Are all split epimorphisms effective? An epimorphism is called split if it has a section (a right inverse).
An epimorphism is called effective if it has a kernel pair and is the coequalizer of its kernel pair.
The ncatlab implies here that every split epimorphism is effective, but I can't find a proof anywhere, nor have I had success in coming up with a proof or counterexample.
There are two parts to this. First, is it true that every split epimorphism has a kernel pair? I haven't managed to come up with an obvious construction of its kernel pair; what strikes me as the natural choice doesn't seem to work. (My inclination would be to try, if $f$ is a split epi with section $s$, the pair $(s \circ f, 1)$, because I think $f$ is definitely the coequalizer of this - but that doesn't seem to work.)
Second: if every split epimorphism does have a kernel pair, then is it further true that this kernel pair must have a coequalizer?
 A: While it definitely suggests it, I don't think the nLab page means that every split epimorphisms is an effective epimorphism but only every split epimorphism is a coequalizer of its kernel pair if it has one. For example, on the split epimorphism page it does not state that a split epimorphism is an effective epimorphism, only that it is a regular epimorphism. There is also this note (emphasis added):

Descent and effective descent morphisms are only defined in a category with pullbacks. The other notions can be defined in any category, although of course for an effective epimorphism one must in general assert the existence of the kernel pair.

Let $r : A \to B$ and $r\circ s = id$. Assume the kernel pair $(R,p,q)$ of $r$ exists, i.e. $$\require{AMScd}\begin{CD}R@>q>> A \\@VpVV@VVrV\\A@>>r> B\end{CD}$$ is a pullback square. We want to show that for any $h:A \to Z$ such that $h\circ p=h\circ q$, there exists a unique $t : B \to Z$ such that $t\circ r = h$. We can immediately see that $t$ can only be $h\circ s$, so it remains to show that $h\circ s\circ r = h$. Since $r\circ s\circ r = r$, we have that there exists a unique $k: B \to R$ such that $p\circ k = s\circ r$ and $q\circ k = id$. This means that $h\circ s \circ r = h\circ p \circ k = h\circ q \circ k = h$ as needed. $\square$
