The Stacks project

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00VE. Beware of the difference between the letter 'O' and the digit '0'.