59.70 Extension by zero

The general material in Modules on Sites, Section 18.19 allows us to make the following definition.

Definition 59.70.1. Let $j : U \to X$ be an étale morphism of schemes.

1. The restriction functor $j^{-1} : \mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale}) \to \mathop{\mathit{Sh}}\nolimits (U_{\acute{e}tale})$ has a left adjoint $j_!^{Sh} : \mathop{\mathit{Sh}}\nolimits (U_{\acute{e}tale}) \to \mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale})$.

2. The restriction functor $j^{-1} : \textit{Ab}(X_{\acute{e}tale}) \to \textit{Ab}(U_{\acute{e}tale})$ has a left adjoint which is denoted $j_! : \textit{Ab}(U_{\acute{e}tale}) \to \textit{Ab}(X_{\acute{e}tale})$ and called extension by zero.

3. Let $\Lambda$ be a ring. The restriction functor $j^{-1} : \textit{Mod}(X_{\acute{e}tale}, \Lambda ) \to \textit{Mod}(U_{\acute{e}tale}, \Lambda )$ has a left adjoint which is denoted $j_! : \textit{Mod}(U_{\acute{e}tale}, \Lambda ) \to \textit{Mod}(X_{\acute{e}tale}, \Lambda )$ and called extension by zero.

If $\mathcal{F}$ is an abelian sheaf on $X_{\acute{e}tale}$, then $j_!\mathcal{F} \not= j_!^{Sh}\mathcal{F}$ in general. On the other hand $j_!$ for sheaves of $\Lambda$-modules agrees with $j_!$ on underlying abelian sheaves (Modules on Sites, Remark 18.19.6). The functor $j_!$ is characterized by the functorial isomorphism

$\mathop{\mathrm{Hom}}\nolimits _ X(j_!\mathcal{F}, \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _ U(\mathcal{F}, j^{-1}\mathcal{G})$

for all $\mathcal{F} \in \textit{Ab}(U_{\acute{e}tale})$ and $\mathcal{G} \in \textit{Ab}(X_{\acute{e}tale})$. Similarly for sheaves of $\Lambda$-modules.

To describe the functors in Definition 59.70.1 more explicitly, recall that $j^{-1}$ is just the restriction via the functor $U_{\acute{e}tale}\to X_{\acute{e}tale}$. In other words, $j^{-1}\mathcal{G}(U') = \mathcal{G}(U')$ for $U'$ étale over $U$. On the other hand, for $\mathcal{F} \in \textit{Ab}(U_{\acute{e}tale})$ we consider the presheaf

59.70.1.1
$$\label{etale-cohomology-equation-j-p-shriek} j_{p!}\mathcal{F} : X_{\acute{e}tale}\longrightarrow \textit{Ab}, \quad V \longmapsto \bigoplus \nolimits _{V \to U} \mathcal{F}(V \to U)$$

Then $j_!\mathcal{F}$ is the sheafification of $j_{p!}\mathcal{F}$. This is proven in Modules on Sites, Lemma 18.19.2; more generally see the discussion in Modules on Sites, Sections 18.19 and 18.16.

Exercise 59.70.2. Prove directly that the functor $j_!$ defined as the sheafification of the functor $j_{p!}$ given in (59.70.1.1) is a left adjoint to $j^{-1}$.

Proposition 59.70.3. Let $j : U \to X$ be an étale morphism of schemes. Let $\mathcal{F}$ in $\textit{Ab}(U_{\acute{e}tale})$. If $\overline{x} : \mathop{\mathrm{Spec}}(k) \to X$ is a geometric point of $X$, then

$(j_!\mathcal{F})_{\overline{x}} = \bigoplus \nolimits _{\overline{u} : \mathop{\mathrm{Spec}}(k) \to U,\ j(\overline{u}) = \overline{x}} \mathcal{F}_{\bar{u}}.$

In particular, $j_!$ is an exact functor.

Proof. Exactness of $j_!$ is very general, see Modules on Sites, Lemma 18.19.3. Of course it does also follow from the description of stalks. The formula for the stalk follows from Modules on Sites, Lemma 18.38.1 and the description of points of the small étale site in terms of geometric points, see Lemma 59.29.12.

