The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

24.7 Covering hypercoverings

Here are some ways to construct hypercoverings. We note that since the category $\text{SR}(\mathcal{C}, X)$ has fibre products the category of simplicial objects of $\text{SR}(\mathcal{C}, X)$ has fibre products as well, see Simplicial, Lemma 14.7.2.

Lemma 24.7.1. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $K, L, M$ be simplicial objects of $\text{SR}(\mathcal{C}, X)$. Let $a : K \to L$, $b : M \to L$ be morphisms. Assume

  1. $K$ is a hypercovering of $X$,

  2. the morphism $M_0 \to L_0$ is a covering, and

  3. for all $n \geq 0$ in the diagram

    \[ \xymatrix{ M_{n + 1} \ar[dd] \ar[rr] \ar[rd]^\gamma & & (\text{cosk}_ n \text{sk}_ n M)_{n + 1} \ar[dd] \\ & L_{n + 1} \times _{(\text{cosk}_ n \text{sk}_ n L)_{n + 1}} (\text{cosk}_ n \text{sk}_ n M)_{n + 1} \ar[ld] \ar[ru] & \\ L_{n + 1} \ar[rr] & & (\text{cosk}_ n \text{sk}_ n L)_{n + 1} } \]

    the arrow $\gamma $ is a covering.

Then the fibre product $K \times _ L M$ is a hypercovering of $X$.

Proof. The morphism $(K \times _ L M)_0 = K_0 \times _{L_0} M_0 \to K_0$ is a base change of a covering by (2), hence a covering, see Lemma 24.3.2. And $K_0 \to \{ X \to X\} $ is a covering by (1). Thus $(K \times _ L M)_0 \to \{ X \to X\} $ is a covering by Lemma 24.3.2. Hence $K \times _ L M$ satisfies the first condition of Definition 24.3.3.

We still have to check that

\[ K_{n + 1} \times _{L_{n + 1}} M_{n + 1} = (K \times _ L M)_{n + 1} \longrightarrow (\text{cosk}_ n \text{sk}_ n (K \times _ L M))_{n + 1} \]

is a covering for all $n \geq 0$. We abbreviate as follows: $A = (\text{cosk}_ n \text{sk}_ n K)_{n + 1}$, $B = (\text{cosk}_ n \text{sk}_ n L)_{n + 1}$, and $C = (\text{cosk}_ n \text{sk}_ n M)_{n + 1}$. The functor $\text{cosk}_ n \text{sk}_ n$ commutes with fibre products, see Simplicial, Lemma 14.19.13. Thus the right hand side above is equal to $A \times _ B C$. Consider the following commutative diagram

\[ \xymatrix{ K_{n + 1} \times _{L_{n + 1}} M_{n + 1} \ar[r] \ar[d] & M_{n + 1} \ar[d] \ar[rd]_\gamma \ar[rrd] & & \\ K_{n + 1} \ar[r] \ar[rd] & L_{n + 1} \ar[rrd] & L_{n + 1} \times _ B C \ar[l] \ar[r] & C \ar[d] \\ & A \ar[rr] & & B } \]

This diagram shows that

\[ K_{n + 1} \times _{L_{n + 1}} M_{n + 1} = (K_{n + 1} \times _ B C) \times _{(L_{n + 1} \times _ B C), \gamma } M_{n + 1} \]

Now, $K_{n + 1} \times _ B C \to A \times _ B C$ is a base change of the covering $K_{n + 1} \to A$ via the morphism $A \times _ B C \to A$, hence is a covering. By assumption (3) the morphism $\gamma $ is a covering. Hence the morphism

\[ (K_{n + 1} \times _ B C) \times _{(L_{n + 1} \times _ B C), \gamma } M_{n + 1} \longrightarrow K_{n + 1} \times _ B C \]

is a covering as a base change of a covering. The lemma follows as a composition of coverings is a covering. $\square$

Lemma 24.7.2. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. If $K, L$ are hypercoverings of $X$, then $K \times L$ is a hypercovering of $X$.

Proof. You can either verify this directly, or use Lemma 24.7.1 above and check that $L \to \{ X \to X\} $ has property (3). $\square$

Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Since the category $\text{SR}(\mathcal{C}, X)$ has coproducts and finite limits, it is permissible to speak about the objects $U \times K$ and $\mathop{\mathrm{Hom}}\nolimits (U, K)$ for certain simplicial sets $U$ (for example those with finitely many nondegenerate simplices) and any simplicial object $K$ of $\text{SR}(\mathcal{C}, X)$. See Simplicial, Sections 14.13 and 14.17.

