Two representatives in a filtered colimit of sets are equal exactly when they become equal at one common later stage

On tagged elements define $(j,x)\approx (k,y)$ when some $j\stackrel{a}{\to }\mathrm{\ell }\stackrel{b}{\leftarrow }k$ satisfies $D(a)x=D(b)y$. Identity arrows prove reflexivity, and exchanging $a,b$ proves symmetry.

For transitivity, suppose the first equality is witnessed in $\mathrm{\ell }$ by $a:j\to \mathrm{\ell }$, $b:k\to \mathrm{\ell }$, and the second in $m$ by $c:k\to m$, $d:n\to m$. Choose $r:\mathrm{\ell }\to p$ and $s:m\to p$ by [F1]. Then $rb,sc:k\rightrightarrows p$ are parallel, so choose $t:p\to q$ with $trb=tsc$. Applying $D$ shows $tra(x)=tsd(z)$, proving $(j,x)\approx (n,z)$.

Thus $\approx$ is an equivalence relation. It contains every generating pair $(j,x)$ and $(k,D(u)x)$ by choosing $\mathrm{\ell }=k$, $a=u$, and $b=1_k$. Conversely, a witness $a,b$ gives a chain of two generating identifications from $(j,x)$ and $(k,y)$ to their equal tagged element in $D(\mathrm{\ell })$. Hence $\approx$ is exactly the equivalence relation in [L1].

By [L1], equality of the two colimit classes is equivalence under that relation. Step 2.1 identifies this with the existence of the displayed common stage, proving both directions of the biconditional.