Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 20 / 11 / i / 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 $t:GZ\to GY$ satisfyThese equations prove that $q$ is a coequalizer: if $hGf=hGg$, then $h=hGft=hGgt=hsq$, 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 $Y\rightrightarrows Z\stackrel{p}{\to }W$ 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 $\mathcal{D}$.