Lemma 66.26.1. Let $S$ be a scheme. Let $f : X \to Y$ be an étale morphism of algebraic spaces over $S$. Then $f^{-1}\mathcal{O}_ Y = \mathcal{O}_ X$, and $f^*\mathcal{G} = f_{small}^{-1}\mathcal{G}$ for any sheaf of $\mathcal{O}_ Y$-modules $\mathcal{G}$. In particular, $f^* : \textit{Mod}(\mathcal{O}_ Y) \to \textit{Mod}(\mathcal{O}_ X)$ is exact.
66.26 Sheaves of modules on algebraic spaces
If $X$ is an algebraic space, then a sheaf of modules on $X$ is a sheaf of $\mathcal{O}_ X$-modules on the small étale site of $X$ where $\mathcal{O}_ X$ is the structure sheaf of $X$. The category of sheaves of modules is denoted $\textit{Mod}(\mathcal{O}_ X)$.
Given a morphism $f : X \to Y$ of algebraic spaces, by Lemma 66.21.3 we get a morphism of ringed topoi and hence by Modules on Sites, Definition 18.13.1 we get well defined pullback and direct image functors
66.26.0.1
\begin{equation} \label{spaces-properties-equation-push-pull} f^* : \textit{Mod}(\mathcal{O}_ Y) \longrightarrow \textit{Mod}(\mathcal{O}_ X), \quad f_* : \textit{Mod}(\mathcal{O}_ X) \longrightarrow \textit{Mod}(\mathcal{O}_ Y) \end{equation}
which are adjoint in the usual way. If $g : Y \to Z$ is another morphism of algebraic spaces over $S$, then we have $(g \circ f)^* = f^* \circ g^*$ and $(g \circ f)_* = g_* \circ f_*$ simply because the morphisms of ringed topoi compose in the corresponding way (by the lemma).
Proof. By the description of inverse image in Lemma 66.18.11 and the definition of the structure sheaves it is clear that $f_{small}^{-1}\mathcal{O}_ Y = \mathcal{O}_ X$. Since the pullback
\[ f^*\mathcal{G} = f_{small}^{-1}\mathcal{G} \otimes _{f_{small}^{-1}\mathcal{O}_ Y} \mathcal{O}_ X \]
by definition we conclude that $f^*\mathcal{G} = f_{small}^{-1}\mathcal{G}$. The exactness is clear because $f_{small}^{-1}$ is exact, as $f_{small}$ is a morphism of topoi. $\square$
We continue our abuse of notation introduced in Equation (66.18.11.1) by writing
66.26.1.1
\begin{equation} \label{spaces-properties-equation-restrict-modules} \mathcal{G}|_{X_{\acute{e}tale}} = f^*\mathcal{G} = f_{small}^{-1}\mathcal{G} \end{equation}
in the situation of the lemma above. We will discuss this in a more technical fashion in Section 66.27.
Lemma 66.26.2. Let $S$ be a scheme. Let be a cartesian square of algebraic spaces over $S$. Let $\mathcal{F} \in \textit{Mod}(\mathcal{O}_ X)$. If $g$ is étale, then $f'_*(\mathcal{F}|_{X'}) = (f_*\mathcal{F})|_{Y'}$1 and $R^ if'_*(\mathcal{F}|_{X'}) = (R^ if_*\mathcal{F})|_{Y'}$ in $\textit{Mod}(\mathcal{O}_{Y'})$.
\[ \xymatrix{ X' \ar[r] \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]
Proof. This is a reformulation of Lemma 66.18.12 in the case of modules. $\square$
Lemma 66.26.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. A sheaf $\mathcal{F}$ of $\mathcal{O}_ X$-modules is given by the following data:
for every $U \in \mathop{\mathrm{Ob}}\nolimits (X_{\acute{e}tale})$ a sheaf $\mathcal{F}_ U$ of $\mathcal{O}_ U$-modules on $U_{\acute{e}tale}$,
for every $f : U' \to U$ in $X_{\acute{e}tale}$ an isomorphism $c_ f : f_{small}^*\mathcal{F}_ U \to \mathcal{F}_{U'}$.
These data are subject to the condition that given any $f : U' \to U$ and $g : U'' \to U'$ in $X_{\acute{e}tale}$ the composition $c_ g \circ g_{small}^*c_ f$ is equal to $c_{f \circ g}$.
Proof. Combine Lemmas 66.26.1 and 66.18.13, and use the fact that any morphism between objects of $X_{\acute{e}tale}$ is an étale morphism of schemes. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)