The Stacks project

Lemma 21.37.2. Let $u : \mathcal{C} \to \mathcal{D}$ be a continuous and cocontinuous functor of sites. Let $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ be the corresponding morphism of topoi. Let $\mathcal{O}_\mathcal {D}$ be a sheaf of rings and set $\mathcal{O}_\mathcal {C} = g^{-1}\mathcal{O}_\mathcal {D}$. The functor $g_! : \textit{Mod}(\mathcal{O}_\mathcal {C}) \to \textit{Mod}(\mathcal{O}_\mathcal {D})$ (see Modules on Sites, Lemma 18.41.1) has a left derived functor

\[ Lg_! : D(\mathcal{O}_\mathcal {C}) \longrightarrow D(\mathcal{O}_\mathcal {D}) \]

which is left adjoint to $g^*$. Moreover, for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we have

\[ Lg_!(j_{U!}\mathcal{O}_ U) = g_!j_{U!}\mathcal{O}_ U = j_{u(U)!} \mathcal{O}_{u(U)}. \]

where $j_{U!}$ and $j_{u(U)!}$ are extension by zero associated to the localization morphism $j_ U : \mathcal{C}/U \to \mathcal{C}$ and $j_{u(U)} : \mathcal{D}/u(U) \to \mathcal{D}$.

Proof. We are going to use Derived Categories, Proposition 13.29.2 to construct $Lg_!$. To do this we have to verify assumptions (1), (2), (3), (4), and (5) of that proposition. First, since $g_!$ is a left adjoint we see that it is right exact and commutes with all colimits, so (5) holds. Conditions (3) and (4) hold because the category of modules on a ringed site is a Grothendieck abelian category. Let $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\textit{Mod}(\mathcal{O}_\mathcal {C}))$ be the collection of $\mathcal{O}_\mathcal {C}$-modules which are direct sums of modules of the form $j_{U!}\mathcal{O}_ U$. Note that $g_!j_{U!}\mathcal{O}_ U = j_{u(U)!} \mathcal{O}_{u(U)}$, see proof of Modules on Sites, Lemma 18.41.1. Every $\mathcal{O}_\mathcal {C}$-module is a quotient of an object of $\mathcal{P}$, see Modules on Sites, Lemma 18.28.8. Thus (1) holds. Finally, we have to prove (2). Let $\mathcal{K}^\bullet $ be a bounded above acyclic complex of $\mathcal{O}_\mathcal {C}$-modules with $\mathcal{K}^ n \in \mathcal{P}$ for all $n$. We have to show that $g_!\mathcal{K}^\bullet $ is exact. To do this it suffices to show, for every injective $\mathcal{O}_\mathcal {D}$-module $\mathcal{I}$ that

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_\mathcal {D})}( g_!\mathcal{K}^\bullet , \mathcal{I}[n]) = 0 \]

for all $n \in \mathbf{Z}$. Since $\mathcal{I}$ is injective we have

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_\mathcal {D})}( g_!\mathcal{K}^\bullet , \mathcal{I}[n]) & = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{O}_\mathcal {D})}( g_!\mathcal{K}^\bullet , \mathcal{I}[n]) \\ & = H^ n(\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_\mathcal {D}}( g_!\mathcal{K}^\bullet , \mathcal{I})) \\ & = H^ n(\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_\mathcal {C}}( \mathcal{K}^\bullet , g^{-1}\mathcal{I})) \end{align*}

the last equality by the adjointness of $g_!$ and $g^{-1}$.

The vanishing of this group would be clear if $g^{-1}\mathcal{I}$ were an injective $\mathcal{O}_\mathcal {C}$-module. But $g^{-1}\mathcal{I}$ isn't necessarily an injective $\mathcal{O}_\mathcal {C}$-module as $g_!$ isn't exact in general. We do know that

\[ \mathop{\mathrm{Ext}}\nolimits ^ p_{\mathcal{O}_\mathcal {C}}( j_{U!}\mathcal{O}_ U, g^{-1}\mathcal{I}) = H^ p(U, g^{-1}\mathcal{I}) = 0 \text{ for }p \geq 1 \]

Here the first equality follows from $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_\mathcal {C}}(j_{U!}\mathcal{O}_ U, \mathcal{H}) = \mathcal{H}(U)$ and taking derived functors and the vanishing of $H^ p(U, g^{-1}\mathcal{I})$ for $p > 0$ and $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ follows from Lemma 21.37.1. Since each $\mathcal{K}^{-q}$ is a direct sum of modules of the form $j_{U!}\mathcal{O}_ U$ we see that

\[ \mathop{\mathrm{Ext}}\nolimits ^ p_{\mathcal{O}_\mathcal {C}}(\mathcal{K}^{-q}, g^{-1}\mathcal{I}) = 0 \text{ for }p \geq 1\text{ and all }q \]

Let us use the spectral sequence (see Example 21.32.1)

\[ E_1^{p, q} = \mathop{\mathrm{Ext}}\nolimits ^ p_{\mathcal{O}_\mathcal {C}}( \mathcal{K}^{-q}, g^{-1}\mathcal{I}) \Rightarrow \mathop{\mathrm{Ext}}\nolimits ^{p + q}_{\mathcal{O}_\mathcal {C}}( \mathcal{K}^\bullet , g^{-1}\mathcal{I}) = 0. \]

Note that the spectral sequence abuts to zero as $\mathcal{K}^\bullet $ is acyclic (hence vanishes in the derived category, hence produces vanishing ext groups). By the vanishing of higher exts proved above the only nonzero terms on the $E_1$ page are the terms $E_1^{0, q} = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_\mathcal {C}}( \mathcal{K}^{-q}, g^{-1}\mathcal{I})$. We conclude that the complex $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_\mathcal {C}}( \mathcal{K}^\bullet , g^{-1}\mathcal{I})$ is acyclic as desired.

Thus the left derived functor $Lg_!$ exists. It is left adjoint to $g^{-1} = g^* = Rg^* = Lg^*$, i.e., we have

21.37.2.1
\begin{equation} \label{sites-cohomology-equation-to-prove} \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_\mathcal {C})}(K, g^*L) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_\mathcal {D})}(Lg_!K, L) \end{equation}

by Derived Categories, Lemma 13.30.3. This finishes the proof. $\square$


Comments (3)

Comment #5430 by CQ on

The equation is due to the definition of , right? I couldn't find reference for this.

Comment #5657 by on

Yes! Because no tensor product is needed when you pullback modules. This of course then also means that pullback is exact on the category of modules and you get the statements on the derived versions too.

Comment #8654 by ZL on

There seems to be typos on the superscript for the -spectral sequence: "" should be "" Similarly the non-zero terms in the -page should be .

There are also:

  • 3 comment(s) on Section 21.37: Derived lower shriek

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