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

\[ R_{-1}\colon \underbrace{\mathcal{P}\webleft (B\webright )}_{\cong \mathrm{Rel}\webleft (\text{pt},B\webright )}\to \underbrace{\mathcal{P}\webleft (A\webright )}_{\cong \mathrm{Rel}\webleft (\text{pt},A\webright )} \]

defined by

and being explicitly computed by

\begin{align*} R_{-1}\webleft (V\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\text{Rift}_{R}\webleft (V\webright )\\ & \cong \int _{x\in B}\textup{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{x}_{-_{1}},V^{x}_{-_{2}}\webright ),\end{align*}

where we have used Proposition 7.2.4.1.1.

We have

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

This finishes the proof.


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