I'm reviewing some notes of mine on regular categories to try and get a better feel for regular, strong, and extremal epimorphisms, and this leads me to ask: is a category $\mathsf C$ regular if and only if $(\mathrm{ExtEpi},\mathrm{Mono})$ is a strong/orthogonal factorization system?

If anyone cares, here's my motivation.

In the category of sets, an image factorization of a function $f:A\to B$ is a composite $A\twoheadrightarrow \operatorname{Im}f\rightarrowtail B$ where $\operatorname{Im}f$ is in bijection with the set of fibers of $f$, the left arrow collapses fibers, and the right arrow takes a fiber to the value of $f$ on it. Thus image factorizations can be constructed functorially by taking for each function the coequalizer of its kernel pair - this collapses the fibers. Travelling further down this path leads to regular epis and the classical approach to regular categories.

Instead, we can define the image of an arrow $f$ (when it exists) to be the smallest subobject of its codomain through which $f$ factors. Often enough the first arrow in the factorization is an epi, so we define an extremal epi to be an arrow which doesn't factor through any proper subobjects of its codomain. Using this as a primitive notion, we can almost define the image of an arrow $f$ to be a subobject of its codomain which factors $f$ through an extremal epi. The only obstruction is the fact extremal epis are in general red herrings - they may not be epimorphisms. However, if $\mathsf C$ has equalizers, they are, and all is well.

Thus, in order for a category with equalizers $\mathsf C$ to have "image factorizations", it seems all we want is for extremal epimorphisms and monomorphisms to constitute a strong factorization system. The nlab writes:

A regular category is a finitely complete category which admits a good notion of image factorization

so tentatively I'm guessing that regular categories are precisely the finitely complete categories in which $(\mathrm{ExtEpi},\mathrm{Mono})$ is a strong/orthogonal factorization system. Is this correct?

Added. Just spilling some more thoughts. In general, if $\mathsf C$ admits intersections and pullbacks of monomorphisms, the prefactorization system $(\mathrm{StrEpi},\mathrm{Mono})$ is a factorization system. Moreover, in complete generality, strong epi implies extremal epi. Thus, if $(\mathrm{StrEpi},\mathrm{Mono})$ is already a strong factorization system, asking for $(\mathrm{StrEpi},\mathrm{Mono})$ to also be a strong factorization system amounts to asking every extremal epi to also be strong. What I'm having trouble seeing is how the notion of regular epi pops out here...


1 Answer 1


It is an old result of Joyal that a category is regular if and only if it is finitely complete and if the classes (strong epi, mono) form a stable (orthogonal) factorisation system, that is, if every morphism can be factorised as a strong epi followed by a mono, and the pullback of a strong epi is a strong epi. For one can show that under these hypotheses, every strong epi is the coequaliser of its kernel pair, and hence a regular epi. For proofs, see

Kelly, G. M. A note on relations relative to a factorization system. Category theory (Como, 1990), 249--261, Lecture Notes in Math., 1488, Springer, Berlin, 1991.

or Proposition A1.3.4 of Johnstone's Sketches of an elephant.

Also note that in a category with finite limits, the classes of strong epis and extremal epis coincide. A good reference for such matters is

Kelly, G. M. Monomorphisms, epimorphisms, and pull-backs. J. Austral. Math. Soc. 9 1969 124--142.

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    $\begingroup$ And if you want an example to show that stability is necessary, the category of topological spaces is a good one: the strong/extremal/regular epis are the topological quotient maps, which form a factorization system with the monomorphisms (arbitrary injective continuous maps) but are not stable under pullback. $\endgroup$ Oct 24, 2016 at 7:01

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