Homotopy pullback of finite groupoids

Identity pairs satisfy the compatibility equation. If $(u,v)$ and $(u^{\mathrm{\prime }},v^{\mathrm{\prime }})$ are composable compatible pairs, then $q(v^{\mathrm{\prime }}\circ v)\circ \alpha =q(v^{\mathrm{\prime }})\circ q(v)\circ \alpha =q(v^{\mathrm{\prime }})\circ {\alpha }^{\mathrm{\prime }}\circ p(u)={\alpha }^{\mathrm{\prime }\mathrm{\prime }}\circ p(u^{\mathrm{\prime }}\circ u)$, so their composite is compatible. Inverting the compatibility equation shows $(u^{-1},v^{-1})$ is compatible with the reversed objects. Thus the stated data form a groupoid.

Choose one representative from each of the finitely many isomorphism classes of $\mathcal{C}$ and $\mathcal{E}$. If $u:c\to c_0$ and $v:e\to e_0$ are isomorphisms to representatives, then $(c,e,\alpha )$ is isomorphic to $(c_0,e_0,q(v)\circ \alpha \circ p(u{)}^{-1})$. For each representative pair $(c_0,e_0)$, the possible isomorphisms $p(c_0)\cong q(e_0)$ are either absent or form a set on which postcomposition by ${\mathrm{Aut}}_{\mathcal{D}}(q(e_0))$ acts freely and transitively; that set has finite size by [F1]. Thus there are finitely many isomorphism classes of triples. The automorphism group of a triple is the subgroup of ${\mathrm{Aut}}_{\mathcal{C}}(c_0)\times {\mathrm{Aut}}_{\mathcal{E}}(e_0)$ satisfying the compatibility equation, and is finite by [F1]. Hence the homotopy pullback is a finite groupoid.