Equalizer of reflexive pair A reflexive pair is a pair of parallel morphisms $f,g:X\to Y$ having a common section, i.e. a map $s:Y\to X$ such that $f\circ s = g\circ s = id_Y$. 
Reflexive maps are famous because their coequalizers are reflexive coequalizers (see link above). 
My question is about the equalizer of a reflexive pair:

Is the equalizer of the pair above given by $(Y,s)$? It clearly makes the right diagram commute. But is it universal?

Mind that I am not talking about the dual of the reflexive equalizer (i.e. the coreflexive equalizer), I want the pair to be reflexive, not coreflexive.
 A: It is not necessarily universal. For example, if $f$ is a split epimorphism, with $s$ a section, then $f,f$ is a reflexive pair, but unless $f$ and $s$ are isomorphisms, $s$ is not the equalizer of $f$ and $f$, since the equalizer of two equal maps is just the identity.
There is however one notable case where $s$ is the equalizer of $f$ and $g$ : when $f$ and $g$ are jointly monic, so that they form a reflexive relation. Then if $f\circ t=g\circ t$, one can show that $t=s\circ f\circ t$. Indeed, it suffices, since $f$ and $g$ are jointly monic, to check that
$$f\circ s\circ f\circ t=f\circ t$$
and
$$g\circ s\circ f \circ t=f\circ t=g\circ t.$$
Moreover, such a factorisation is clearly unique, since $s$ is a (split) monomorphism.
A: Let $\mathbf{AG}$ denote the category of abelian groups.
Let $Y = \mathbb{Z}$, $X =  \mathbb{Z} \oplus \mathbb{Z}$, $f: X \to Y, f(z_1,z_2) = z_1$, $g = f$ and $s : Y \to X, s(z) = (z,0)$.
Then $s$ is a common section for $f, g$.
The equalizer of $f, g$ is the kernel of $f-g$. Since $f-g =0$ this kernel is the identity $id_X$. This shows that $s$ is not the equalizer of $f,g$.
