Finite groupoid push-pull satisfies Fubini and adjunction

For fixed $z\in \mathcal{Z}$, an object of the iterated fiber is $(y,\beta :q(y)\to z,x,\alpha :p(x)\to y)$. The functor $(y,\beta ,x,\alpha )\mapsto (x,\beta \circ q(\alpha ))$ maps it to $(\mathcal{X}{)}_z$. A quasi-inverse sends $(x,\gamma :q(p(x))\to z)$ to $(p(x),\gamma ,x,{\mathrm{id}}_{p(x)})$. The composite on $(\mathcal{X}{)}_z$ is the identity. For any iterated object, the isomorphism $\alpha :p(x)\to y$ and ${\mathrm{id}}_x$ give a natural isomorphism from the quasi-inverse image back to that object. Hence the functor is an equivalence.

Fix an isomorphism class $[e]$ of $\mathcal{E}$, set $y=b(e)$, and write $A={\mathrm{Aut}}_{\mathcal{Y}}(y)$, $B={\mathrm{Aut}}_{\mathcal{E}}(e)$, $I=\mathrm{im}(B\to A)$, and $N=\ker (B\to A)$. The classes in the homotopy fiber ${\mathcal{E}}_y$ represented by this $[e]$ are the orbits of $I$ on the $A$-torsor of isomorphisms $b(e)\cong y$, so there are $\mathrm{\mid }A\mathrm{\mid }\mathrm{/}\mathrm{\mid }I\mathrm{\mid }$ of them; each has automorphism group $N$. Their contribution after the outer factor $1\mathrm{/}\mathrm{\mid }A\mathrm{\mid }$ is $f([e])\mathrm{/}(\mathrm{\mid }I\mathrm{\mid }\mathrm{\mid }N\mathrm{\mid })=f([e])\mathrm{/}\mathrm{\mid }B\mathrm{\mid }$, by the finite homomorphism counting identity $\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }I\mathrm{\mid }\mathrm{\mid }N\mathrm{\mid }$. Summing over $[e]$ proves ${\int }_{y\in \mathcal{Y}}{b}_{!}f(y)={\int }_{e\in \mathcal{E}}f(e)$.

Compose spans $\mathcal{X}\stackrel{a}{\leftarrow }\mathcal{E}\stackrel{b}{\to }\mathcal{Y}$ and $\mathcal{Y}\stackrel{c}{\leftarrow }\mathcal{F}\stackrel{d}{\to }\mathcal{Z}$, and let $P=\mathcal{E}{\times }_{\mathcal{Y}}^h\mathcal{F}$ with projections ${\pi }_E,{\pi }_F$. For fixed $z\in \mathcal{F}$, the homotopy fiber of ${\pi }_F$ over $z$ has objects $((e,f^{\mathrm{\prime }},\alpha ),\tau :f^{\mathrm{\prime }}\cong z)$. The functor to ${\mathcal{E}}_{c(z)}$ sends this object to $(e,c(\tau )\circ \alpha )$ and sends an arrow $(u,v)$ to $u$. Such an arrow satisfies ${\tau }^{\mathrm{\prime }}\circ v=\tau$ and $c(v)\circ \alpha ={\alpha }^{\mathrm{\prime }}\circ b(u)$, so its image satisfies ${\beta }^{\mathrm{\prime }}\circ b(u)=\beta$. Conversely, for any such $u$ between images $(e,\beta )$ and $(e^{\mathrm{\prime }},{\beta }^{\mathrm{\prime }})$, the only possible lift has $v=({\tau }^{\mathrm{\prime }}{)}^{-1}\circ \tau :f^{\mathrm{\prime }}\to f^{\mathrm{\prime }\mathrm{\prime }}$; the equation ${\beta }^{\mathrm{\prime }}\circ b(u)=\beta$ implies $c(v)\circ \alpha ={\alpha }^{\mathrm{\prime }}\circ b(u)$, so this lift exists and is unique. The functor is fully faithful; every object $(e,\beta )$ is its image from $((e,z,\beta ),{\mathrm{id}}_z)$, so it is essentially surjective. It is therefore an equivalence. By [F1], equivalent fibers have equal finite integrals, giving $c^{\ast }{b}_{!}g=({\pi }_F{)}_{!}{\pi }_E^{\ast }g$ for every invariant function $g$ on $\mathcal{E}$. Applying step 1.1 to ${\pi }_F$ and $d$ now gives $d_{!}{c}^{\ast }{b}_{!}{a}^{\ast }f=(d\circ {\pi }_F{)}_{!}(a\circ {\pi }_E{)}^{\ast }f$, which is the push-pull map of the homotopy-pullback composite span.

Put $\varphi ([e]):=\overline{f([a(e)])}g([b(e)])$, which is invariant on ${\pi }_0(\mathcal{E})$. Expanding the left inner product by the definition of $b_{!}$ gives $⟨b_{!}{a}^{\ast }f,g{⟩}_{\mathcal{Y}}={\sum }_{[y]}\frac{1}{\mathrm{\mid }{\mathrm{Aut}}_{\mathcal{Y}}(y)\mathrm{\mid }}{\sum }_{[(e,\eta )]\in {\pi }_0({\mathcal{E}}_y)}\frac{\overline{f([a(e)])}g([y])}{\mathrm{\mid }{\mathrm{Aut}}_{{\mathcal{E}}_y}(e,\eta )\mathrm{\mid }}$. Since $\eta :b(e)\cong y$ and $g$ is invariant, $g([y])=g([b(e)])$, so this is ${\int }_{y\in \mathcal{Y}}{b}_{!}\varphi (y)={\int }_{e\in \mathcal{E}}\varphi (e)={\sum }_{[e]}\frac{\varphi ([e])}{\mathrm{\mid }{\mathrm{Aut}}_{\mathcal{E}}(e)\mathrm{\mid }}$ by step 1.2. On the other side, expanding $a_{!}{b}^{\ast }g$ and using $\xi :a(e)\cong x$ to replace $\overline{f([x])}$ by $\overline{f([a(e)])}$ gives $⟨f,a_{!}{b}^{\ast }g{⟩}_{\mathcal{X}}={\sum }_{[x]}\frac{1}{\mathrm{\mid }{\mathrm{Aut}}_{\mathcal{X}}(x)\mathrm{\mid }}{\sum }_{[(e,\xi )]\in {\pi }_0({\mathcal{E}}_x)}\frac{\varphi ([e])}{\mathrm{\mid }{\mathrm{Aut}}_{{\mathcal{E}}_x}(e,\xi )\mathrm{\mid }}={\int }_{x\in \mathcal{X}}{a}_{!}\varphi (x)={\int }_{e\in \mathcal{E}}\varphi (e)$, again by step 1.2. Both inner products therefore equal the same finite weighted sum, proving the adjoint identity.

Step 2.2 proves that the adjoint of $b_{!}{a}^{\ast }$ is $a_{!}{b}^{\ast }$, and steps 1.1 and 2.1 prove composition through homotopy-pullback spans.