My question is about this accepted answer from MathOverflow. Everything in it makes sense to me except for the following definition of a certain functor.
If $$ is a presheaf on $\int_{}{}$, then define a presheaf $$ on $$ as follows: For $\in$, let $()=\coprod_{\in()}(,)$.
I just can't see how this would act on morphisms in $C$. Maybe I am missing something obvious, but I'm not sure. Any help is much appreciated.
Write out what you have and see what you can do with it.
Given an arrow of $\mathcal C$, $f:X\to Y$, you want to construct a function $F(f):F(Y)\to F(X)$. $F(f)$ looks like $F(f)(s,g)=(s',g')$ where $s\in P(Y)$ and $g\in G(Y,s)$. We need to figure out a $s'\in P(X)$ and a $g'\in G(X,s')$. $P(f):P(Y)\to P(X)$ so that suggests $s'=P(f)(s)$. This would make $f$ exactly an arrow from $(Y,s')\to(X,s)$ which we can apply $G$ to to get $g'=G(f)(g)$. Putting it all together, $$F(f)(s, g)=(P(f)(s), G(f)(g))$$ Unsurprisingly, this is just the obvious thing. We have two contravariant functors, $P$ and $G$, and we want to produce a contravariant functor so we apply them component-wise. The only complexity here is the dependency of the second component on the first. In the case where we didn't have this dependency, $\coprod_{s\in P(X)}H(X)\cong P(X)\times H(X)$ for which the answer would have been very obvious.