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 $p_1,p_2 😛 \to A$ 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?)
Best Answer
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
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.
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
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.
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 $K[f]\rightrightarrows X$.