Local cartesian closure is equivalent to every pullback functor having a right adjoint

Assume condition (1), fix $f:X\to Y$, and put $\mathcal{K}=\mathcal{C}\mathrm{/}Y$. For an object $a:A\to X$ of $\mathcal{C}\mathrm{/}X$, regard $a$ as a morphism ${\mathrm{\Sigma }}_{f}a\to f$ in $\mathcal{K}$. By [L1] and [L3], the category $\mathcal{K}$ is cartesian closed and has pullbacks. Form in $\mathcal{K}$ the pullback ${\mathrm{\Pi }}_f(a):=({\mathrm{\Sigma }}_{f}a{)}^f{\times }_{f^f}\mathbf{1},$ where $({\mathrm{\Sigma }}_{f}a{)}^f\to f^f$ is induced by $a$ and $\mathbf{1}\to f^f$ is the transpose of $1_f:f\to f$. For $B\to Y$, currying identifies a map $B\to ({\mathrm{\Sigma }}_{f}a{)}^f$ with a map $h:B{\times }_{Y}X\to A$ over $Y$; the pullback equation defining ${\mathrm{\Pi }}_f(a)$ says exactly that $ah$ is the projection $B{\times }_{Y}X\to X$. Hence $(\mathcal{C}\mathrm{/}Y)(B,{\mathrm{\Pi }}_f(a))\cong (\mathcal{C}\mathrm{/}X)(f^{\ast }B,a)$ naturally in $B$ and $a$. Thus $f^{\ast }$ has right adjoint ${\mathrm{\Pi }}_f$.

Assume condition (2), and fix an object $X$. Chosen pullbacks give binary products in $\mathcal{C}\mathrm{/}X$, and $1_X:X\to X$ is terminal there. For $a:A\to X$, the pullback universal property gives ${\mathrm{\Sigma }}_a⊣a^{\ast }$, while condition (2) gives $a^{\ast }⊣{\mathrm{\Pi }}_a$. The product functor with $a$ on $\mathcal{C}\mathrm{/}X$ is the composite $-{\times }_{X}a={\mathrm{\Sigma }}_a{a}^{\ast }\mathrm{.}$ Consequently it has right adjoint ${\mathrm{\Pi }}_a{a}^{\ast }$. Since this holds for every $a$, the slice $\mathcal{C}\mathrm{/}X$ is cartesian closed.

Step 1.1 proves that condition (1) implies condition (2). Step 1.2 proves that each slice $\mathcal{C}\mathrm{/}X$ is cartesian closed, so condition (2) implies condition (1). Therefore the two conditions are equivalent.