That is, unpacking the definitions, are all morphisms $f : A \rightarrow B$ with retraction $r : B \rightarrow A$ the equalizer of some morphism $k : B \rightarrow C$ with the unique zero morphism $0_{BC} : B \rightarrow C$?

The ncatlab says yes [edit: apparently no it doesn't, as pointed out below], as they assert that the types of mono and epi form a total order with no qualifiers, but I haven't seen any proof of this - which is weird, because it seems an essential thing to prove (as proving that split monos are all normal would prove that they are also regular, strong, extremal, etc). Nor is a proof completely obvious to me - the proof that all split monos are regular involves proving that they equalize $f \circ r$ with the identity $1_B$, but both of these seem far from being zero morphisms.

I suspect that the solution might be that if $f \circ r$ is not the identity map in a category with zero morphisms, i.e. $f$ is not an isomorphism, then it necessarily follows that it must be a zero morphism itself, but I haven't had any luck proving that or coming up with counterexamples. It's clear to me that if $f$ is not an isomorphism, then $ f \circ r$ cannot be a monomorphism or an epimorphism either, which is suggestive.

  • 1
    $\begingroup$ The nLab page you linked to says nothing about normal epis. $\endgroup$
    – Arnaud D.
    Feb 25 '19 at 13:19
  • $\begingroup$ That's true, it doesn't actually. They're mentioned on the page monomorphisms, but that page doesn't assert that all the variations are totally ordered. Still, it would be quite an omission to not mention that they are not totally ordered on the one page, and to not mention the one kind of epimorphism that doesn't fall under the total order on the other page. $\endgroup$ Feb 25 '19 at 13:22

It is not true. For example, in the category of groups normal monomorphisms are just injective morphisms with normal image. Now if $G$ is a nonabelian group, the diagonal $$\delta:G\to G\times G:g\mapsto (g,g)$$ is obviously a split monomorphism, but its image is not a normal subgroup of $G\times G$, so it is not a normal monomorphism.

  • $\begingroup$ Thank you! I guess I had assumed that this was the case because on ncatlab they didn't explicitly mention that it wasn't, because every other type of mono/epi seems to fall into this pattern, and I thought it would be very strange for ncatlab to leave out the one kind of epi/mono that defies the standard ordering. I read between the lines a bit too much, then. On the meta level, I guess that it makes sense to immediately reach for non-Abelian groups when trying to come up with a counterexample, as Abelian categories are (bi)normal but Grp isn't. $\endgroup$ Feb 25 '19 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.