Lemma 7.5.4. The functor $u_ p$ is a left adjoint to the functor $u^ p$. In other words the formula

$\mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(\mathcal{F}, u^ p\mathcal{G}) = \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{D})}(u_ p\mathcal{F}, \mathcal{G})$

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

Proof. Let $\mathcal{G}$ be a presheaf on $\mathcal{D}$ and let $\mathcal{F}$ be a presheaf on $\mathcal{C}$. We will show that the displayed formula holds by constructing maps either way. We will leave it to the reader to verify they are each others inverse.

Given a map $\alpha : u_ p \mathcal{F} \to \mathcal{G}$ we get $u^ p\alpha : u^ p u_ p \mathcal{F} \to u^ p \mathcal{G}$. Lemma 7.5.3 says that there is a map $\mathcal{F} \to u^ p u_ p \mathcal{F}$. The composition of the two gives the desired map. (The good thing about this construction is that it is clearly functorial in everything in sight.)

Conversely, given a map $\beta : \mathcal{F} \to u^ p\mathcal{G}$ we get a map $u_ p\beta : u_ p\mathcal{F} \to u_ p u^ p\mathcal{G}$. We claim that the functor $u^ p\mathcal{G}_ Y$ on $\mathcal{I}_ Y$ has a canonical map to the constant functor with value $\mathcal{G}(Y)$. Namely, for every object $(X, \phi )$ of $\mathcal{I}_ Y$, the value of $u^ p\mathcal{G}_ Y$ on this object is $\mathcal{G}(u(X))$ which maps to $\mathcal{G}(Y)$ by $\mathcal{G}(\phi ) = \phi ^*$. This is a transformation of functors because $\mathcal{G}$ is a functor itself. This leads to a map $u_ p u^ p \mathcal{G}(Y) \to \mathcal{G}(Y)$. Another trivial verification shows that this is functorial in $Y$ leading to a map of presheaves $u_ p u^ p \mathcal{G} \to \mathcal{G}$. The composition $u_ p\mathcal{F} \to u_ p u^ p\mathcal{G} \to \mathcal{G}$ is the desired map. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).