Identifying subsets of $B$ with relations from $\text{pt}$ to $B$ via Chapter 2: Constructions With Sets, of , we see that the direct image with compact support function associated to $R$ is equivalently the function

\[ R_{!}\colon \underbrace{\mathcal{P}\webleft (A\webright )}_{\cong \mathrm{Rel}\webleft (A,\text{pt}\webright )}\to \underbrace{\mathcal{P}\webleft (B\webright )}_{\cong \mathrm{Rel}\webleft (B,\text{pt}\webright )} \]

defined by

being explicitly computed by

\begin{align*} R^{*}\webleft (U\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\text{Ran}_{R}\webleft (U\webright )\\ & \cong \int _{a\in A}\textup{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{-_{2}}_{a},U^{-_{1}}_{a}\webright ), \end{align*}

where we have used Proposition 7.2.3.1.1.

We have

\begin{align*} \text{Ran}_{R}\webleft (V\webright )& \cong \int _{a\in A}\textup{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{-_{2}}_{a},U^{-_{1}}_{a}\webright )\\ & =\webleft\{ b\in B\ \middle |\ \int _{a\in A}\textup{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{b}_{a},U^{\star }_{a}\webright )=\mathsf{true}\webright\} \\ & = \webleft\{ b\in B\ \middle |\ \begin{aligned} & \text{for each $a\in A$, at least one of the}\\[-2.5pt]& \text{following conditions hold:}\\[7.5pt]& \mspace {25mu}\rlap {\text{1.}}\mspace {22.5mu}\text{We have $R^{b}_{a}=\mathsf{false}$}\\ & \mspace {25mu}\rlap {\text{2.}}\mspace {22.5mu}\text{The following conditions hold:}\\[7.5pt]& \mspace {50mu}\rlap {\text{(a)}}\mspace {30mu}\text{We have $R^{b}_{a}=\mathsf{true}$}\\ & \mspace {50mu}\rlap {\text{(b)}}\mspace {30mu}\text{We have $U^{\star }_{a}=\mathsf{true}$}\\[10pt]\end{aligned} \webright\} \\ & = \webleft\{ b\in B\ \middle |\ \begin{aligned} & \text{for each $a\in A$, at least one of the}\\[-2.5pt]& \text{following conditions hold:}\\[7.5pt]& \mspace {25mu}\rlap {\text{1.}}\mspace {22.5mu}\text{We have $b\not\in R\webleft (A\webright )$}\\ & \mspace {25mu}\rlap {\text{2.}}\mspace {22.5mu}\text{The following conditions hold:}\\[7.5pt]& \mspace {50mu}\rlap {\text{(a)}}\mspace {30mu}\text{We have $b\in R\webleft (a\webright )$}\\ & \mspace {50mu}\rlap {\text{(b)}}\mspace {30mu}\text{We have $a\in U$}\\[10pt]\end{aligned} \webright\} \\ & = \webleft\{ b\in B\ \middle |\ \begin{aligned} & \text{for each $a\in A$, if we have}\\ & \text{$b\in R\webleft (a\webright )$, then $a\in U$}\end{aligned} \webright\} \\ & = \webleft\{ b\in B\ \middle |\ R^{-1}\webleft (b\webright )\subset U\webright\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R^{-1}\webleft (U\webright ).\end{align*}

This finishes the proof.


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