# In an algebraic category a morphism is a regular epi iff it is surjective

According to the nLab (see the 4th point under "Examples") in an "algebraic category" a morphism is a regular epi if and only, if it is surjective. Here a morphism $e$ is said to be surjective, if its underlying function $U(e)$ is.

There is a definition (or perhaps multiple definitions?) given for the term "algebraic category". I want to prove the above theorem, where I replace "is an algebraic category" by some "reasonable" and "reasonably weak assumptions" for a concrete category $(\mathcal{A}, U : \mathcal{A}\to \mathsf{Set})$.

Here are some ideas:

Let $e : A \to E$ be a regular epi and assume, that there is an object $S$ in $\mathcal{A}$ which is free on the singleton $1$, that is: There is a map $\eta : 1 \to U(S)$ such that for all objects $A$ in $\mathcal{A}$ and maps $y : 1 \to U(A)$, there is a unique morphism $\bar{y} : S \to A$ in $\mathcal{A}$, such that $U(\bar{y}) \circ \eta = y$. Let $y\in U(E)$. Then $y$ "is" a morphism $y : 1 \to U(E)$, hence there is a corresponding morphism $\bar{y} : S\to E$. Under certain assumptions (axiom of choice?) $S$ is in fact regularly projective, in particular there exists a morphism $x : S \to A$, such that $e\circ x = \bar{y}$. Then $U(x) \circ \eta : 1\to U(A)$ "is" an element of $U(A)$ and furthermore $U(e)\circ (U(x) \circ \eta) = U(e\circ x)\circ \eta = U(\bar{y})\circ \eta = y$, hence $U(e)$ is surjective.

Conversily, let $U(e)$ be surjective. Suppose $e$ has a kernel pair $(P,p_1,p_2)$ and that $U$ preserves limits ($U$ is represented by $S$). Let $q : A \to Q$ be such that $q\circ p_1 = q\circ p_2$. Then $(U(P),U(p_1),U(p_2))$ is a kernel pair of $U(e)$, because $U$ preserves kernel pairs. Furthermore $U(e)$ is surjective, hence a regular epi and therefore the coequalizer of $U(p_1),U(p_2)$, and $U(q)\circ U(p_1) = U(q)\circ U(p_2)$, so there exists a unique map $\tilde{u} : U(E) \to U(Q)$, such that $\tilde{u} \circ U(e) = U(q)$. A morphism $u : E \to Q$ with $u\circ e = q$ is unique, since then $U(u) = \tilde{u}$. Now showing that a $u$ with $U(u) = \tilde{u}$ exists remains.

How can I fill the holes of this proof idea to make it an actual proof?

A part of this question is, what assumption I really need, and in particular the following things are unclear:

• if or when $S$ is regularly projective and whether I actually need that
• if or when the $\tilde{u}$ can be extended to a morphism $u$

Any alternative suggestions are welcome as well.

A reasonable set of assumptions is that

1. $U$ is representable by an object $s$ (the free object on one element),
2. $U$ is conservative, and
3. $A$ has, and $U$ preserves, reflexive coequalizers.

This is notably the case if $A$ is the category of models of a Lawvere theory, or somewhat more generally the category of algebras of a monad preserving reflexive coequalizers. Conversely, by the crude monadicity theorem, with the above hypotheses (together with the hypothesis that $A$ has all coproducts), the functor $U$ is monadic, and the monad preserves reflexive coequalizers.

An example where the above hypotheses don't hold is $A = \text{Top}$, where the only hypothesis that fails is conservativity. Here the regular epis are the quotient maps, but there are many surjective continuous maps that aren't quotient maps.

From here the important technical fact is the following.

Lemma: A conservative functor reflects any limits or colimits it preserves.

Hence, with the above hypotheses, $U$ preserves and reflects kernel pairs and reflexive coequalizers, which implies that it both preserves and reflects regular epimorphisms (because the coequalizer of a kernel pair is always reflexive).

Of course to apply this statement to a familiar case like groups you need a way of verifying that $U$ preserves reflexive coequalizers. A more general result you can try to prove is that the forgetful functor from models of a Lawvere theory to $\text{Set}$ preserves sifted colimits (basically because these commute with finite products in $\text{Set}$).

As far as regular projectivity goes, $s$ is regular projective if and only if $U$ preserves regular epimorphisms more or less by definition, so that's just a restatement of the question in one direction.

If you want a reference, Corollary 3.5.3 in Borceux Volume II covers the case of Lawvere theories.

• Nice answer. Out of curiosity: Do you know what (again: "reasonable") conditions a category $\mathcal{X}$ (in place of $\mathsf{Set}$) needs to have, such that the forgetful functor $U$ from a category of models in $\mathcal{X}$ of some Lawvere theory, is conservative and preserves sifted colimits? (that is: what kind of category should $\mathcal{X}$ be, such that this theorem still "works"?) – Stefan Perko May 29 '16 at 18:26
• @Stefan: off the top of my head, I would guess you need sifted colimits and finite products to commute in $X$. This is the case e.g. if $X$ is itself the category of models of a Lawvere theory in $\text{Set}$. – Qiaochu Yuan May 29 '16 at 18:29