Lemma 24.7.3. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $K$ be a hypercovering of $X$. Let $k \geq 0$ be an integer. Let $u : Z \to K_ k$ be a covering in $\text{SR}(\mathcal{C}, X)$. Then there exists a morphism of hypercoverings $f: L \to K$ such that $L_ k \to K_ k$ factors through $u$.

Proof. Denote $Y = K_ k$. Let $C[k]$ be the cosimplicial set defined in Simplicial, Example 14.5.6. We will use the description of $\mathop{\mathrm{Hom}}\nolimits (C[k], Y)$ and $\mathop{\mathrm{Hom}}\nolimits (C[k], Z)$ given in Simplicial, Lemma 14.15.2. There is a canonical morphism $K \to \mathop{\mathrm{Hom}}\nolimits (C[k], Y)$ corresponding to $\text{id} : K_ k = Y \to Y$. Consider the morphism $\mathop{\mathrm{Hom}}\nolimits (C[k], Z) \to \mathop{\mathrm{Hom}}\nolimits (C[k], Y)$ which on degree $n$ terms is the morphism

\[ \prod \nolimits _{\alpha : [k] \to [n]} Z \longrightarrow \prod \nolimits _{\alpha : [k] \to [n]} Y \]

using the given morphism $Z \to Y$ on each factor. Set

\[ L = K \times _{\mathop{\mathrm{Hom}}\nolimits (C[k], Y)} \mathop{\mathrm{Hom}}\nolimits (C[k], Z). \]

The morphism $L_ k \to K_ k$ sits in to a commutative diagram

\[ \xymatrix{ L_ k \ar[r] \ar[d] & \prod _{\alpha : [k] \to [k]} Z \ar[r]^-{\text{pr}_{\text{id}_{[k]}}} \ar[d] & Z \ar[d] \\ K_ k \ar[r] & \prod _{\alpha : [k] \to [k]} Y \ar[r]^-{\text{pr}_{\text{id}_{[k]}}} & Y } \]

Since the composition of the two bottom arrows is the identity we conclude that we have the desired factorization.

We still have to show that $L$ is a hypercovering of $X$. To see this we will use Lemma 24.7.1. Condition (1) is satisfied by assumption. For (2), the morphism

\[ \mathop{\mathrm{Hom}}\nolimits (C[k], Z)_0 \to \mathop{\mathrm{Hom}}\nolimits (C[k], Y)_0 \]

is a covering because it is isomorphic to $Z \to Y$ as there is only one morphism $[k] \to [0]$.

Let us consider condition (3) for $n = 0$. Then, since $(\text{cosk}_0 T)_1 = T \times T$ (Simplicial, Example 14.19.1) and since $\mathop{\mathrm{Hom}}\nolimits (C[k], Z)_1 = \prod _{\alpha : [k] \to [1]} Z$ we obtain the diagram

\[ \xymatrix{ \prod \nolimits _{\alpha : [k] \to [1]} Z \ar[r] \ar[d] & Z \times Z \ar[d] \\ \prod \nolimits _{\alpha : [k] \to [1]} Y \ar[r] & Y \times Y } \]

with horizontal arrows corresponding to the projection onto the factors corresponding to the two nonsurjective $\alpha $. Thus the arrow $\gamma $ is the morphism

\[ \prod \nolimits _{\alpha : [k] \to [1]} Z \longrightarrow \prod \nolimits _{\alpha : [k] \to [1]\text{ not onto}} Z \times \prod \nolimits _{\alpha : [k] \to [1]\text{ onto}} Y \]

which is a product of coverings and hence a covering by Lemma 24.3.2.

Let us consider condition (3) for $n > 0$. We claim there is an injective map $\tau : S' \to S$ of finite sets, such that for any object $T$ of $\text{SR}(\mathcal{C}, X)$ the morphism

