Item 1: The Beck–Chevalley Condition
We have
\begin{align*} \webleft [g^{-1}\circ f_{!}\webright ]\webleft (U\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}g^{-1}\webleft (f_{!}\webleft (U\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ y\in Y\ \middle |\ g\webleft (y\webright )\in f_{!}\webleft (U\webright )\right\} \\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{there exists some $x\in U$}\\ & \text{such that $f\webleft (x\webright )=g\webleft (y\webright )$}\end{aligned}\right\} \\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ x\in U\right\} $}\end{aligned}\right\} \\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ x\in U\right\} $}\\ & \text{such that $y=y$}\end{aligned}\right\} \\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ x\in U\right\} $}\\ & \text{such that $\operatorname {\mathrm{\mathrm{pr}}}_{2}\webleft (x,y\webright )=y$}\end{aligned}\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft (\operatorname {\mathrm{\mathrm{pr}}}_{2}\webright )_{!}\webleft (\left\{ \webleft (x,y\webright )\in X\times _{Z}Y\ \middle |\ x\in U\right\} \webright )\\ & = \webleft (\operatorname {\mathrm{\mathrm{pr}}}_{2}\webright )_{!}\webleft (\left\{ \webleft (x,y\webright )\in X\times _{Z}Y\ \middle |\ \operatorname {\mathrm{\mathrm{pr}}}_{1}\webleft (x,y\webright )\in U\right\} \webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft (\operatorname {\mathrm{\mathrm{pr}}}_{2}\webright )_{!}\webleft (\operatorname {\mathrm{\mathrm{pr}}}^{-1}_{1}\webleft (U\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\webleft (\operatorname {\mathrm{\mathrm{pr}}}_{2}\webright )_{!}\circ \operatorname {\mathrm{\mathrm{pr}}}^{-1}_{1}\webright ]\webleft (U\webright ) \end{align*}
for each $U\in \mathcal{P}\webleft (X\webright )$. Therefore, we have
\[ g^{-1}\circ f_{!}=\webleft (\operatorname {\mathrm{\mathrm{pr}}}_{2}\webright )_{!}\circ \operatorname {\mathrm{\mathrm{pr}}}^{-1}_{1}. \]
For the second equality, we have
\begin{align*} \webleft [f^{-1}\circ g_{!}\webright ]\webleft (U\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f^{-1}\webleft (g_{!}\webleft (U\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ f\webleft (x\webright )\in g_{!}\webleft (V\webright )\right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $y\in V$}\\ & \text{such that $f\webleft (x\webright )=g\webleft (y\webright )$}\end{aligned}\right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ y\in V\right\} $}\end{aligned}\right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ y\in V\right\} $}\\ & \text{such that $x=x$}\end{aligned}\right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ y\in V\right\} $}\\ & \text{such that $\operatorname {\mathrm{\mathrm{pr}}}_{1}\webleft (x,y\webright )=x$}\end{aligned}\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft (\operatorname {\mathrm{\mathrm{pr}}}_{1}\webright )_{!}\webleft (\left\{ \webleft (x,y\webright )\in X\times _{Z}Y\ \middle |\ y\in V\right\} \webright )\\ & = \webleft (\operatorname {\mathrm{\mathrm{pr}}}_{1}\webright )_{!}\webleft (\left\{ \webleft (x,y\webright )\in X\times _{Z}Y\ \middle |\ \operatorname {\mathrm{\mathrm{pr}}}_{2}\webleft (x,y\webright )\in V\right\} \webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft (\operatorname {\mathrm{\mathrm{pr}}}_{1}\webright )_{!}\webleft (\operatorname {\mathrm{\mathrm{pr}}}^{-1}_{2}\webleft (V\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\webleft (\operatorname {\mathrm{\mathrm{pr}}}_{1}\webright )_{!}\circ \operatorname {\mathrm{\mathrm{pr}}}^{-1}_{2}\webright ]\webleft (V\webright ) \end{align*}
for each $V\in \mathcal{P}\webleft (Y\webright )$. Therefore, we have
\[ f^{-1}\circ g_{!}=\webleft (\operatorname {\mathrm{\mathrm{pr}}}_{1}\webright )_{!}\circ \operatorname {\mathrm{\mathrm{pr}}}^{-1}_{2}. \]
This finishes the proof.
