nLab
regular epimorphism

Idea

An epimorphism is regular if it behaves like a covering.

• effective epimorphism $\Rightarrow$ regular epimorphism $\iff$ covering

• effective monomorphism $\Rightarrow$ regular monomorphism $\iff$ embedding .

Definition

A regular epimorphism is a morphism $f:c\to d$ (in a given category) that is the coequalizer of some parallel pair of morphisms, i.e. if there exists some colimit diagram of the form

$$a\overrightarrow{\to }c\stackrel{f}{\to }d.$$a \stackrel{\to}{\to} c \stackrel{f}{\to} d \,.

The dual concept is that of regular monomorphism. A morphism having a kernel pair (such as any morphism in a category with pullbacks) is a regular epimorphism if and only if it is the quotient object of its kernel pair (see for instance Lemma 5.6.6 in Practical Foundations; this also follows from the theory of generalized kernels). In general, a regular epimorphism with a kernel pair is an effective epimorphism.

Although the definition doesn't state so explicitly, it is true that any regular epimorphism is an epimorphism. In fact, every regular epimorphism is a strong epimorphism. On the other hand, every split epimorphism is regular.

Examples

• In the category of sets, every epimorphism is regular. Thus, it can be hard to know, when generalising concepts from $Set$ to other categories, what kind of epimorphism to use. In particular, one may define a projective object (and hence the axiom of choice) using regular epimorphisms.

• In the category of monoids, the inclusion $\mathbb{N}\hookrightarrow \mathbb{Z}$ is an epimorphism, even though it is far from a surjection. But in this or any other algebraic category (a category of models of an algebraic theory), the morphisms whose underlying function is surjective are precisely the regular epimorphisms.

In the context of higher category theory

Comparing the above to the defintion of effective epimorphism, we propose the following definition in an (∞,1)-category:

A regular epimorphism $f:c\to d$ in an (∞,1)-category $C$ is a morphism with a simplicial resolution:

one for which there exists a functor $c:{\Delta }^{\mathrm{op}}\to C$, such that $f$ is the colimiting cocone

$$\cdots c_2\overrightarrow{\overrightarrow{\to }}{c}_1\overrightarrow{\to }c\stackrel{f}{\to }d$$\cdots c_2 \stackrel{\to}{\stackrel{\to}{\to}} c_1 \stackrel{\to}{\to} c \stackrel{f}{\to} d

over this diagram.

This is a morphism such that for all objects $e\in C$ the induced morphism $f^{*}:C(d,e)\to C(c,e)$ is a regular monomorphism in the (∞,1)-category ∞Grpd.

Compare this to the stronger notion of an effective ∞-epimorphism.

Warning. Such a morphism may fail to satisfy some condition for being a plain epimorphism in an (∞,1)-category that you might think of. The idea is that there may not be a good notion of epimorphism in an (∞,1)-category apart from regular epimorphism.

a previous version of this entry caused the following discussion

Mike Shulman: I would expect a “regular epimorphism” to be, in particular, an “epimorphism,” and I would expect an “epimorphism” $e:A\to B$ to mean that $\mathrm{hom}(e,X):\mathrm{hom}(B,X)\to \mathrm{hom}(A,X)$ is a “monomorphism” in some sense. But it’s unclear to me whether morphisms with a simplicial resolution have this property. In a 1-category, it’s true of course that if $e$ is a coequalizer then $\mathrm{hom}(e,X)$ is monic (injective), but in a (2,1)-category, all you can say if $e$ has a (truncated) simplicial resolution is that $\mathrm{hom}(e,X)$ is faithful (i.e. precomposing with $e$ is injective on the 2-cells of your (2,1)-category, but not necessarily any more than that). It seems fairly likely to me that as you go up to the $(\mathrm{\infty }\mathrm{,1})$-context the injectivity will be “pushed off to infinity” and there will be no monic-like property of $\mathrm{hom}(e,X)$ left.

I’m not necessarily saying we shouldn’t use the term “regular epimorphism” in this sense, since as we know the notion of “plain epimorphism” in 1-categories is not usually very useful anyway. But if I’m right, I think it would be good to alert the reader that the red herring principle is coming into play.

Urs Schreiber: following the discussion at the $n$Forum I put in the above warning.