The Stacks project

59.49 Exactness of big lower shriek

This is just the following technical result. Note that the functor $f_{big!}$ has nothing whatsoever to do with cohomology with compact support in general.

Lemma 59.49.1. Let $\tau \in \{ Zariski, {\acute{e}tale}, smooth, syntomic, fppf\} $. Let $f : X \to Y$ be a morphism of schemes. Let

\[ f_{big} : \mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/X)_\tau ) \longrightarrow \mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/Y)_\tau ) \]

be the corresponding morphism of topoi as in Topologies, Lemma 34.3.16, 34.4.16, 34.5.10, 34.6.10, or 34.7.12.

  1. The functor $f_{big}^{-1} : \textit{Ab}((\mathit{Sch}/Y)_\tau ) \to \textit{Ab}((\mathit{Sch}/X)_\tau )$ has a left adjoint

    \[ f_{big!} : \textit{Ab}((\mathit{Sch}/X)_\tau ) \to \textit{Ab}((\mathit{Sch}/Y)_\tau ) \]

    which is exact.

  2. The functor $f_{big}^* : \textit{Mod}((\mathit{Sch}/Y)_\tau , \mathcal{O}) \to \textit{Mod}((\mathit{Sch}/X)_\tau , \mathcal{O})$ has a left adjoint

    \[ f_{big!} : \textit{Mod}((\mathit{Sch}/X)_\tau , \mathcal{O}) \to \textit{Mod}((\mathit{Sch}/Y)_\tau , \mathcal{O}) \]

    which is exact.

Moreover, the two functors $f_{big!}$ agree on underlying sheaves of abelian groups.

Proof. Recall that $f_{big}$ is the morphism of topoi associated to the continuous and cocontinuous functor $u : (\mathit{Sch}/X)_\tau \to (\mathit{Sch}/Y)_\tau $, $U/X \mapsto U/Y$. Moreover, we have $f_{big}^{-1}\mathcal{O} = \mathcal{O}$. Hence the existence of $f_{big!}$ follows from Modules on Sites, Lemma 18.16.2, respectively Modules on Sites, Lemma 18.41.1. Note that if $U$ is an object of $(\mathit{Sch}/X)_\tau $ then the functor $u$ induces an equivalence of categories

\[ u' : (\mathit{Sch}/X)_\tau /U \longrightarrow (\mathit{Sch}/Y)_\tau /U \]

because both sides of the arrow are equal to $(\mathit{Sch}/U)_\tau $. Hence the agreement of $f_{big!}$ on underlying abelian sheaves follows from the discussion in Modules on Sites, Remark 18.41.2. The exactness of $f_{big!}$ follows from Modules on Sites, Lemma 18.16.3 as the functor $u$ above which commutes with fibre products and equalizers. $\square$

Next, we prove a technical lemma that will be useful later when comparing sheaves of modules on different sites associated to algebraic stacks.

Lemma 59.49.2. Let $X$ be a scheme. Let $\tau \in \{ Zariski, {\acute{e}tale}, smooth, syntomic, fppf\} $. Let $\mathcal{C}_1 \subset \mathcal{C}_2 \subset (\mathit{Sch}/X)_\tau $ be full subcategories with the following properties:

  1. For an object $U/X$ of $\mathcal{C}_ t$,

    1. if $\{ U_ i \to U\} $ is a covering of $(\mathit{Sch}/X)_\tau $, then $U_ i/X$ is an object of $\mathcal{C}_ t$,

    2. $U \times \mathbf{A}^1/X$ is an object of $\mathcal{C}_ t$.

  2. $X/X$ is an object of $\mathcal{C}_ t$.

We endow $\mathcal{C}_ t$ with the structure of a site whose coverings are exactly those coverings $\{ U_ i \to U\} $ of $(\mathit{Sch}/X)_\tau $ with $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ t)$. Then

  1. The functor $\mathcal{C}_1 \to \mathcal{C}_2$ is fully faithful, continuous, and cocontinuous.

Denote $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_1) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_2)$ the corresponding morphism of topoi. Denote $\mathcal{O}_ t$ the restriction of $\mathcal{O}$ to $\mathcal{C}_ t$. Denote $g_!$ the functor of Modules on Sites, Definition 18.16.1.

  1. The canonical map $g_!\mathcal{O}_1 \to \mathcal{O}_2$ is an isomorphism.

Proof. Assertion (a) is immediate from the definitions. In this proof all schemes are schemes over $X$ and all morphisms of schemes are morphisms of schemes over $X$. Note that $g^{-1}$ is given by restriction, so that for an object $U$ of $\mathcal{C}_1$ we have $\mathcal{O}_1(U) = \mathcal{O}_2(U) = \mathcal{O}(U)$. Recall that $g_!\mathcal{O}_1$ is the sheaf associated to the presheaf $g_{p!}\mathcal{O}_1$ which associates to $V$ in $\mathcal{C}_2$ the group

\[ \mathop{\mathrm{colim}}\nolimits _{V \to U} \mathcal{O}(U) \]

where $U$ runs over the objects of $\mathcal{C}_1$ and the colimit is taken in the category of abelian groups. Below we will use frequently that if

\[ V \to U \to U' \]

are morphisms with $U, U' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_1)$ and if $f' \in \mathcal{O}(U')$ restricts to $f \in \mathcal{O}(U)$, then $(V \to U, f)$ and $(V \to U', f')$ define the same element of the colimit. Also, $g_!\mathcal{O}_1 \to \mathcal{O}_2$ maps the element $(V \to U, f)$ simply to the pullback of $f$ to $V$.

Surjectivity. Let $V$ be a scheme and let $h \in \mathcal{O}(V)$. Then we obtain a morphism $V \to X \times \mathbf{A}^1$ induced by $h$ and the structure morphism $V \to X$. Writing $\mathbf{A}^1 = \mathop{\mathrm{Spec}}(\mathbf{Z}[x])$ we see the element $x \in \mathcal{O}(X \times \mathbf{A}^1)$ pulls back to $h$. Since $X \times \mathbf{A}^1$ is an object of $\mathcal{C}_1$ by assumptions (1)(b) and (2) we obtain the desired surjectivity.

Injectivity. Let $V$ be a scheme. Let $s = \sum _{i = 1, \ldots , n} (V \to U_ i, f_ i)$ be an element of the colimit displayed above. For any $i$ we can use the morphism $f_ i : U_ i \to X \times \mathbf{A}^1$ to see that $(V \to U_ i, f_ i)$ defines the same element of the colimit as $(f_ i : V \to X \times \mathbf{A}^1, x)$. Then we can consider

\[ f_1 \times \ldots \times f_ n : V \to X \times \mathbf{A}^ n \]

and we see that $s$ is equivalent in the colimit to

\[ \sum \nolimits _{i = 1, \ldots , n} (f_1 \times \ldots \times f_ n : V \to X \times \mathbf{A}^ n, x_ i) = (f_1 \times \ldots \times f_ n : V \to X \times \mathbf{A}^ n, x_1 + \ldots + x_ n) \]

Now, if $x_1 + \ldots + x_ n$ restricts to zero on $V$, then we see that $f_1 \times \ldots \times f_ n$ factors through $X \times \mathbf{A}^{n - 1} = V(x_1 + \ldots + x_ n)$. Hence we see that $s$ is equivalent to zero in the colimit. $\square$

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 04CB. Beware of the difference between the letter 'O' and the digit '0'.