$g\leq e$. Then ${\mathbf d}(c)=g$. This and $c\leq t$ yield that $c=tg$. It follows that there is an arrow $$(g,x)\stackrel{(e,g)}{\longrightarrow} (e,x)$$ in the category $\int_{L(S)}\Phi(X,\mu)$. Since $(f,s)(e,g)=(f,c)=(f,t)(e,g)$ in $L(S)$, the diagram $$(g,x) \stackrel{(e,g)}{\longrightarrow}(e,x) {\mathrel{
  \settowidth{\@tempdima}{$\scriptstyle(f,s)$}
  \settowidth{\@tempdimb}{$\scriptstyle(f,t)$}
  \ifdim\@tempdimb>\@tempdima \@tempdima=\@tempdimb\fi
  \mathop{\vcenter{
    \offinterlineskip\ialign{\hbox to\dimexpr\@tempdima+2em{##}\cr
    \rightarrowfill\cr\noalign{\kern.3ex}
    \rightarrowfill\cr}}}\limits^{\!(f,s)}_{\!(f,t)}}} (f,y)$$ is commutative. Therefore, the functor $\Phi(X,\mu)$ satisfies axiom (F3) from the definition of a filtered functor (see Subsection \[subs:2.5\]). It satisfies (F1) and (F2) due to Proposition \[prop:trans\], since universal $S$-sets are transitive.

Conversely, let $(X,\mu)$ be an $S$-set such that the functor $\Phi(X,\mu)$ is filtered. Assume that $s,t\in S$ and $x\in X$ are such that $s\cdot x=t\cdot x$. Let $e={\mathbf d}(s){\mathbf d}(t)$ and $h={\mathbf r}(s){\mathbf r}(t)$. Then $hse\cdot x=hte\cdot x$ and also ${\mathbf d}(hse)={\mathbf d}(hte)$, ${\mathbf r}(hse)={\mathbf r}(hte)$. We put $p={\mathbf d}(hse)$ and $q={\mathbf r}(hse)$. It follows that in the category $\int_{L(S)}\Phi(X,\mu)$ we have two parallel arrows $$(p,x)  {\mathrel{
  \settowidth{\@tempdima}{$\scriptstyle(q,hse)$}
  \settowidth{\@tempdimb}{$\scriptstyle(q,hte)$}
  \ifdim\@tempdimb>\@tempdima \@tempdima=\@tempdimb\fi
  \mathop{\vcenter{
    \offinterlineskip\ialign{\hbox to\dimexpr\@tempdima+2em{##}\cr
    \rightarrowfill\cr\noalign{\kern.3ex}
    \rightarrowfill\cr}}}\limits^{\!(q,hse)}_{\!(q,hte)}}} (q,y).$$ By axiom (F3), there is a commutative diagram $$(r,z) \stackrel{(p,a)}\longrightarrow (p,x)  {\mathrel{
  \settowidth{\@tempdima}{$\scriptstyle(q,hse)$}
  \settowidth{\@tempdimb}{$\scriptstyle(q,hte)$}
  \ifdim\@tempdimb>\@tempdima \@tempdima=\@tempdimb\fi
  \mathop{\vcenter{
    \of