The Stacks project

96.17 Injective sheaves

The pushforward of an injective abelian sheaf or module is injective.

Lemma 96.17.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Let $\tau \in \{ Zar, {\acute{e}tale}, smooth, syntomic, fppf\} $.

  1. $f_*\mathcal{I}$ is injective in $\textit{Ab}(\mathcal{Y}_\tau )$ for $\mathcal{I}$ injective in $\textit{Ab}(\mathcal{X}_\tau )$, and

  2. $f_*\mathcal{I}$ is injective in $\textit{Mod}(\mathcal{Y}_\tau , \mathcal{O}_\mathcal {Y})$ for $\mathcal{I}$ injective in $\textit{Mod}(\mathcal{X}_\tau , \mathcal{O}_\mathcal {X})$.

Proof. This follows formally from the fact that $f^{-1}$ is an exact left adjoint of $f_*$, see Homology, Lemma 12.29.1. $\square$

In the rest of this section we prove that pullback $f^{-1}$ has a left adjoint $f_!$ on abelian sheaves and modules. If $f$ is representable (by schemes or by algebraic spaces), then it will turn out that $f_!$ is exact and $f^{-1}$ will preserve injectives. We first prove a few preliminary lemmas about fibre products and equalizers in categories fibred in groupoids and their behaviour with respect to morphisms.

Lemma 96.17.2. Let $p : \mathcal{X} \to (\mathit{Sch}/S)_{fppf}$ be a category fibred in groupoids.

  1. The category $\mathcal{X}$ has fibre products.

  2. If the $\mathit{Isom}$-presheaves of $\mathcal{X}$ are representable by algebraic spaces, then $\mathcal{X}$ has equalizers.

  3. If $\mathcal{X}$ is an algebraic stack (or more generally a quotient stack), then $\mathcal{X}$ has equalizers.

Proof. Part (1) follows Categories, Lemma 4.35.15 as $(\mathit{Sch}/S)_{fppf}$ has fibre products.

Let $a, b : x \to y$ be morphisms of $\mathcal{X}$. Set $U = p(x)$ and $V = p(y)$. The category of schemes has equalizers hence we can let $W \to U$ be the equalizer of $p(a)$ and $p(b)$. Denote $c : z \to x$ a morphism of $\mathcal{X}$ lying over $W \to U$. The equalizer of $a$ and $b$, if it exists, is the equalizer of $a \circ c$ and $b \circ c$. Thus we may assume that $p(a) = p(b) = f : U \to V$. As $\mathcal{X}$ is fibred in groupoids, there exists a unique automorphism $i : x \to x$ in the fibre category of $\mathcal{X}$ over $U$ such that $a \circ i = b$. Again the equalizer of $a$ and $b$ is the equalizer of $\text{id}_ x$ and $i$. Recall that the $\mathit{Isom}_\mathcal {X}(x)$ is the presheaf on $(\mathit{Sch}/U)_{fppf}$ which to $T/U$ associates the set of automorphisms of $x|_ T$ in the fibre category of $\mathcal{X}$ over $T$, see Stacks, Definition 8.2.2. If $\mathit{Isom}_\mathcal {X}(x)$ is representable by an algebraic space $G \to U$, then we see that $\text{id}_ x$ and $i$ define morphisms $e, i : U \to G$ over $U$. Set $M = U \times _{e, G, i} U$, which by Morphisms of Spaces, Lemma 67.4.7 is a scheme. Then it is clear that $x|_ M \to x$ is the equalizer of the maps $\text{id}_ x$ and $i$ in $\mathcal{X}$. This proves (2).

If $\mathcal{X} = [U/R]$ for some groupoid in algebraic spaces $(U, R, s, t, c)$ over $S$, then the hypothesis of (2) holds by Bootstrap, Lemma 80.11.5. If $\mathcal{X}$ is an algebraic stack, then we can choose a presentation $[U/R] \cong \mathcal{X}$ by Algebraic Stacks, Lemma 94.16.2. $\square$

Lemma 96.17.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$.

  1. The functor $f$ transforms fibre products into fibre products.

  2. If $f$ is faithful, then $f$ transforms equalizers into equalizers.

Proof. By Categories, Lemma 4.35.15 we see that a fibre product in $\mathcal{X}$ is any commutative square lying over a fibre product diagram in $(\mathit{Sch}/S)_{fppf}$. Similarly for $\mathcal{Y}$. Hence (1) is clear.

Let $x \to x'$ be the equalizer of two morphisms $a, b : x' \to x''$ in $\mathcal{X}$. We will show that $f(x) \to f(x')$ is the equalizer of $f(a)$ and $f(b)$. Let $y \to f(x)$ be a morphism of $\mathcal{Y}$ equalizing $f(a)$ and $f(b)$. Say $x, x', x''$ lie over the schemes $U, U', U''$ and $y$ lies over $V$. Denote $h : V \to U'$ the image of $y \to f(x)$ in the category of schemes. The morphism $y \to f(x)$ is isomorphic to $f(h^*x') \to f(x')$ by the axioms of fibred categories. Hence, as $f$ is faithful, we see that $h^*x' \to x'$ equalizes $a$ and $b$. Thus we obtain a unique morphism $h^*x' \to x$ whose image $y = f(h^*x') \to f(x)$ is the desired morphism in $\mathcal{Y}$. $\square$

Lemma 96.17.4. Let $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Z} \to \mathcal{Y}$ be faithful $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$.

  1. the functor $\mathcal{X} \times _\mathcal {Y} \mathcal{Z} \to \mathcal{Y}$ is faithful, and

  2. if $\mathcal{X}, \mathcal{Z}$ have equalizers, so does $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$.

Proof. We think of objects in $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ as quadruples $(U, x, z, \alpha )$ where $\alpha : f(x) \to g(z)$ is an isomorphism over $U$, see Categories, Lemma 4.32.3. A morphism $(U, x, z, \alpha ) \to (U', x', z', \alpha ')$ is a pair of morphisms $a : x \to x'$ and $b : z \to z'$ compatible with $\alpha $ and $\alpha '$. Thus it is clear that if $f$ and $g$ are faithful, so is the functor $\mathcal{X} \times _\mathcal {Y} \mathcal{Z} \to \mathcal{Y}$. Now, suppose that $(a, b), (a', b') : (U, x, z, \alpha ) \to (U', x', z', \alpha ')$ are two morphisms of the $2$-fibre product. Then consider the equalizer $x'' \to x$ of $a$ and $a'$ and the equalizer $z'' \to z$ of $b$ and $b'$. Since $f$ commutes with equalizers (by Lemma 96.17.3) we see that $f(x'') \to f(x)$ is the equalizer of $f(a)$ and $f(a')$. Similarly, $g(z'') \to g(z)$ is the equalizer of $g(b)$ and $g(b')$. Picture

\[ \xymatrix{ f(x'') \ar[r] \ar@{..>}[d]_{\alpha ''}& f(x) \ar[d]_\alpha \ar@<0.5ex>[r]^{f(a)} \ar@<-0.5ex>[r]_{f(a')} & f(x') \ar[d]^{\alpha '} \\ g(z'') \ar[r] & g(z) \ar@<0.5ex>[r]^{g(b)} \ar@<-0.5ex>[r]_{g(b')} & g(z') } \]

It is clear that the dotted arrow exists and is an isomorphism. However, it is not a priori the case that the image of $\alpha ''$ in the category of schemes is the identity of its source. On the other hand, the existence of $\alpha ''$ means that we can assume that $x''$ and $z''$ are defined over the same scheme and that the morphisms $x'' \to x$ and $z'' \to z$ have the same image in the category of schemes. Redoing the diagram above we see that the dotted arrow now does project to an identity morphism and we win. Some details omitted. $\square$

As we are working with big sites we have the following somewhat counter intuitive result (which also holds for morphisms of big sites of schemes). Warning: This result isn't true if we drop the hypothesis that $f$ is faithful.

Lemma 96.17.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Let $\tau \in \{ Zar, {\acute{e}tale}, smooth, syntomic, fppf\} $. The functor $f^{-1} : \textit{Ab}(\mathcal{Y}_\tau ) \to \textit{Ab}(\mathcal{X}_\tau )$ has a left adjoint $f_! : \textit{Ab}(\mathcal{X}_\tau ) \to \textit{Ab}(\mathcal{Y}_\tau )$. If $f$ is faithful and $\mathcal{X}$ has equalizers, then

  1. $f_!$ is exact, and

  2. $f^{-1}\mathcal{I}$ is injective in $\textit{Ab}(\mathcal{X}_\tau )$ for $\mathcal{I}$ injective in $\textit{Ab}(\mathcal{Y}_\tau )$.

Proof. By Stacks, Lemma 8.10.3 the functor $f$ is continuous and cocontinuous. Hence by Modules on Sites, Lemma 18.16.2 the functor $f^{-1} : \textit{Ab}(\mathcal{Y}_\tau ) \to \textit{Ab}(\mathcal{X}_\tau )$ has a left adjoint $f_! : \textit{Ab}(\mathcal{X}_\tau ) \to \textit{Ab}(\mathcal{Y}_\tau )$. To see (1) we apply Modules on Sites, Lemma 18.16.3 and to see that the hypotheses of that lemma are satisfied use Lemmas 96.17.2 and 96.17.3 above. Part (2) follows from this formally, see Homology, Lemma 12.29.1. $\square$

Lemma 96.17.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Let $\tau \in \{ Zar, {\acute{e}tale}, smooth, syntomic, fppf\} $. The functor $f^* : \textit{Mod}(\mathcal{Y}_\tau , \mathcal{O}_\mathcal {Y}) \to \textit{Mod}(\mathcal{X}_\tau , \mathcal{O}_\mathcal {X})$ has a left adjoint $f_! : \textit{Mod}(\mathcal{X}_\tau , \mathcal{O}_\mathcal {X}) \to \textit{Mod}(\mathcal{Y}_\tau , \mathcal{O}_\mathcal {Y})$ which agrees with the functor $f_!$ of Lemma 96.17.5 on underlying abelian sheaves. If $f$ is faithful and $\mathcal{X}$ has equalizers, then

  1. $f_!$ is exact, and

  2. $f^{-1}\mathcal{I}$ is injective in $\textit{Mod}(\mathcal{X}_\tau , \mathcal{O}_\mathcal {X})$ for $\mathcal{I}$ injective in $\textit{Mod}(\mathcal{Y}_\tau , \mathcal{O}_\mathcal {X})$.

Proof. Recall that $f$ is a continuous and cocontinuous functor of sites and that $f^{-1}\mathcal{O}_\mathcal {Y} = \mathcal{O}_\mathcal {X}$. Hence Modules on Sites, Lemma 18.41.1 implies $f^*$ has a left adjoint $f_!^{Mod}$. Let $x$ be an object of $\mathcal{X}$ lying over the scheme $U$. Then $f$ induces an equivalence of ringed sites

\[ \mathcal{X}/x \longrightarrow \mathcal{Y}/f(x) \]

as both sides are equivalent to $(\mathit{Sch}/U)_\tau $, see Lemma 96.9.4. Modules on Sites, Remark 18.41.2 shows that $f_!$ agrees with the functor on abelian sheaves.

Assume now that $\mathcal{X}$ has equalizers and that $f$ is faithful. Lemma 96.17.5 tells us that $f_!$ is exact. Finally, Homology, Lemma 12.29.1 implies the statement on pullbacks of injective modules. $\square$


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