The Stacks project

19.7 Abelian Sheaves on a site

Let $\mathcal{C}$ be a site. In this section we prove that there are enough injectives for abelian sheaves on $\mathcal{C}$.

Denote $i : \textit{Ab}(\mathcal{C}) \longrightarrow \textit{PAb}(\mathcal{C})$ the forgetful functor from abelian sheaves to abelian presheaves. Let ${}^\# : \textit{PAb}(\mathcal{C}) \longrightarrow \textit{Ab}(\mathcal{C})$ denote the sheafification functor. Recall that ${}^\# $ is a left adjoint to $i$, that ${}^\# $ is exact, and that $i\mathcal{F}^\# = \mathcal{F}$ for any abelian sheaf $\mathcal{F}$. Finally, let $\mathcal{G} \to J(\mathcal{G})$ denote the canonical embedding into an injective presheaf we found in Section 19.6.

For any sheaf $\mathcal{F}$ in $\textit{Ab}(\mathcal{C})$ and any ordinal $\beta $ we define a sheaf $J_\beta (\mathcal{F})$ by transfinite recursion. We set $J_0(\mathcal{F}) = \mathcal{F}$. We define $J_1(\mathcal{F}) = J(i\mathcal{F})^\# $. Sheafification of the canonical map $i\mathcal{F} \to J(i\mathcal{F})$ gives a functorial map

\[ \mathcal{F} \longrightarrow J_1(\mathcal{F}) \]

which is injective as $\# $ is exact. We set $J_{\alpha + 1}(\mathcal{F}) = J_1(J_\alpha (\mathcal{F}))$. So that there are canonical injective maps $J_\alpha (\mathcal{F}) \to J_{\alpha + 1}(\mathcal{F})$. For a limit ordinal $\beta $, we define

\[ J_\beta (\mathcal{F}) = \mathop{\mathrm{colim}}\nolimits _{\alpha < \beta } J_\alpha (\mathcal{F}). \]

Note that this is a directed colimit. Hence for any ordinals $\alpha < \beta $ we have an injective map $J_\alpha (\mathcal{F}) \to J_\beta (\mathcal{F})$.

Lemma 19.7.1. With notation as above. Suppose that $\mathcal{G}_1 \to \mathcal{G}_2$ is an injective map of abelian sheaves on $\mathcal{C}$. Let $\alpha $ be an ordinal and let $\mathcal{G}_1 \to J_\alpha (\mathcal{F})$ be a morphism of sheaves. There exists a morphism $\mathcal{G}_2 \to J_{\alpha + 1}(\mathcal{F})$ such that the following diagram commutes

\[ \xymatrix{ \mathcal{G}_1 \ar[d] \ar[r] & \mathcal{G}_2 \ar[d] \\ J_{\alpha }(\mathcal{F}) \ar[r] & J_{\alpha + 1}(\mathcal{F}) } \]

Proof. This is because the map $i\mathcal{G}_1 \to i\mathcal{G}_2$ is injective and hence $i\mathcal{G}_1 \to iJ_\alpha (\mathcal{F})$ extends to $i\mathcal{G}_2 \to J(iJ_\alpha (\mathcal{F}))$ which gives the desired map after applying the sheafification functor. $\square$

This lemma says that somehow the system $\{ J_{\alpha }(\mathcal{F})\} $ is an injective embedding of $\mathcal{F}$. Of course we cannot take the limit over all $\alpha $ because they form a class and not a set. However, the idea is now that you don't have to check injectivity on all injections $\mathcal{G}_1 \to \mathcal{G}_2$, plus the following lemma.

Lemma 19.7.2. Suppose that $\mathcal{G}_ i$, $i\in I$ is set of abelian sheaves on $\mathcal{C}$. There exists an ordinal $\beta $ such that for any sheaf $\mathcal{F}$, any $i\in I$, and any map $\varphi : \mathcal{G}_ i \to J_\beta (\mathcal{F})$ there exists an $\alpha < \beta $ such that $ \varphi $ factors through $J_\alpha (\mathcal{F})$.

Proof. This reduces to the case of a single sheaf $\mathcal{G}$ by taking the direct sum of all the $\mathcal{G}_ i$.

Consider the sets

\[ S = \coprod \nolimits _{U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})} \mathcal{G}(U). \]


\[ T_\beta = \coprod \nolimits _{U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})} J_\beta (\mathcal{F})(U) \]

The transition maps between the sets $T_\beta $ are injective. If the cofinality of $\beta $ is large enough, then $T_\beta = \mathop{\mathrm{colim}}\nolimits _{\alpha < \beta } T_\alpha $, see Sites, Lemma 7.17.10. A morphism $\mathcal{G} \to J_\beta (\mathcal{F})$ factors through $J_\alpha (\mathcal{F})$ if and only if the associated map $S \to T_\beta $ factors through $T_\alpha $. By Sets, Lemma 3.7.1 if the cofinality of $\beta $ is bigger than the cardinality of $S$, then the result of the lemma is true. Hence the lemma follows from the fact that there are ordinals with arbitrarily large cofinality, see Sets, Proposition 3.7.2. $\square$

