Identifying subsets of $B$ with relations from $B$ to $\text{pt}$ via Chapter 2: Constructions With Sets, Item 3 of Proposition 2.4.3.1.6, we see that the weak inverse image function associated to $R$ is equivalently the function

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

defined by

\[ R^{-1}\webleft (V\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}V\mathbin {\diamond }R \]

for each $V\in \mathcal{P}\webleft (A\webright )$, where $R\mathbin {\diamond }V$ is the composition

\[ A\mathbin {\overset {R}{\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}}}B \mathbin {\overset {V}{\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}}}\text{pt}. \]

Explicitly, we have

\begin{align*} R^{-1}\webleft (V\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}V\mathbin {\diamond }R\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\int ^{b\in B}V^{-_{1}}_{b}\times R^{b}_{-_{2}}. \end{align*}

We have

\begin{align*} V\mathbin {\diamond }R& \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\int ^{b\in B}V^{-_{1}}_{b}\times R^{b}_{-_{2}}\\ & =\webleft\{ a\in A\ \middle |\ \int ^{b\in B}V^{\star }_{b}\times R^{b}_{a}=\mathsf{true}\webright\} \\ & = \webleft\{ a\in A\ \middle |\ \begin{aligned} & \text{there exists $b\in B$ such that the}\\[-2.5pt]& \text{following conditions hold:}\\[7.5pt]& \mspace {25mu}\rlap {\text{1.}}\mspace {22.5mu}\text{We have $V^{\star }_{b}=\mathsf{true}$}\\ & \mspace {25mu}\rlap {\text{2.}}\mspace {22.5mu}\text{We have $R^{b}_{a}=\mathsf{true}$}\\[10pt]\end{aligned} \webright\} \\ & = \webleft\{ a\in A\ \middle |\ \begin{aligned} & \text{there exists $b\in B$ such that the}\\[-2.5pt]& \text{following conditions hold:}\\[7.5pt]& \mspace {25mu}\rlap {\text{1.}}\mspace {22.5mu}\text{We have $b\in V$}\\ & \mspace {25mu}\rlap {\text{2.}}\mspace {22.5mu}\text{We have $b\in R\webleft (a\webright )$}\\[10pt]\end{aligned} \webright\} \\ & = \webleft\{ a\in A\ \middle |\ \text{there exists $b\in V$ such that $b\in R\webleft (a\webright )$}\webright\} \\ & = \webleft\{ a\in A\ \middle |\ R\webleft (a\webright )\cap V\neq \emptyset \webright\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R^{-1}\webleft (V\webright )\end{align*}

This finishes the proof.


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