Regular coimages in a finitely complete and finitely cocomplete category According to the nLab, in a finitely complete and finitely cocomplete category $\mathcal{A}$, for every morphism $f : A \to B$ a regular coimage exists and is given by the coequalizer of $f$'s kernel pair. 
A regular coimage of $f$ is (if I am not mistaken) a regular epi $e : A\to E$ through which $f$ factors by some morphism $m$, such that for all regular epis $e' : A\to E'$ through which $f$ factors by a morphism $m'$, there is a (necessarily unique) morphism $k : E'\to E$, such that $k\circ e' = e$ and $m\circ k = m'$.
Even with the proof sketch (of the dual statement) given on the nLab I do not now how to show this. I have two questions, which can be answered simultaneously:


*

*How do I prove this statement?

*What is a "minimal" assumption one has to make with regard to the existence of limits and colimits in $\mathcal{A}$? In particular: Does it suffice, that only kernel pairs and coequalizers of kernel pairs exist in $\mathcal{A}$?


Let $p_1, p_2: A\times_B A\rightrightarrows A$, which exist because the pullback is a finite limit. Let $A\to E:=\text{coker}(p_1,p_2)$, which exists as the colimit of the finite diagram $(p_1,p_2)$ above. By definition, $fp_1=fp_2$, so $f$ factors as $A\to E\xrightarrow{f^\prime} B$ for some $f^\prime$. 
Now suppose that $g_1, g_2: C\rightrightarrows A\xrightarrow{q} F$ is another regular cokernel diagram such that $f$ factors as $A\to F\xrightarrow{a} B$. Then $q g_1 = qg_2$ implies $fg_1 = aqg_1 = aqg_2 = fg_2$, so $(g_1,g_2)$ gives rise to a morphism $C\to A\times_B A$ which intertwines between $(g_1,g_2)$ and $(p_1,p_2)$. By the functoriality of colimits, this yields a morphism $F\to E$ which satisfies the desired compatibilities.
