The Stacks project

25.3 Hypercoverings

If we assume our category is a site, then we can make the following definition.

Definition 25.3.1. Let $\mathcal{C}$ be a site. Let $f = (\alpha , f_ i) : \{ U_ i\} _{i \in I} \to \{ V_ j\} _{j \in J}$ be a morphism in the category $\text{SR}(\mathcal{C})$. We say that $f$ is a covering if for every $j \in J$ the family of morphisms $\{ U_ i \to V_ j\} _{i \in I, \alpha (i) = j}$ is a covering for the site $\mathcal{C}$. Let $X$ be an object of $\mathcal{C}$. A morphism $K \to L$ in $\text{SR}(\mathcal{C}, X)$ is a covering if its image in $\text{SR}(\mathcal{C})$ is a covering.

Lemma 25.3.2. Let $\mathcal{C}$ be a site.

  1. A composition of coverings in $\text{SR}(\mathcal{C})$ is a covering.

  2. If $K \to L$ is a covering in $\text{SR}(\mathcal{C})$ and $L' \to L$ is a morphism, then $L' \times _ L K$ exists and $L' \times _ L K \to L'$ is a covering.

  3. If $\mathcal{C}$ has products of pairs, and $A \to B$ and $K \to L$ are coverings in $\text{SR}(\mathcal{C})$, then $A \times K \to B \times L$ is a covering.

Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Then (1) and (2) holds for $\text{SR}(\mathcal{C}, X)$ and (3) holds if $\mathcal{C}$ has fibre products.

Proof. Part (1) is immediate from the axioms of a site. Part (2) follows by the construction of fibre products in $\text{SR}(\mathcal{C})$ in the proof of Lemma 25.2.3 and the requirement that the morphisms in a covering of $\mathcal{C}$ are representable. Part (3) follows by thinking of $A \times K \to B \times L$ as the composition $A \times K \to B \times K \to B \times L$ and hence a composition of basechanges of coverings. The final statement follows because $\text{SR}(\mathcal{C}, X) = \text{SR}(\mathcal{C}/X)$. $\square$

By Lemma 25.2.3 and Simplicial, Lemma 14.19.2 the coskeleton of a truncated simplicial object of $\text{SR}(\mathcal{C}, X)$ exists if $\mathcal{C}$ has fibre products. Hence the following definition makes sense.

Definition 25.3.3. Let $\mathcal{C}$ be a site. Assume $\mathcal{C}$ has fibre products. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be an object of $\mathcal{C}$. A hypercovering of $X$ is a simplicial object $K$ of $\text{SR}(\mathcal{C}, X)$ such that

  1. The object $K_0$ is a covering of $X$ for the site $\mathcal{C}$.

  2. For every $n \geq 0$ the canonical morphism

    \[ K_{n + 1} \longrightarrow (\text{cosk}_ n \text{sk}_ n K)_{n + 1} \]

    is a covering in the sense defined above.

Condition (1) makes sense since each object of $\text{SR}(\mathcal{C}, X)$ is after all a family of morphisms with target $X$. It could also be formulated as saying that the morphism of $K_0$ to the final object of $\text{SR}(\mathcal{C}, X)$ is a covering.

Example 25.3.4 (Čech hypercoverings). Let $\mathcal{C}$ be a site with fibre products. Let $\{ U_ i \to X\} _{i \in I}$ be a covering of $\mathcal{C}$. Set $K_0 = \{ U_ i \to X\} _{i \in I}$. Then $K_0$ is a $0$-truncated simplicial object of $\text{SR}(\mathcal{C}, X)$. Hence we may form

\[ K = \text{cosk}_0 K_0. \]

Clearly $K$ passes condition (1) of Definition 25.3.3. Since all the morphisms $K_{n + 1} \to (\text{cosk}_ n \text{sk}_ n K)_{n + 1}$ are isomorphisms by Simplicial, Lemma 14.19.10 it also passes condition (2). Note that the terms $K_ n$ are the usual

\[ K_ n = \{ U_{i_0} \times _ X U_{i_1} \times _ X \ldots \times _ X U_{i_ n} \to X \} _{(i_0, i_1, \ldots , i_ n) \in I^{n + 1}} \]

A hypercovering of $X$ of this form is called a Čech hypercovering of $X$.

Example 25.3.5 (Hypercovering by a simplicial object of the site). Let $\mathcal{C}$ be a site with fibre products. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Let $U$ be a simplicial object of $\mathcal{C}$. As usual we denote $U_ n = U([n])$. Finally, assume given an augmentation

\[ a : U \to X \]

In this situation we can consider the simplicial object $K$ of $\text{SR}(\mathcal{C}, X)$ with terms $K_ n = \{ U_ n \to X\} $. Then $K$ is a hypercovering of $X$ in the sense of Definition 25.3.3 if and only if the following three conditions1 hold:

  1. $\{ U_0 \to X\} $ is a covering of $\mathcal{C}$,

  2. $\{ U_1 \to U_0 \times _ X U_0\} $ is a covering of $\mathcal{C}$,

  3. $\{ U_{n + 1} \to (\text{cosk}_ n\text{sk}_ n U)_{n + 1}\} $ is a covering of $\mathcal{C}$ for $n \geq 1$.

We omit the straightforward verification.

