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
Here is an annotated version of the diagram:
$$ \require{AMScd} \begin{CD} X\times_Y X@>\alpha>>C\times_Y X @>\zeta>> X\\ @V\beta VV @VV\gamma V @VV{c}V\\ X\times_C C @>\delta>> C\times_Y C @>>{q_2}> C\\ @V\epsilon VV @VV{q_1}V @VV{i}V\\ X @>>{c}> C @>>{i}> Y \end{CD} $$
In this diagram there is a lot left implicit. Of course, the bottom right square is a pullback square by definition. The bottom left and top right squares are also defined (or "created") as pullbacks (well, you could define $\delta$ and $\epsilon$ from the pullback of $i$ and $ci$, justifying the notation $X\times_C C$, but this is equivalent by the pullback lemma). By axiom, since $c$ is obviously a regular epimorphism, both $\gamma$ and $\delta$ are (regular) epimorphisms.
The arrow $\phi:X\times_Y X\to C\times_Y C$ satisfies $q_1\phi=cp_1$ therefore $\langle p_1,\phi\rangle$ defines a unique arrow $\beta:X\times_Y X\to X\times_C C$ making $\epsilon\beta=p_1$ and $\delta\beta=\phi$. Similarly, $\langle\phi,p_2\rangle$ defines a unique arrow $\alpha:X\times_YX\to C\times_Y X$ such that $\zeta\alpha=p_2,\,\gamma\alpha=\phi=\delta\beta$.
In order to conclude $\phi=\delta\beta=\gamma\alpha$ is an epimorphism, knowing that $\delta,\gamma$ are regular epimorphisms means it suffices to show the top left corner is a pullback square since this would imply $\alpha,\beta$ are (regular) epimorphisms.
So, say, $u:Z\to X\times_C C$ and $v:Z\to C\times_Y X$ is such that $\gamma v=\delta u$. Using $f=ci$ and repeatedly moving through the commutative squares shows $f\circ\epsilon u=f\circ\zeta v$, so these induce a unique $w:Z\to X\times_Y X$ with $\epsilon\beta w=p_1w=\epsilon u$ and $\zeta\alpha w=p_2w=\zeta v$. This observation gives the uniqueness criterion of a pullback square so it remains to show $\beta w=u$ and $\alpha w=v$. Since $\beta w,u:Z\to X\times_C C$ are determined by their components on each factor of the fibred product (pullback) it suffices to show $\epsilon\circ\beta w=\epsilon\circ u$ (which we already know) and $\delta\circ\beta w=\delta\circ u$. $\delta\beta w=\phi w=\gamma\alpha w$ is an arrow $Z\to C\times_Y C$ and this is further determined by its components on each factor.
Observe $q_1\phi w=cp_1 w=c\epsilon u=q_1\delta u$ and $q_2\gamma\alpha w=c\zeta\alpha w=c\zeta v=q_2\gamma v=q_2\delta u$ thus $\delta\circ\beta w$ and $\delta\circ u$ have the same $q_1,q_2$-components and are equal; thus $\beta w,u$ have the same $\epsilon,\delta$-components and are equal.
A similar chase shows $\alpha w=v$ as desired. I guess this is like a $2$-dimensional pullback lemma, not one which I've ever seen before. The top left corner is then a pullback, hence $\alpha,\beta$ are regular epimorphisms and $\phi$ is a composite of epimorphisms hence an epimorphism itself.