For later use we note that the isomorphism

\begin{align*} (j_!\mathcal{F})_{\overline{x}} & = (j_{p!}\mathcal{F})_{\overline{x}} \\ & = \mathop{\mathrm{colim}}\nolimits _{(V, \overline{v})} j_{p!}\mathcal{F}(V) \\ & = \mathop{\mathrm{colim}}\nolimits _{(V, \overline{v})} \bigoplus \nolimits _{\varphi : V \to U} \mathcal{F}(V \xrightarrow {\varphi } U) \\ & \to \bigoplus \nolimits _{\overline{u} : \mathop{\mathrm{Spec}}(k) \to U,\ j(\overline{u}) = \overline{x}} \mathcal{F}_{\bar{u}}. \end{align*}

constructed in Modules on Sites, Lemma 18.38.1 sends $(V, \overline{v}, \varphi , s)$ to the class of $s$ in the stalk of $\mathcal{F}$ at $\overline{u} = \varphi (\overline{v})$. $\square$

Lemma 59.70.4. Let $j : U \to X$ be an open immersion of schemes. For any abelian sheaf $\mathcal{F}$ on $U_{\acute{e}tale}$, the adjunction mappings $j^{-1}j_*\mathcal{F} \to \mathcal{F}$ and $\mathcal{F} \to j^{-1}j_!\mathcal{F}$ are isomorphisms. In fact, $j_!\mathcal{F}$ is the unique abelian sheaf on $X_{\acute{e}tale}$ whose restriction to $U$ is $\mathcal{F}$ and whose stalks at geometric points of $X \setminus U$ are zero.

Proof. We encourage the reader to prove the first statement by working through the definitions, but here we just use that it is a special case of the very general Modules on Sites, Lemma 18.19.8. For the second statement, observe that if $\mathcal{G}$ is an abelian sheaf on $X_{\acute{e}tale}$ whose restriction to $U$ is $\mathcal{F}$, then we obtain by adjointness a map $j_!\mathcal{F} \to \mathcal{G}$. This map is then an isomorphism at stalks of geometric points of $U$ by Proposition 59.70.3. Thus if $\mathcal{G}$ has vanishing stalks at geometric points of $X \setminus U$, then $j_!\mathcal{F} \to \mathcal{G}$ is an isomorphism by Theorem 59.29.10. $\square$

Lemma 59.70.5 (Extension by zero commutes with base change). Let $f: Y \to X$ be a morphism of schemes. Let $j: V \to X$ be an étale morphism. Consider the fibre product

$\xymatrix{ V' = Y \times _ X V \ar[d]_{f'} \ar[r]_-{j'} & Y \ar[d]^ f \\ V \ar[r]^ j & X }$

Then we have $j'_! f'^{-1} = f^{-1} j_!$ on abelian sheaves and on sheaves of modules.

Proof. This is true because $j'_! f'^{-1}$ is left adjoint to $f'_* (j')^{-1}$ and $f^{-1} j_!$ is left adjoint to $j^{-1}f_*$. Further $f'_* (j')^{-1} = j^{-1}f_*$ because $f_*$ commutes with étale localization (by construction). In fact, the lemma holds very generally in the setting of a morphism of sites, see Modules on Sites, Lemma 18.20.1. $\square$

Lemma 59.70.6. Let $j : U \to X$ be separated and étale. Then there is a functorial injective map $j_!\mathcal{F} \to j_*\mathcal{F}$ on abelian sheaves and sheaves of $\Lambda$-modules.

Proof. We prove this in the case of abelian sheaves. Let us construct a canonical map

$j_{p!}\mathcal{F} \to j_*\mathcal{F}$

of abelian presheaves on $X_{\acute{e}tale}$ for any abelian sheaf $\mathcal{F}$ on $U_{\acute{e}tale}$ where $j_{p!}$ is as in (59.70.1.1). Sheafification of this map will be the desired map $j_!\mathcal{F} \to j_*\mathcal{F}$. Evaluating both sides on $V \to X$ étale we obtain

$j_{p!}\mathcal{F}(V) = \bigoplus \nolimits _{\varphi : V \to U} \mathcal{F}(V \xrightarrow {\varphi } U) \quad \text{and}\quad j_*\mathcal{F}(V) = \mathcal{F}(V \times _ X U)$

For each $\varphi$ we have an open and closed immersion

$\Gamma _\varphi = (1, \varphi ) : V \longrightarrow V \times _ X U$

over $U$. It is open as it is a morphism between schemes étale over $U$ and it is closed as it is a section of a scheme separated over $V$ (Schemes, Lemma 26.21.11). Thus for a section $s_\varphi \in \mathcal{F}(V \xrightarrow {\varphi } U)$ there exists a unique section $s'_\varphi$ in $\mathcal{F}(V \times _ X U)$ which pulls back to $s_\varphi$ by $\Gamma _\varphi$ and which restricts to zero on the complement of the image of $\Gamma _\varphi$.