24.7.3.1
\begin{equation} \label{hypercovering-equation-map} \mathop{\mathrm{Hom}}\nolimits (C[k], T)_{n + 1} \to (\text{cosk}_ n \text{sk}_ n \mathop{\mathrm{Hom}}\nolimits (C[k], T))_{n + 1} \end{equation}

is isomorphic to the projection $\prod _{s \in S} T \to \prod _{s' \in S'} T$ functorially in $T$. If this is true, then we see, arguing as in the previous paragraph, that the arrow $\gamma $ is the morphism

\[ \prod \nolimits _{s \in S} Z \longrightarrow \prod \nolimits _{s \in S'} Z \times \prod \nolimits _{s \not\in \tau (S')} Y \]

which is a product of coverings and hence a covering by Lemma 24.3.2. By construction, we have $\mathop{\mathrm{Hom}}\nolimits (C[k], T)_{n + 1} = \prod _{\alpha : [k] \to [n + 1]} T$ (see Simplicial, Lemma 14.15.2). Correspondingly we take $S = \text{Map}([k], [n + 1])$. On the other hand, Simplicial, Lemma 14.19.5, provides a description of points of $(\text{cosk}_ n \text{sk}_ n \mathop{\mathrm{Hom}}\nolimits (C[k], T))_{n + 1}$ as sequences $(f_0, \ldots , f_{n + 1})$ of points of $\mathop{\mathrm{Hom}}\nolimits (C[k], T)_ n$ satisfying $d^ n_{j - 1} f_ i = d^ n_ i f_ j$ for $0 \leq i < j \leq n + 1$. We can write $f_ i = (f_{i, \alpha })$ with $f_{i, \alpha }$ a point of $T$ and $\alpha \in \text{Map}([k], [n])$. The conditions translate into

\[ f_{i, \delta ^ n_{j - 1} \circ \beta } = f_{j, \delta _ i^ n \circ \beta } \]

for any $0 \leq i < j \leq n + 1$ and $\beta : [k] \to [n - 1]$. Thus we see that

\[ S' = \{ 0, \ldots , n + 1\} \times \text{Map}([k], [n]) / \sim \]

where the equivalence relation is generated by the equivalences

\[ (i, \delta ^ n_{j - 1} \circ \beta ) \sim (j, \delta _ i^ n \circ \beta ) \]

for $0 \leq i < j \leq n + 1$ and $\beta : [k] \to [n - 1]$. A computation (omitted) shows that the morphism (24.7.3.1) corresponds to the map $S' \to S$ which sends $(i, \alpha )$ to $\delta ^{n + 1}_ i \circ \alpha \in S$. (It may be a comfort to the reader to see that this map is well defined by part (1) of Simplicial, Lemma 14.2.3.) To finish the proof it suffices to show that if $\alpha , \alpha ' : [k] \to [n]$ and $0 \leq i < j \leq n + 1$ are such that

\[ \delta ^{n + 1}_ i \circ \alpha = \delta ^{n + 1}_ j \circ \alpha ' \]

then we have $\alpha = \delta ^ n_{j - 1} \circ \beta $ and $\alpha ' = \delta _ i^ n \circ \beta $ for some $\beta : [k] \to [n - 1]$. This is easy to see and omitted. $\square$

Lemma 24.7.4. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $K$ be a hypercovering of $X$. Let $n \geq 0$ be an integer. Let $u : \mathcal{F} \to F(K_ n)$ be a morphism of presheaves which becomes surjective on sheafification. Then there exists a morphism of hypercoverings $f: L \to K$ such that $F(f_ n) : F(L_ n) \to F(K_ n)$ factors through $u$.

Proof. Write $K_ n = \{ U_ i \to X\} _{i \in I}$. Thus the map $u$ is a morphism of presheaves of sets $u : \mathcal{F} \to \amalg h_{u_ i}$. The assumption on $u$ means that for every $i \in I$ there exists a covering $\{ U_{ij} \to U_ i\} _{j \in I_ i}$ of the site $\mathcal{C}$ and a morphism of presheaves $t_{ij} : h_{U_{ij}} \to \mathcal{F}$ such that $u \circ t_{ij}$ is the map $h_{U_{ij}} \to h_{U_ i}$ coming from the morphism $U_{ij} \to U_ i$. Set $J = \amalg _{i \in I} I_ i$, and let $\alpha : J \to I$ be the obvious map. For $j \in J$ denote $V_ j = U_{\alpha (j)j}$. Set $Z = \{ V_ j \to X\} _{j \in J}$. Finally, consider the morphism $u' : Z \to K_ n$ given by $\alpha : J \to I$ and the morphisms $V_ j = U_{\alpha (j)j} \to U_{\alpha (j)}$ above. Clearly, this is a covering in the category $\text{SR}(\mathcal{C}, X)$, and by construction $F(u') : F(Z) \to F(K_ n)$ factors through $u$. Thus the result follows from Lemma 24.7.3 above. $\square$


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