Lemma 7.19.2. The functor ${}_ pu$ is a right adjoint to the functor $u^ p$. In other words the formula

holds bifunctorially in $\mathcal{F}$ and $\mathcal{G}$.

Lemma 7.19.2. The functor ${}_ pu$ is a right adjoint to the functor $u^ p$. In other words the formula

\[ \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(u^ p\mathcal{G}, \mathcal{F}) = \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{D})}(\mathcal{G}, {}_ pu\mathcal{F}) \]

holds bifunctorially in $\mathcal{F}$ and $\mathcal{G}$.

**Proof.**
This is proved in exactly the same way as the proof of Lemma 7.5.4. We note that the map $u^ p{}_ pu \mathcal{F} \to \mathcal{F}$ from Lemma 7.19.1 is the map that is used to go from the right to the left.

Alternately, think of a presheaf of sets $\mathcal{F}$ on $\mathcal{C}$ as a presheaf $\mathcal{F}'$ on $\mathcal{C}^{opp}$ with values in $\textit{Sets}^{opp}$, and similarly on $\mathcal{D}$. Check that $({}_ pu \mathcal{F})' = u_ p(\mathcal{F}')$, and that $(u^ p\mathcal{G})' = u^ p(\mathcal{G}')$. By Remark 7.5.5 we have the adjointness of $u_ p$ and $u^ p$ for presheaves with values in $\textit{Sets}^{opp}$. The result then follows formally from this. $\square$

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