Example 25.3.6 (Čech hypercovering associated to a cover). Let $\mathcal{C}$ be a site with fibre products. Let $U \to X$ be a morphism of $\mathcal{C}$ such that $\{ U \to X\} $ is a covering of $\mathcal{C}$2. Consider the simplical object $K$ of $\text{SR}(\mathcal{C}, X)$ with terms

\[ K_ n = \{ U \times _ X U \times _ X \ldots \times _ X U \to X\} \quad (n + 1 \text{ factors}) \]

Then $K$ is a hypercovering of $X$. This example is a special case of both Example 25.3.4 and of Example 25.3.5.

Lemma 25.3.7. Let $\mathcal{C}$ be a site with fibre products. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be an object of $\mathcal{C}$. The collection of all hypercoverings of $X$ forms a set.

Proof. Since $\mathcal{C}$ is a site, the set of all coverings of $X$ forms a set. Thus we see that the collection of possible $K_0$ forms a set. Suppose we have shown that the collection of all possible $K_0, \ldots , K_ n$ form a set. Then it is enough to show that given $K_0, \ldots , K_ n$ the collection of all possible $K_{n + 1}$ forms a set. And this is clearly true since we have to choose $K_{n + 1}$ among all possible coverings of $(\text{cosk}_ n \text{sk}_ n K)_{n + 1}$. $\square$

Remark 25.3.8. The lemma does not just say that there is a cofinal system of choices of hypercoverings that is a set, but that really the hypercoverings form a set.

The category of presheaves on $\mathcal{C}$ has finite (co)limits. Hence the functors $\text{cosk}_ n$ exists for presheaves of sets.

Lemma 25.3.9. Let $\mathcal{C}$ be a site with fibre products. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be an object of $\mathcal{C}$. Let $K$ be a hypercovering of $X$. Consider the simplicial object $F(K)$ of $\textit{PSh}(\mathcal{C})$, endowed with its augmentation to the constant simplicial presheaf $h_ X$.

  1. The morphism of presheaves $F(K)_0 \to h_ X$ becomes a surjection after sheafification.

  2. The morphism

    \[ (d^1_0, d^1_1) : F(K)_1 \longrightarrow F(K)_0 \times _{h_ X} F(K)_0 \]

    becomes a surjection after sheafification.

  3. For every $n \geq 1$ the morphism

    \[ F(K)_{n + 1} \longrightarrow (\text{cosk}_ n \text{sk}_ n F(K))_{n + 1} \]

    turns into a surjection after sheafification.

Proof. We will use the fact that if $\{ U_ i \to U\} _{i \in I}$ is a covering of the site $\mathcal{C}$, then the morphism

\[ \amalg _{i \in I} h_{U_ i} \to h_ U \]

becomes surjective after sheafification, see Sites, Lemma 7.12.4. Thus the first assertion follows immediately.

For the second assertion, note that according to Simplicial, Example 14.19.1 the simplicial object $\text{cosk}_0 \text{sk}_0 K$ has terms $K_0 \times \ldots \times K_0$. Thus according to the definition of a hypercovering we see that $(d^1_0, d^1_1) : K_1 \to K_0 \times K_0$ is a covering. Hence (2) follows from the claim above and the fact that $F$ transforms products into fibred products over $h_ X$.

For the third, we claim that $\text{cosk}_ n \text{sk}_ n F(K) = F(\text{cosk}_ n \text{sk}_ n K)$ for $n \geq 1$. To prove this, denote temporarily $F'$ the functor $\text{SR}(\mathcal{C}, X) \to \textit{PSh}(\mathcal{C})/h_ X$. By Lemma 25.2.3 the functor $F'$ commutes with finite limits. By our description of the $\text{cosk}_ n$ functor in Simplicial, Section 14.12 we see that $\text{cosk}_ n \text{sk}_ n F'(K) = F'(\text{cosk}_ n \text{sk}_ n K)$. Recall that the category used in the description of $(\text{cosk}_ n U)_ m$ in Simplicial, Lemma 14.19.2 is the category $(\Delta /[m])^{opp}_{\leq n}$. It is an amusing exercise to show that $(\Delta /[m])_{\leq n}$ is a connected category (see Categories, Definition 4.16.1) as soon as $n \geq 1$. Hence, Categories, Lemma 4.16.2 shows that $\text{cosk}_ n \text{sk}_ n F'(K) = \text{cosk}_ n \text{sk}_ n F(K)$. Whence the claim. Property (2) follows from this, because now we see that the morphism in (2) is the result of applying the functor $F$ to a covering as in Definition 25.3.1, and the result follows from the first fact mentioned in this proof. $\square$

[1] As $\mathcal{C}$ has fibre products, the category $\mathcal{C}/X$ has all finite limits. Hence the required coskeleta exist by Simplicial, Lemma 14.19.2.
[2] A morphism of $\mathcal{C}$ with this property is sometimes called a “cover”.

Comments (3)

Comment #478 by a on

Def. 24.2.2 say associates a "sheaf", but does one mean pre sheaf?

Lemma 24.2.3 proof of part 1 says the coproduct of and is ... what does this last expression mean? Should it be ?

proof of part 3 of same lemma: line 222 of the code says but should it be ?

line 289: missing right parenthesis )

In tag 017Z the definition of coskelet says it goes from but I think it should go the other way.

Comment #490 by on

Thanks very much! Made the corresponding edits here.

If you'd like to be listed among the contributors, then please sign off your comments with your name.

Comment #1024 by correction_bot on

In the comment before Definition 24.2.6 (tag 01G5), perhaps note that the existence of coskeleton functors for in case has fibre products follows from Lemma 24.2.3 (tag 01G2) and Lemma 14.17.3 (tag 0183).

Directly after Definition 24.2.6, in the sentence "Condition (2) makes sense since…" it looks like (2) should be changed to (1).

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