Recall that for an object $X$ of $\mathcal{C}$ we denote $\mathbf{Z}_ X$ the presheaf of abelian groups $\Gamma (U, \mathbf{Z}_ X) = \oplus _{U \to X} \mathbf{Z}$, see Modules on Sites, Section 18.4. The sheaf associated to this presheaf is denoted $\mathbf{Z}_ X^\# $, see Modules on Sites, Section 18.5. It can be characterized by the property
\begin{equation} \label{injectives-equation-free-sheaf-on} \mathop{\mathrm{Mor}}\nolimits _{\textit{Ab}(\mathcal{C})}(\mathbf{Z}_ X^\# , \mathcal{G}) = \mathcal{G}(X) \end{equation}

where the element $\varphi $ of the left hand side is mapped to $\varphi (1 \cdot \text{id}_ X)$ in the right hand side. We can use these sheaves to characterize injective abelian sheaves.

Lemma 19.7.3. Suppose $\mathcal{J}$ is a sheaf of abelian groups with the following property: For all $X\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, for any abelian subsheaf $\mathcal{S} \subset \mathbf{Z}_ X^\# $ and any morphism $\varphi : \mathcal{S} \to \mathcal{J}$, there exists a morphism $\mathbf{Z}_ X^\# \to \mathcal{J}$ extending $\varphi $. Then $\mathcal{J}$ is an injective sheaf of abelian groups.

Proof. Let $\mathcal{F} \to \mathcal{G}$ be an injective map of abelian sheaves. Suppose $\varphi : \mathcal{F} \to \mathcal{J}$ is a morphism. Arguing as in the proof of More on Algebra, Lemma 15.54.1 we see that it suffices to prove that if $\mathcal{F} \not= \mathcal{G}$, then we can find an abelian sheaf $\mathcal{F}'$, $\mathcal{F} \subset \mathcal{F}' \subset \mathcal{G}$ such that (a) the inclusion $\mathcal{F} \subset \mathcal{F}'$ is strict, and (b) $\varphi $ can be extended to $\mathcal{F}'$. To find $\mathcal{F}'$, let $X$ be an object of $\mathcal{C}$ such that the inclusion $\mathcal{F}(X) \subset \mathcal{G}(X)$ is strict. Pick $s \in \mathcal{G}(X)$, $s \not\in \mathcal{F}(X)$. Let $\psi : \mathbf{Z}_ X^\# \to \mathcal{G}$ be the morphism corresponding to the section $s$ via ( Set $\mathcal{S} = \psi ^{-1}(\mathcal{F})$. By assumption the morphism

\[ \mathcal{S} \xrightarrow {\psi } \mathcal{F} \xrightarrow {\varphi } \mathcal{J} \]

can be extended to a morphism $\varphi ' : \mathbf{Z}_ X^\# \to \mathcal{J}$. Note that $\varphi '$ annihilates the kernel of $\psi $ (as this is true for $\varphi $). Thus $\varphi '$ gives rise to a morphism $\varphi '' : \mathop{\mathrm{Im}}(\psi ) \to \mathcal{J}$ which agrees with $\varphi $ on the intersection $\mathcal{F} \cap \mathop{\mathrm{Im}}(\psi )$ by construction. Thus $\varphi $ and $\varphi ''$ glue to give an extension of $\varphi $ to the strictly bigger subsheaf $\mathcal{F}' = \mathcal{F} + \mathop{\mathrm{Im}}(\psi )$. $\square$

Theorem 19.7.4. The category of sheaves of abelian groups on a site has enough injectives. In fact there exists a functorial injective embedding, see Homology, Definition 12.27.5.

Proof. Let $\mathcal{G}_ i$, $i \in I$ be a set of abelian sheaves such that every subsheaf of every $\mathbf{Z}_ X^\# $ occurs as one of the $\mathcal{G}_ i$. Apply Lemma 19.7.2 to this collection to get an ordinal $\beta $. We claim that for any sheaf of abelian groups $\mathcal{F}$ the map $\mathcal{F} \to J_\beta (\mathcal{F})$ is an injection of $\mathcal{F}$ into an injective. Note that by construction the assignment $\mathcal{F} \mapsto \big (\mathcal{F} \to J_\beta (\mathcal{F})\big )$ is indeed functorial.

The proof of the claim comes from the fact that by Lemma 19.7.3 it suffices to extend any morphism $\gamma : \mathcal{G} \to J_\beta (\mathcal{F})$ from a subsheaf $\mathcal{G}$ of some $\mathbf{Z}_ X^\# $ to all of $\mathbf{Z}_ X^\# $. Then by Lemma 19.7.2 the map $\gamma $ lifts into $J_\alpha (\mathcal{F})$ for some $\alpha < \beta $. Finally, we apply Lemma 19.7.1 to get the desired extension of $\gamma $ to a morphism into $J_{\alpha + 1}(\mathcal{F}) \to J_\beta (\mathcal{F})$. $\square$

Comments (3)

Comment #6625 by J. des Esseintes on

In the proof of Lemma 01DN, it is claimed that . I do not see how this follows without assuming some lower bound on the cofinality of (larger than the size of any open cover, say).

Comment #6626 by on

OK, yes, good catch! I will fix this the next time I go through all the comments.

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