The Stacks project

Lemma 21.38.5. Assumption and notation as in Situation 21.38.3.

  1. There are left adjoints $g_! : \textit{Mod}(\mathcal{O}_{\mathcal{C}'}) \to \textit{Mod}(\mathcal{O}_\mathcal {C})$ and $g_!^{\textit{Ab}} : \textit{Ab}(\mathcal{C}') \to \textit{Ab}(\mathcal{C})$ to $g^* = g^{-1}$ on modules and on abelian sheaves.

  2. The diagram

    \[ \xymatrix{ \textit{Mod}(\mathcal{O}_{\mathcal{C}'}) \ar[d] \ar[r]_{g_!} & \textit{Mod}(\mathcal{O}_\mathcal {C}) \ar[d] \\ \textit{Ab}(\mathcal{C}') \ar[r]^{g_!^{\textit{Ab}}} & \textit{Ab}(\mathcal{C}) } \]

    commutes.

  3. There are left adjoints $Lg_! : D(\mathcal{O}_{\mathcal{C}'}) \to D(\mathcal{O}_\mathcal {C})$ and $Lg_!^{\textit{Ab}} : D(\mathcal{C}') \to D(\mathcal{C})$ to $g^* = g^{-1}$ on derived categories of modules and abelian sheaves.

  4. The diagram

    \[ \xymatrix{ D(\mathcal{O}_{\mathcal{C}'}) \ar[d] \ar[r]_{Lg_!} & D(\mathcal{O}_\mathcal {C}) \ar[d] \\ D(\mathcal{C}') \ar[r]^{Lg_!^{\textit{Ab}}} & D(\mathcal{C}) } \]

    commutes.

Proof. The functor $u$ is continuous and cocontinuous Stacks, Lemma 8.10.3. Hence the existence of the functors $g_!$, $g_!^{\textit{Ab}}$, $Lg_!$, and $Lg_!^{\textit{Ab}}$ can be found in Modules on Sites, Sections 18.16 and 18.41 and Section 21.37.

To prove (2) it suffices to show that the canonical map

\[ g_!^{\textit{Ab}}j_{U'!}\mathcal{O}_{U'} \to j_{u(U')!}\mathcal{O}_{u(U')} \]

is an isomorphism for all objects $U'$ of $\mathcal{C}'$, see Modules on Sites, Remark 18.41.2. Similarly, to prove (4) it suffices to show that the canonical map

\[ Lg_!^{\textit{Ab}}j_{U'!}\mathcal{O}_{U'} \to j_{u(U')!}\mathcal{O}_{u(U')} \]

is an isomorphism in $D(\mathcal{C})$ for all objects $U'$ of $\mathcal{C}'$, see Remark 21.37.3. This will also imply the previous formula hence this is what we will show.

We will use that for a localization morphism $j$ the functors $j_!$ and $j_!^{\textit{Ab}}$ agree (see Modules on Sites, Remark 18.19.6) and that $j_!$ is exact (Modules on Sites, Lemma 18.19.3). Let us adopt the notation of Lemma 21.38.4. Since $Lg_!^{\textit{Ab}} \circ j_{U'!} = j_{U!} \circ L(g')^{\textit{Ab}}_!$ (by commutativity of Sites, Lemma 7.28.4 and uniqueness of adjoint functors) it suffices to prove that $L(g')^{\textit{Ab}}_!\mathcal{O}_{U'} = \mathcal{O}_ U$. Using the results of Lemma 21.38.4 we have for any object $E$ of $D(\mathcal{C}/u(U'))$ the following sequence of equalities

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{C}/U)}(L(g')_!^{\textit{Ab}}\mathcal{O}_{U'}, E) & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{C}'/U')}(\mathcal{O}_{U'}, (g')^{-1}E) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{C}'/U')}((\pi '_{U'})^{-1}\mathcal{O}_ V, (g')^{-1}E) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{D}/V)}(\mathcal{O}_ V, R\pi '_{U', *}(g')^{-1}E) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{D}/V)}(\mathcal{O}_ V, (\sigma ')^{-1}(g')^{-1}E) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{D}/V)}(\mathcal{O}_ V, \sigma ^{-1}E) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{D}/V)}(\mathcal{O}_ V, \pi _{U, *}E) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{C}/U)}(\pi _ U^{-1}\mathcal{O}_ V, E) \\ & = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{C}/U)}(\mathcal{O}_ U, E) \end{align*}

By Yoneda's lemma we conclude. $\square$


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 08PC. Beware of the difference between the letter 'O' and the digit '0'.