows X$ by definition consists of the projection morphisms of a limiting cone $K[f]\rightrightarrows X\xrightarrow{f}Y$ over two copies of $X\xrightarrow{f}Y$ in $\mathcal C/Y$. Let $\mathcal C/Y_K$ be the full subcategory of $\mathcal C/Y$ of objects that have squares/morphisms $X\xrightarrow{f}Y$ that have kernel pairs. From the universal property of products/kernel pairs, we have a functor $K\colon\mathcal C/Y_K\to\mathcal C/Y$.
Explicitly, on objects it takes a morphism $X\xrightarrow{f}Y$ with a kernel pair to the equal composites $K[f]\rightrightarrows X\xrightarrow{f}Y$. On morphisms, if $X\xrightarrow{f}Y$ factors as $X\xrightarrow{q}I\xrightarrow{m}Y$ and both $f$ and $m$ have kernel pairs, then the composites $K[f]\rightrightarrows X\xrightarrow{q}I\xrightarrow{m}Y$ are equal hence they factor uniquely as the composites $K[f]\xrightarrow{K(q)}K[m]\rightrightarrows I$.
The morphism $K[f]\xrightarrow{K(q)}K[m]$ is a morphism in $\mathcal C/Y$ from $K[f]\rightrightarrows X\xrightarrow{f}Y$ to $K[m]\rightrightarrows I\xrightarrow{m} Y$, and we declare it the image of $X\xrightarrow{q} I$ under the functor $K$.
Lemma. Suppose that $X\xrightarrow{f}Y$ factors as $X\xrightarrow{q}I\xrightarrow{m}I$ and both $X\xrightarrow{f}Y$ and $I\xrightarrow{m}Y$ have kernel pairs.
If $q$ coequalizes the pair $K[f]\rightrightarrows X$ (but not necessarily a coequalizer), then the composites $K[f]\xrightarrow{K(q)}K[m]\rightrightarrows I$ are equal; if $q$ is an epimorphism in $\mathcal C$ then $K[f]\xrightarrow{K(q)}K[m]\rightrightarrows I$ are epimorphisms.
Proof.
Since the composites $K[f]\xrightarrow{K(q)}K[m]\rightrightarrows I$ are by definition equal to the composites $K[f]\rightrightarrows X\xrightarrow{q}I$, both claims follow.
Theorem. Suppose $X\xrightarrow{f}Y$ has a kernel pair and factors as $X\xrightarrow{q}I\xrightarrow{m}Y$. Then we have $(1)\Rightarrow(2)\Rightarrow(3)$ for the conditions below.
- The induced morphism $K[f]\xrightarrow{K(q)}K[m]$ is an epimorphism in $\mathcal C/Y$.
- $I\xrightarrow{m}Y$ is a monomorphism.
- The kernel pair $K[m]\rightrightarrows I$ exists and $X\xrightarrow{q}I$ coequalizes (but is not necessarily the coequalizer of) the kernel pair $K[f]\rightrightarrows X$.
If $X\xrightarrow{q}I$ is an epimorphism, then we also have $(2)\Rightarrow(1)$.
Proof.
Certainly the existence of the kernel pair of $I\xrightarrow{m} Y$ is necessary. Assuming that it exists, the fact that the composites $K[f]\xrightarrow{K(q)}K[m]\rightrightarrows I$ are equal to the composites $K[f]\rightrightarrows X\xrightarrow{q}I$ gives us
- If $I\xrightarrow{m}Y$ is a monomorphism and $X\xrightarrow{q}I$ is an epimorphism, $K[m]\rightrightarrows I$ are isomorphisms, hence by the lemma $K[f]\xrightarrow{K(q)}K[m]$ is an epimorphism in $\mathcal C$.
- If $K[f]\xrightarrow{K(q)}K[m]$ is an epimorphism in $\mathcal C/Y$, the projections $K[m]\rightrightarrows I$ are equal, hence $I\xrightarrow{m}Y$ is a monomorphism.
Corollary 1. If $\mathcal C$ has kernel pairs and stable pullbacks of regular epimorphisms, then for any factorization $X\xrightarrow{q}I\xrightarrow{m}Y$ of $X\xrightarrow{f}Y$ through the coequalizer of the kernel pair, $m$ is a monomorphism.
Proof. If $X\xrightarrow{q}I$ has pullbacks, then $K[f]\xrightarrow{K(q)}K[m]$ is the composite of pullbacks of $X\xrightarrow{q}I$.
Lemma.