To show that our map is injective suppose that $\sum _{i = 1, \ldots , n} s_{\varphi _ i}$ is an element of $j_{p!}\mathcal{F}(V)$ in the formula above maps to zero in $j_*\mathcal{F}(V)$. Our task is to show that $\sum _{i = 1, \ldots , n} s_{\varphi _ i}$ restricts to zero on the members of an étale covering of $V$. Looking at all pairwise equalizers (which are open and closed in $V$) of the morphisms $\varphi _ i : V \to U$ and working locally on $V$, we may assume the images of the morphisms $\Gamma _{\varphi _1}, \ldots , \Gamma _{\varphi _ n}$ are pairwise disjoint. Since our assumption is that $\sum _{i = 1, \ldots , n} s'_{\varphi _ i} = 0$ we then immediately conclude that $s'_{\varphi _ i} = 0$ for each $i$ (by the disjointness of the supports of these sections), whence $s_{\varphi _ i} = 0$ for all $i$ as desired. $\square$

Lemma 59.70.7. Let $j : U \to X$ be finite and étale. Then the map $j_! \to j_*$ of Lemma 59.70.6 is an isomorphism on abelian sheaves and sheaves of $\Lambda$-modules.

Proof. It suffices to check $j_!\mathcal{F} \to j_*\mathcal{F}$ is an isomorphism étale locally on $X$. Thus we may assume $U \to X$ is a finite disjoint union of isomorphisms, see Étale Morphisms, Lemma 41.18.3. We omit the proof in this case. $\square$

Lemma 59.70.8. Let $X$ be a scheme. Let $Z \subset X$ be a closed subscheme and let $U \subset X$ be the complement. Denote $i : Z \to X$ and $j : U \to X$ the inclusion morphisms. For every abelian sheaf $\mathcal{F}$ on $X_{\acute{e}tale}$ there is a canonical short exact sequence

$0 \to j_!j^{-1}\mathcal{F} \to \mathcal{F} \to i_*i^{-1}\mathcal{F} \to 0$

on $X_{\acute{e}tale}$.

Proof. We obtain the maps by the adjointness properties of the functors involved. For a geometric point $\overline{x}$ in $X$ we have either $\overline{x} \in U$ in which case the map on the left hand side is an isomorphism on stalks and the stalk of $i_*i^{-1}\mathcal{F}$ is zero or $\overline{x} \in Z$ in which case the map on the right hand side is an isomorphism on stalks and the stalk of $j_!j^{-1}\mathcal{F}$ is zero. Here we have used the description of stalks of Lemma 59.46.3 and Proposition 59.70.3. $\square$

Lemma 59.70.9. Consider a cartesian diagram of schemes

$\xymatrix{ U \ar[d]_ g \ar[r]_{j'} & X \ar[d]^ f \\ V \ar[r]^ j & Y }$

where $f$ is finite, $g$ is étale, and $j$ is an open immersion. Then $f_* \circ j'_! = j_! \circ g_*$ as functors $\textit{Ab}(U_{\acute{e}tale}) \to \textit{Ab}(Y_{\acute{e}tale})$.

Proof. Let $\mathcal{F}$ be an object of $\textit{Ab}(U_{\acute{e}tale})$. Let $\overline{y}$ be a geometric point of $Y$ not contained in the open $V$. Then

$(f_*j'_!\mathcal{F})_{\overline{y}} = \bigoplus \nolimits _{\overline{x},\ f(\overline{x}) = \overline{y}} (j'_!\mathcal{F})_{\overline{x}} = 0$

by Proposition 59.55.2 and because the stalk of $j'_!\mathcal{F}$ at $\overline{x} \not\in U$ are zero by Lemma 59.70.4. On the other hand, we have

$j^{-1}f_*j'_!\mathcal{F} = g_*(j')^{-1}j'_!\mathcal{F} = g_*\mathcal{F}$

by Lemmas 59.55.3 and Lemma 59.70.4. Hence by the characterization of $j_!$ in Lemma 59.70.4 we see that $f_*j'_!\mathcal{F} = j_!g_*\mathcal{F}$. We omit the verification that this identification is functorial in $\mathcal{F}$. $\square$

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).