Definition 24.3.1. 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.

## 24.3 Hypercoverings

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

Lemma 24.3.2. Let $\mathcal{C}$ be a site.

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

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.

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 24.2.3 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 24.2.3 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 24.3.3. 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

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

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 24.3.4. Let $\{ U_ i \to X\} _{i \in I}$ be a covering of the site $\mathcal{C}$. 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

Clearly $K$ passes condition (1) of Definition 24.3.3. 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

Lemma 24.3.5. 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 24.3.6. 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 24.3.7. Let $\mathcal{C}$ be a site with fibre products. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be an object of $\mathcal{C}$. 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 $h_ X$.

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

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.

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

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 24.2.3 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 24.3.1, and the result follows from the first fact mentioned in this proof. $\square$

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