# Canonical isomorphism from coimage to image iff regular epi - regular mono factorization?

Let $\mathsf C$ be a finitely bicomplete category. Define the regular coimage of an arrow to be the coequalizer of its kernel pair, and dually the regular image to be the equalizer of the cokernel pair.

Universal properties induce a canonical arrow $\ell:\operatorname{Coim}f\to \operatorname{Im}f$ and a factorization $f=\operatorname{im}f\circ \ell\circ \operatorname{coim}f$.

In this note by Marta Bunge, proposition 0.8 states that in a finitely bicomplete category, TFAE.

1. The canonical arrow $\operatorname{Coim}f\to \operatorname{Im}f$ is an isomorphism.
2. Every arrow factors essentially uniquely as a regular epi followed by a regular mono.

Remark. I think the uniqueness in (2) follows from the fact regular epis/monos are strong epis/monos.

At any rate, (1)$\implies$(2) is clear, but I don't understand how to prove the converse. For instance, by assumption there's an essentially unique factorization $f=m\circ e$, but it's only uniquely isomorphic to the regular coimage/image factorization if we know apriori the canonical arrow is an iso.

I thought perhaps that $\operatorname{im}f\cong\operatorname{im}m$ and $\operatorname{coim}f\cong\operatorname{coim}e$. If so, I can use the fact an arrow is a regular epi/mono iff it's isomorphic to it's regular coimage/image which gives $\operatorname{im}m\cong m$ and $\operatorname{coim}e\cong e$. This would give some coherent isomorphism between the coimage and the image, and by universal properties is must be the only one. However Bunge doesn't write anything similar.

How to prove (2)$\implies$(1)?

If $f$ factors as $m\circ e$ with $m$ a regular mono and $e$ a regular epi, then the kernel pair of $f$ is also the kernel pair of $e$. Since every regular epimorphism is the coequalizer of its kernel pair, $e$ must be the regular image of $f$ and dually, $m$ must be its coimage. Hence the canonical arrow $l$ must be an isomorphism.
• You can prove it like this, but I think that the easiest way to prove it is to notice that $m$ being a mono implies that $f\circ u=f\circ v \Leftrightarrow e\circ u = e\circ v$, so the universal properties are exactly the same. – Arnaud D. May 16 '17 at 14:18
• That's neater. In particular all that matters is that $m$ is a mono - being strong is unnecessary and the properties of $e$ don't matter. – Arrow May 16 '17 at 14:19