= Solution
For $G:\mathcal D\to\mathcal C$, a <functor-split coequalizer pair> is a parallel pair $f,g:Y\rightrightarrows Z$ such that its image admits a <split coequalizer> in $\mathcal C$. In one orientation this means <morphisms> $q:GZ\to Q$, $s:Q\to GZ$ and $t:GZ\to GY$ satisfy
$$
qGf=qGg,\quad qs=1_Q,\quad (Gf)t=1_{GZ},\quad(Gg)t=sq.
$$
These equations prove that $q$ is a <coequalizer>: if $hGf=hGg$, then $h=hGf\,t=hGg\,t=hsq$, so $hs$ factors $h$; uniqueness follows from $qs=1$. They remain true under every <functor>, making a <split coequalizer> an absolute <colimit>.
To say that $G$ reflects such <coequalizers> means that any fork $Y\rightrightarrows Z\xrightarrow{p}W$ in $\mathcal D$ whose image is a <split coequalizer> is itself a <coequalizer> in $\mathcal D$. This is a reflection assertion about an existing fork; it does not by itself require lifting every split fork from $\mathcal C$, or creating its splitting maps in $\mathcal D$.
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