A categorical first isomorphism theorem It is known, that for a morphism of universal algebras $f : A \to B$, if $R$ is the congruence relation given by $xRy \Leftrightarrow fx=fy$, then $\operatorname{im} f \cong A/R $.
Here is an idea for a categorical version of this statement:

Let $f : A\to B$ be a morphism with a kernel pair $p_1,p_2 :P \to A$ and a coequalizer $e : A \to E$ of $p_1$ and $p_2$. Then there exists a unique morphism $m : E\to B$, such that $f = m\circ e$ and $m$ is an image of $f$ in this case.

Put simply: "$A/R = E \cong \operatorname{im} f$".
Such a morphism exists, as $e$ is a coequalizer and $f\circ p_1 = f\circ p_2$. If $m$ happend to be mono, then $m$ is an image of $f$, since $e$ is a regular epi and hence a strong epi.
What kind of reasonable properties should my ambient category have, such that $m$ is mono? Does this work for all regular categories or maybe even for all categories with (strong epi-mono)-factorizations or something like that? (Am I missing something simple?)
 A: 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 $K[f]\rightrightarrows X$.
A: Using the information given by Vej Kse in the answer he linked, I'll answer this question as follows:
Let $\mathcal{A}$ have kernel pairs and coequalizers of kernel pairs. Then every morphism $f$ in $\mathcal{A}$ has a regular coimage $\operatorname{coim}_\text{R} f$ and it is given by the coequalizer its kernel pair (as shown here).
If $\operatorname{coim}_\text{R} f$ is mono then we have $p_1 = p_2$ for the kernel pair $(p_1,p_2)$ of $f$, hence $f$ is mono.
The "first isomorphism theorem" now says: 

Every morphism $f$ factors as its regular coimage followed by a mono $m$.

As already mentoined, $m$ is then an image of $f$. So if the first isomorphism theorem holds, then images also exist in $\mathcal{A}$ and every morphism factors as its regular coimage followed by its image.
The following are equivalent (in $\mathcal{A}$):


*

*The first isomorphism theorem.

*Every morphism factors as a regular epi followed by a mono.

*Every extremal epi is regular.

*Every strong epi is regular.

*Regular epis are stable under composition.


Proof:


*

*$\Rightarrow$ 2.: Obvious.

*$\Rightarrow$ 3.: An extremal epi $e$ factors as $e = m'\circ e'$ with a regular epi $e'$ and a mono $m'$ then $m'$ iso, hence $e$ is a regular epi.

*$\Rightarrow$ 4.: Every strong epi is extremal.

*$\Rightarrow$ 5.: Strong and regular epis coincide and the former are stable under composition.

*$\Rightarrow$ 1.:
Let $f = m \circ \operatorname{coim}_\text{R} f$ and $m = m' \circ \operatorname{coim}_\text{R} m$. Then $e' := \operatorname{coim}_\text{R} m \circ \operatorname{coim}_\text{R} f$ is a regular epi (since they are stable under composition) and $f = m' \circ e'$. Because of the property regular coimages have, there is a morphism $k$ such that $\operatorname{coim}_\text{R} f = k\circ e'$. Then $k\circ \operatorname{coim}_\text{R} m = \operatorname{id}$, in particular $\operatorname{coim}_\text{R} m$ is mono, hence $m$ is mono. $\checkmark$

A remark:
The first isomorphism theorem holds in every regular category, even in the weak sense that:


*

*kernel pairs exist

*coequalizers of kernel pair exist

*the pullback of a regular epi along any morphism exists and is again a regular epi


This of course includes all categories of models of a Lawvere theory in $\mathsf{Set}$, including the category of sets, groups, rings, modules, etc., and all abelian categories.