If $X\xrightarrow{q}I\xrightarrow{m}Y$ is a factorization of $X\xrightarrow{f}Y$ through a (weak) coequalizer of $W\rightrightarrows X$ that coequalizes $K[f]\rightrightarrows X$, then $X\xrightarrow{q}I$ is a (weak) coequalizer of $K[f]\rightrightarrows X$.
Proof.
Since $X\xrightarrow{q}I$ coequalizes $W\rightrightarrows X$, $f=m\circ q$ also coequalizes them, hence $W\rightrightarrows X$ factor uniquely as $W\to K[f]\rightrightarrows X$. Consequently, if $X\xrightarrow{j}Z$ coequalizes $K[f]\rightrightarrows X$, it also coequalizes $W\rightrightarrows X$, hence admits a factorization through $X\xrightarrow{q}I$.
Corollary 2. Suppose $\mathcal C$ has kernel pairs and consider the following three conditions.
- For each (strong epi-*) factorization $X\xrightarrow{q'}J\xrightarrow{n'}Y$ of $X\xrightarrow{f}Y$ such that $X\xrightarrow{q'}J$ coequalizes $K[f]\rightrightarrows X$, the induced morphism $K[f]\xrightarrow{K(q')}K[m]$ is an epimorphism in $\mathcal C/Y$.
- Every (strong epi-*)-factorization such that the strong epimorphism coequalizes the kernel pair of the morphism is actually a (strong epi-mono) factorization.
- Every strong epimorphism is a weak coequalizer.
We always have $(1)\Rightarrow(2)$. If $\mathcal C$ has finite products and (strong epi-*)-factorizations, then we have $(2)\Rightarrow (3)$.
If all strong epimorphisms are actually epimorphisms, we always have $(2)\Rightarrow (1)$. If $\mathcal C$ has (strong epi-mono)-factorizations and every strong epimorphism is actually an epimorphism, we also have $(3)\Rightarrow(2)$, in which case all strong epimorphisms are actually regular epimorphisms.
Proof.
$(1)\Rightarrow(2)$ and $(2)\Rightarrow(1)$ when strong epimorphisms are epimorphisms follow directly from the theorem and uses only the kernel pairs in $\mathcal C$.
From $(3)$ and strong epimorphisms being actually epimorphisms, we can use the lemma to conclude that both the $X\xrightarrow{q}I$ in the (strong epi-mono)-factorization of $X\xrightarrow{f}Y$ and $X\xrightarrow{q'}J$ in a (strong epi-*) factorization of $X\xrightarrow{f}Y$ that coequalizes $K[f]\rightrightarrows X$ are coequalizers of $K[f]\rightrightarrows X$. Hence the two factorizations are isomorphic, so both are (strong epi-mono)-factorizations of $X\xrightarrow{f}Y$.
It remains to check $(2)\Rightarrow(3)$.
Let $X\xrightarrow{q}I\xrightarrow{m}Y$ be the (strong epi-mono)-factorization of $X\xrightarrow{f}Y$ and let $X\xrightarrow{g} Z$ be any other morphism that coequalizes the kernel pair $K[f]\rightrightarrows X$.
Then $X\xrightarrow{(f,g)}Y\times Z$ also coequalizes $K[f]\rightrightarrows X$. Let $X\xrightarrow{q'}J\xrightarrow{m'}Y$ be the (strong epi-mono)-factorization of the product $X\xrightarrow{(f,g)}Y\times Z$; note that $X\xrightarrow{q'}J$ also coequalizes $K[f]\rightrightarrows X$.
From $(2)$ we can conclude that $J\xrightarrow{m}Y\times Z\xrightarrow{\pi_1} Y$ is a monomorphism. This gives a (*-mono)-factorization $X\xrightarrow{q'}J\xrightarrow{m'}Y\times Z\xrightarrow{\pi_1}Y$ in addition to the (strong epi-mono)-factorization $X\xrightarrow{q}I\xrightarrow{m}Y$ of $X\xrightarrow{f}Y$. Therefore, by definition of strong epi, there is a morphism $I\xrightarrow{d}J$ so that $X\xrightarrow{q'}J$ factors as $X\xrightarrow{q}I\xrightarrow{d}J$, and hence $X\xrightarrow{g}Z$ factors as $X\xrightarrow{q}I\xrightarrow{d}J\xrightarrow{m'}Y\times Z\xrightarrow{\pi_2}Z$. This shows that the strong epimorphism $X\xrightarrow{q}I$ in the (strong epi-mono)-factorization of $X\xrightarrow{f}Y$ is a weak coequalizer of $K[f]\rightrightarrows X$.
Best Answer
$\require{AMScd}$ 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.