Item 2: The Projection Formula I
We claim that
\[ f_{!}\webleft (U\webright )\cap V\subset f_{!}\webleft (U\cap f^{-1}\webleft (V\webright )\webright ). \]
Indeed, we have
\begin{align*} f_{!}\webleft (U\webright )\cap V & \subset f_{!}\webleft (U\webright )\cap f_{!}\webleft (f^{-1}\webleft (V\webright )\webright )\\ & = f_{!}\webleft (U\cap f^{-1}\webleft (V\webright )\webright ), \end{align*}
where we have used:
-
1.
Item 2 of Proposition 4.6.1.1.5 for the inclusion.
-
2.
Item 6 of Proposition 4.6.1.1.5 for the equality.
Conversely, we claim that
\[ f_{!}\webleft (U\cap f^{-1}\webleft (V\webright )\webright )\subset f_{!}\webleft (U\webright )\cap V. \]
Indeed:
-
1.
Let $y\in f_{!}\webleft (U\cap f^{-1}\webleft (V\webright )\webright )$.
-
2.
Since $y\in f_{!}\webleft (U\cap f^{-1}\webleft (V\webright )\webright )$, there exists some $x\in U\cap f^{-1}\webleft (V\webright )$ such that $f\webleft (x\webright )=y$.
-
3.
Since $x\in U\cap f^{-1}\webleft (V\webright )$, we have $x\in U$, and thus $f\webleft (x\webright )\in f_{!}\webleft (U\webright )$.
-
4.
Since $x\in U\cap f^{-1}\webleft (V\webright )$, we have $x\in f^{-1}\webleft (V\webright )$, and thus $f\webleft (x\webright )\in V$.
-
5.
Since $f\webleft (x\webright )\in f_{!}\webleft (U\webright )$ and $f\webleft (x\webright )\in V$, we have $f\webleft (x\webright )\in f_{!}\webleft (U\webright )\cap V$.
-
6.
But $y=f\webleft (x\webright )$, so $y\in f_{!}\webleft (U\webright )\cap V$.
-
7.
Thus $f_{!}\webleft (U\cap f^{-1}\webleft (V\webright )\webright )\subset f_{!}\webleft (U\webright )\cap V$.
This finishes the proof.
Item 3: The Projection Formula II
We have
\begin{align*} f_{*}\webleft (U\webright )\cap V & \subset f_{*}\webleft (U\webright )\cap f_{*}\webleft (f^{-1}\webleft (V\webright )\webright )\\ & = f_{*}\webleft (U\cap f^{-1}\webleft (V\webright )\webright ), \end{align*}
where we have used:
-
1.
Item 2 of Proposition 4.6.3.1.7 for the inclusion.
-
2.
Item 6 of Proposition 4.6.3.1.7 for the equality.
Since $\mathcal{P}\webleft (Y\webright )$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2).
Item 4: Strong Closed Monoidality
This is a repetition of Item 19 of Proposition 4.4.7.1.3 and is proved there.
Item 5: The External Tensor Product
We have
\begin{align*} U\boxtimes _{X\times Y}V & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{\mathrm{pr}}}^{-1}_{1}\webleft (U\webright )\cap \operatorname {\mathrm{\mathrm{pr}}}^{-1}_{2}\webleft (V\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ \webleft (x,y\webright )\in X\times Y\ \middle |\ \operatorname {\mathrm{\mathrm{pr}}}_{1}\webleft (x,y\webright )\in U\right\} \\ & \phantom{={}} \mkern 4mu\cup \left\{ \webleft (x,y\webright )\in X\times Y\ \middle |\ \operatorname {\mathrm{\mathrm{pr}}}_{2}\webleft (x,y\webright )\in V\right\} \\ & = \left\{ \webleft (x,y\webright )\in X\times Y\ \middle |\ x\in U\right\} \\ & \phantom{={}} \mkern 4mu\cup \left\{ \webleft (x,y\webright )\in X\times Y\ \middle |\ y\in V\right\} \\ & = \left\{ \webleft (x,y\webright )\in X\times Y\ \middle |\ \text{$x\in U$ and $y\in V$}\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}U\times V. \end{align*}
Next, we claim that Item 5a, Item 5b, Item 5c, and Item 5d are indeed true: