The Stacks project

83.12 Cartesian sheaves and modules

Here is the definition.

Definition 83.12.1. In Situation 83.3.3.

  1. A sheaf $\mathcal{F}$ of sets or of abelian groups on $\mathcal{C}$ is cartesian if the maps $\mathcal{F}(\varphi ) : f_\varphi ^{-1}\mathcal{F}_ m \to \mathcal{F}_ n$ are isomorphisms for all $\varphi : [m] \to [n]$.

  2. If $\mathcal{O}$ is a sheaf of rings on $\mathcal{C}_{total}$, then a sheaf $\mathcal{F}$ of $\mathcal{O}$-modules is cartesian if the maps $f_\varphi ^*\mathcal{F}_ m \to \mathcal{F}_ n$ are isomorphisms for all $\varphi : [m] \to [n]$.

  3. An object $K$ of $D(\mathcal{C}_{total})$ is cartesian if the maps $f_\varphi ^{-1}K_ m \to K_ n$ are isomorphisms for all $\varphi : [m] \to [n]$.

  4. If $\mathcal{O}$ is a sheaf of rings on $\mathcal{C}_{total}$, then an object $K$ of $D(\mathcal{O})$ is cartesian if the maps $Lf_\varphi ^*K_ m \to K_ n$ are isomorphisms for all $\varphi : [m] \to [n]$.

Of course there is a general notion of a cartesian section of a fibred category and the above are merely examples of this. The property on pullbacks needs only be checked for the degeneracies.

Lemma 83.12.2. In Situation 83.3.3.

  1. A sheaf $\mathcal{F}$ of sets or abelian groups is cartesian if and only if the maps $(f_{\delta ^ n_ j})^{-1}\mathcal{F}_{n - 1} \to \mathcal{F}_ n$ are isomorphisms.

  2. An object $K$ of $D(\mathcal{C}_{total})$ is cartesian if and only if the maps $(f_{\delta ^ n_ j})^{-1}K_{n - 1} \to K_ n$ are isomorphisms.

  3. If $\mathcal{O}$ is a sheaf of rings on $\mathcal{C}_{total}$ a sheaf $\mathcal{F}$ of $\mathcal{O}$-modules is cartesian if and only if the maps $(f_{\delta ^ n_ j})^*\mathcal{F}_{n - 1} \to \mathcal{F}_ n$ are isomorphisms.

  4. If $\mathcal{O}$ is a sheaf of rings on $\mathcal{C}_{total}$ an object $K$ of $D(\mathcal{O})$ is cartesian if and only if the maps $L(f_{\delta ^ n_ j})^*K_{n - 1} \to K_ n$ are isomorphisms.

  5. Add more here.

Proof. In each case the key is that the pullback functors compose to pullback functor; for part (4) see Cohomology on Sites, Lemma 21.18.3. We show how the argument works in case (1) and omit the proof in the other cases. The category $\Delta $ is generated by the morphisms the morphisms $\delta ^ n_ j$ and $\sigma ^ n_ j$, see Simplicial, Lemma 14.2.2. Hence we only need to check the maps $(f_{\delta ^ n_ j})^{-1}\mathcal{F}_{n - 1} \to \mathcal{F}_ n$ and $(f_{\sigma ^ n_ j})^{-1}\mathcal{F}_{n + 1} \to \mathcal{F}_ n$ are isomorphisms, see Simplicial, Lemma 14.3.2 for notation. Since $\sigma ^ n_ j \circ \delta _ j^{n + 1} = \text{id}_{[n]}$ the composition

\[ \mathcal{F}_ n = (f_{\sigma ^ n_ j})^{-1} (f_{\delta _ j^{n + 1}})^{-1} \mathcal{F}_ n \to (f_{\sigma ^ n_ j})^{-1} \mathcal{F}_{n + 1} \to \mathcal{F}_ n \]

is the identity. Thus the result for $\delta ^{n + 1}_ j$ implies the result for $\sigma ^ n_ j$. $\square$

Lemma 83.12.3. In Situation 83.3.3 let $a_0$ be an augmentation towards a site $\mathcal{D}$ as in Remark 83.4.1.

  1. The pullback $a^{-1}\mathcal{G}$ of a sheaf of sets or abelian groups on $\mathcal{D}$ is cartesian.

  2. The pullback $a^{-1}K$ of an object $K$ of $D(\mathcal{D})$ is cartesian.

Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}_{total}$ and $\mathcal{O}_\mathcal {D}$ a sheaf of rings on $\mathcal{D}$ and $a^\sharp : \mathcal{O}_\mathcal {D} \to a_*\mathcal{O}$ a morphism as in Section 83.11.

  1. The pullback $a^*\mathcal{F}$ of a sheaf of $\mathcal{O}_\mathcal {D}$-modules is cartesian.

  2. The derived pullback $La^*K$ of an object $K$ of $D(\mathcal{O}_\mathcal {D})$ is cartesian.

Proof. This follows immediately from the identities $a_ m \circ f_\varphi = a_ n$ for all $\varphi : [m] \to [n]$. See Lemma 83.4.2 and the discussion in Section 83.11. $\square$

Lemma 83.12.4. In Situation 83.3.3. The category of cartesian sheaves of sets (resp. abelian groups) is equivalent to the category of pairs $(\mathcal{F}, \alpha )$ where $\mathcal{F}$ is a sheaf of sets (resp. abelian groups) on $\mathcal{C}_0$ and

\[ \alpha : (f_{\delta _1^1})^{-1}\mathcal{F} \longrightarrow (f_{\delta _0^1})^{-1}\mathcal{F} \]

is an isomorphism of sheaves of sets (resp. abelian groups) on $\mathcal{C}_1$ such that $(f_{\delta ^2_1})^{-1}\alpha = (f_{\delta ^2_0})^{-1}\alpha \circ (f_{\delta ^2_2})^{-1}\alpha $ as maps of sheaves on $\mathcal{C}_2$.

Proof. We abbreviate $d^ n_ j = f_{\delta ^ n_ j} : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{n - 1})$. The condition on $\alpha $ in the statement of the lemma makes sense because

\[ d^1_1 \circ d^2_2 = d^1_1 \circ d^2_1, \quad d^1_1 \circ d^2_0 = d^1_0 \circ d^2_2, \quad d^1_0 \circ d^2_0 = d^1_0 \circ d^2_1 \]

as morphisms of topoi $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_2) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_0)$, see Simplicial, Remark 14.3.3. Hence we can picture these maps as follows

\[ \xymatrix{ & (d^2_0)^{-1}(d^1_1)^{-1}\mathcal{F} \ar[r]_-{(d^2_0)^{-1}\alpha } & (d^2_0)^{-1}(d^1_0)^{-1}\mathcal{F} \ar@{=}[rd] & \\ (d^2_2)^{-1}(d^1_0)^{-1}\mathcal{F} \ar@{=}[ru] & & & (d^2_1)^{-1}(d^1_0)^{-1}\mathcal{F} \\ & (d^2_2)^{-1}(d^1_1)^{-1}\mathcal{F} \ar[lu]^{(d^2_2)^{-1}\alpha } \ar@{=}[r] & (d^2_1)^{-1}(d^1_1)^{-1}\mathcal{F} \ar[ru]_{(d^2_1)^{-1}\alpha } } \]

and the condition signifies the diagram is commutative. It is clear that given a cartesian sheaf $\mathcal{G}$ of sets (resp. abelian groups) on $\mathcal{C}_{total}$ we can set $\mathcal{F} = \mathcal{G}_0$ and $\alpha $ equal to the composition

\[ (d_1^1)^{-1}\mathcal{G}_0 \to \mathcal{G}_1 \leftarrow (d_1^0)^{-1}\mathcal{G}_0 \]

where the arrows are invertible as $\mathcal{G}$ is cartesian. To prove this functor is an equivalence we construct a quasi-inverse. The construction of the quasi-inverse is analogous to the construction discussed in Descent, Section 35.3 from which we borrow the notation $\tau ^ n_ i : [0] \to [n]$, $0 \mapsto i$ and $\tau ^ n_{ij} : [1] \to [n]$, $0 \mapsto i$, $1 \mapsto j$. Namely, given a pair $(\mathcal{F}, \alpha )$ as in the lemma we set $\mathcal{G}_ n = (f_{\tau ^ n_ n})^{-1}\mathcal{F}$. Given $\varphi : [n] \to [m]$ we define $\mathcal{G}(\varphi ) : (f_\varphi )^{-1}\mathcal{G}_ n \to \mathcal{G}_ m$ using

\[ \xymatrix{ (f_\varphi )^{-1}\mathcal{G}_ n \ar@{=}[r] & (f_\varphi )^{-1}(f_{\tau ^ n_ n})^{-1}\mathcal{F} \ar@{=}[r] & (f_{\tau ^ m_{\varphi (n)}})^{-1}\mathcal{F} \ar@{=}[r] & (f_{\tau ^ m_{\varphi (n)m}})^{-1}(d^1_1)^{-1}\mathcal{F} \ar[d]^{(f_{\tau ^ m_{\varphi (n)m}})^{-1}\alpha } \\ & \mathcal{G}_ m \ar@{=}[r] & (f_{\tau ^ m_ m})^{-1}\mathcal{F} \ar@{=}[r] & (f_{\tau ^ m_{\varphi (n)m}})^{-1}(d^1_0)^{-1}\mathcal{F} } \]

We omit the verification that the commutativity of the displayed diagram above implies the maps compose correctly and hence give rise to a sheaf on $\mathcal{C}_{total}$, see Lemma 83.3.4. We also omit the verification that the two functors are quasi-inverse to each other. $\square$

Lemma 83.12.5. In Situation 83.3.3 let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}_{total}$. The category of cartesian $\mathcal{O}$-modules is equivalent to the category of pairs $(\mathcal{F}, \alpha )$ where $\mathcal{F}$ is a $\mathcal{O}_0$-module and

\[ \alpha : (f_{\delta _1^1})^*\mathcal{F} \longrightarrow (f_{\delta _0^1})^*\mathcal{F} \]

is an isomorphism of $\mathcal{O}_1$-modules such that $(f_{\delta ^2_1})^*\alpha = (f_{\delta ^2_0})^*\alpha \circ (f_{\delta ^2_2})^*\alpha $ as $\mathcal{O}_2$-module maps.

Proof. The proof is identical to the proof of Lemma 83.12.4 with pullback of sheaves of abelian groups replaced by pullback of modules. $\square$

Lemma 83.12.6. In Situation 83.3.3.

  1. The full subcategory of cartesian abelian sheaves forms a weak Serre subcategory of $\textit{Ab}(\mathcal{C}_{total})$. Colimits of systems of cartesian abelian sheaves are cartesian.

  2. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}_{total}$ such that the morphisms

    \[ f_{\delta ^ n_ j} : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n), \mathcal{O}_ n) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{n - 1}), \mathcal{O}_{n - 1}) \]

    are flat. The full subcategory of cartesian $\mathcal{O}$-modules forms a weak Serre subcategory of $\textit{Mod}(\mathcal{O})$. Colimits of systems of cartesian $\mathcal{O}$-modules are cartesian.

Proof. To see we obtain a weak Serre subcategory in (1) we check the conditions listed in Homology, Lemma 12.10.3. First, if $\varphi : \mathcal{F} \to \mathcal{G}$ is a map between cartesian abelian sheaves, then $\mathop{\mathrm{Ker}}(\varphi )$ and $\mathop{\mathrm{Coker}}(\varphi )$ are cartesian too because the restriction functors $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n)$ and the functors $f_\varphi ^{-1}$ are exact. Similarly, if

\[ 0 \to \mathcal{F} \to \mathcal{H} \to \mathcal{G} \to 0 \]

is a short exact sequence of abelian sheaves on $\mathcal{C}_{total}$ with $\mathcal{F}$ and $\mathcal{G}$ cartesian, then it follows that $\mathcal{H}$ is cartesian from the 5-lemma. To see the property of colimits, use that colimits commute with pullback as pullback is a left adjoint. In the case of modules we argue in the same manner, using the exactness of flat pullback (Modules on Sites, Lemma 18.31.2) and the fact that it suffices to check the condition for $f_{\delta ^ n_ j}$, see Lemma 83.12.2. $\square$

Remark 83.12.7 (Warning). Lemma 83.12.6 notwithstanding, it can happen that the category of cartesian $\mathcal{O}$-modules is abelian without being a Serre subcategory of $\textit{Mod}(\mathcal{O})$. Namely, suppose that we only know that $f_{\delta _1^1}$ and $f_{\delta _0^1}$ are flat. Then it follows easily from Lemma 83.12.5 that the category of cartesian $\mathcal{O}$-modules is abelian. But if $f_{\delta _0^2}$ is not flat (for example), there is no reason for the inclusion functor from the category of cartesian $\mathcal{O}$-modules to all $\mathcal{O}$-modules to be exact.

Lemma 83.12.8. In Situation 83.3.3.

  1. An object $K$ of $D(\mathcal{C}_{total})$ is cartesian if and only if $H^ q(K)$ is a cartesian abelian sheaf for all $q$.

  2. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}_{total}$ such that the morphisms $f_{\delta ^ n_ j} : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n), \mathcal{O}_ n) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{n - 1}), \mathcal{O}_{n - 1})$ are flat. Then an object $K$ of $D(\mathcal{O})$ is cartesian if and only if $H^ q(K)$ is a cartesian $\mathcal{O}$-module for all $q$.

Proof. Part (1) is true because the pullback functors $(f_\varphi )^{-1}$ are exact. Part (2) follows from the characterization in Lemma 83.12.2 and the fact that $L(f_{\delta ^ n_ j})^* = (f_{\delta ^ n_ j})^*$ by flatness. $\square$

Lemma 83.12.9. In Situation 83.3.3.

  1. An object $K$ of $D(\mathcal{C}_{total})$ is cartesian if and only the canonical map

    \[ g_{n!}K_ n \longrightarrow g_{n!}\mathbf{Z} \otimes ^\mathbf {L}_\mathbf {Z} K \]

    is an isomorphism for all $n$.

  2. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}_{total}$ such that the morphisms $f_\varphi ^{-1}\mathcal{O}_ n \to \mathcal{O}_ m$ are flat for all $\varphi : [n] \to [m]$. Then an object $K$ of $D(\mathcal{O})$ is cartesian if and only if the canonical map

    \[ g_{n!}K_ n \longrightarrow g_{n!}\mathcal{O}_ n \otimes ^\mathbf {L}_\mathcal {O} K \]

    is an isomorphism for all $n$.

Proof. Proof of (1). Since $g_{n!}$ is exact, it induces a functor on derived categories adjoint to $g_ n^{-1}$. The map is the adjoint of the map $K_ n \to (g_ n^{-1}g_{n!}\mathbf{Z}) \otimes ^\mathbf {L}_\mathbf {Z} K_ n$ corresponding to $\mathbf{Z} \to g_ n^{-1}g_{n!}\mathbf{Z}$ which in turn is adjoint to $\text{id} : g_{n!}\mathbf{Z} \to g_{n!}\mathbf{Z}$. Using the description of $g_{n!}$ given in Lemma 83.3.5 we see that the restriction to $\mathcal{C}_ m$ of this map is

\[ \bigoplus \nolimits _{\varphi : [n] \to [m]} f_\varphi ^{-1}K_ n \longrightarrow \bigoplus \nolimits _{\varphi : [n] \to [m]} K_ m \]

Thus the statement is clear.

Proof of (2). Since $g_{n!}$ is exact (Lemma 83.6.3), it induces a functor on derived categories adjoint to $g_ n^*$ (also exact). The map is the adjoint of the map $K_ n \to (g_ n^*g_{n!}\mathcal{O}_ n) \otimes ^\mathbf {L}_{\mathcal{O}_ n} K_ n$ corresponding to $\mathcal{O}_ n \to g_ n^*g_{n!}\mathcal{O}_ n$ which in turn is adjoint to $\text{id} : g_{n!}\mathcal{O}_ n \to g_{n!}\mathcal{O}_ n$. Using the description of $g_{n!}$ given in Lemma 83.6.1 we see that the restriction to $\mathcal{C}_ m$ of this map is

\[ \bigoplus \nolimits _{\varphi : [n] \to [m]} f_\varphi ^*K_ n \longrightarrow \bigoplus \nolimits _{\varphi : [n] \to [m]} f_\varphi ^*\mathcal{O}_ n \otimes _{\mathcal{O}_ m} K_ m = \bigoplus \nolimits _{\varphi : [n] \to [m]} K_ m \]

Thus the statement is clear. $\square$

Lemma 83.12.10. In Situation 83.3.3 let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}_{total}$. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules. Then $\mathcal{F}$ is quasi-coherent in the sense of Modules on Sites, Definition 18.23.1 if and only if $\mathcal{F}$ is cartesian and $\mathcal{F}_ n$ is a quasi-coherent $\mathcal{O}_ n$-module for all $n$.

Proof. Assume $\mathcal{F}$ is quasi-coherent. Since pullbacks of quasi-coherent modules are quasi-coherent (Modules on Sites, Lemma 18.23.4) we see that $\mathcal{F}_ n$ is a quasi-coherent $\mathcal{O}_ n$-module for all $n$. To show that $\mathcal{F}$ is cartesian, let $U$ be an object of $\mathcal{C}_ n$ for some $n$. Let us view $U$ as an object of $\mathcal{C}_{total}$. Because $\mathcal{F}$ is quasi-coherent there exists a covering $\{ U_ i \to U\} $ and for each $i$ a presentation

\[ \bigoplus \nolimits _{j \in J_ i} \mathcal{O}_{\mathcal{C}_{total}/U_ i} \to \bigoplus \nolimits _{k \in K_ i} \mathcal{O}_{\mathcal{C}_{total}/U_ i} \to \mathcal{F}|_{\mathcal{C}_{total}/U_ i} \to 0 \]

Observe that $\{ U_ i \to U\} $ is a covering of $\mathcal{C}_ n$ by the construction of the site $\mathcal{C}_{total}$. Next, let $V$ be an object of $\mathcal{C}_ m$ for some $m$ and let $V \to U$ be a morphism of $\mathcal{C}_{total}$ lying over $\varphi : [n] \to [m]$. The fibre products $V_ i = V \times _ U U_ i$ exist and we get an induced covering $\{ V_ i \to V\} $ in $\mathcal{C}_ m$. Restricting the presentation above to the sites $\mathcal{C}_ n/U_ i$ and $\mathcal{C}_ m/V_ i$ we obtain presentations

\[ \bigoplus \nolimits _{j \in J_ i} \mathcal{O}_{\mathcal{C}_ m/U_ i} \to \bigoplus \nolimits _{k \in K_ i} \mathcal{O}_{\mathcal{C}_ m/U_ i} \to \mathcal{F}_ n|_{\mathcal{C}_ n/U_ i} \to 0 \]

and

\[ \bigoplus \nolimits _{j \in J_ i} \mathcal{O}_{\mathcal{C}_ m/V_ i} \to \bigoplus \nolimits _{k \in K_ i} \mathcal{O}_{\mathcal{C}_ m/V_ i} \to \mathcal{F}_ m|_{\mathcal{C}_ m/V_ i} \to 0 \]

These presentations are compatible with the map $\mathcal{F}(\varphi ) : f_\varphi ^*\mathcal{F}_ n \to \mathcal{F}_ m$ (as this map is defined using the restriction maps of $\mathcal{F}$ along morphisms of $\mathcal{C}_{total}$ lying over $\varphi $). We conclude that $\mathcal{F}(\varphi )|_{\mathcal{C}_ m/V_ i}$ is an isomorphism. As $\{ V_ i \to V\} $ is a covering we conclude $\mathcal{F}(\varphi )|_{\mathcal{C}_ m/V}$ is an isomorphism. Since $V$ and $U$ were arbitrary this proves that $\mathcal{F}$ is cartesian. (In case A use Sites, Lemma 7.14.10.)

Conversely, assume $\mathcal{F}_ n$ is quasi-coherent for all $n$ and that $\mathcal{F}$ is cartesian. Then for any $n$ and object $U$ of $\mathcal{C}_ n$ we can choose a covering $\{ U_ i \to U\} $ of $\mathcal{C}_ n$ and for each $i$ a presentation

\[ \bigoplus \nolimits _{j \in J_ i} \mathcal{O}_{\mathcal{C}_ m/U_ i} \to \bigoplus \nolimits _{k \in K_ i} \mathcal{O}_{\mathcal{C}_ m/U_ i} \to \mathcal{F}_ n|_{\mathcal{C}_ n/U_ i} \to 0 \]

Pulling back to $\mathcal{C}_{total}/U_ i$ we obtain complexes

\[ \bigoplus \nolimits _{j \in J_ i} \mathcal{O}_{\mathcal{C}_{total}/U_ i} \to \bigoplus \nolimits _{k \in K_ i} \mathcal{O}_{\mathcal{C}_{total}/U_ i} \to \mathcal{F}|_{\mathcal{C}_{total}/U_ i} \to 0 \]

of modules on $\mathcal{C}_{total}/U_ i$. Then the property that $\mathcal{F}$ is cartesian implies that this is exact. We omit the details. $\square$


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