## 24.12 Constructing hypercoverings

Let $\mathcal{C}$ be a site. In this section we will think of a simplicial object of $\text{SR}(\mathcal{C})$ as follows. As usual, we set $K_ n = K([n])$ and we denote $K(\varphi ) : K_ n \to K_ m$ the morphism associated to $\varphi : [m] \to [n]$. We may write $K_ n = \{ U_{n, i}\} _{i \in I_ n}$. For $\varphi : [m] \to [n]$ the morphism $K(\varphi ) : K_ n \to K_ m$ is given by a map $\alpha (\varphi ) : I_ n \to I_ m$ and morphisms $f_{\varphi , i} : U_{n, i} \to U_{m, \alpha (\varphi )(i)}$ for $i \in I_ n$. The fact that $K$ is a simplicial object of $\text{SR}(\mathcal{C})$ implies that $(I_ n, \alpha (\varphi ))$ is a simplicial set and that $f_{\psi , \alpha (\varphi )(i)} \circ f_{\varphi , i} = f_{\varphi \circ \psi , i}$ when $\psi : [l] \to [m]$.

Lemma 24.12.1. Let $\mathcal{C}$ be a site. Let $K$ be an $r$-truncated simplicial object of $\text{SR}(\mathcal{C})$. The following are equivalent

$K$ is split (Simplicial, Definition 14.18.1),

$f_{\varphi , i} : U_{n, i} \to U_{m, \alpha (\varphi )(i)}$ is an isomorphism for $r \geq n \geq 0$, $\varphi : [m] \to [n]$ surjective, $i \in I_ n$, and

$f_{\sigma ^ n_ j, i} : U_{n, i} \to U_{n + 1, \alpha (\sigma ^ n_ j)(i)}$ is an isomorphism for $0 \leq j \leq n < r$, $i \in I_ n$.

The same holds for simplicial objects if in (2) and (3) we set $r = \infty $.

**Proof.**
The splitting of a simplicial set is unique and is given by the nondegenerate indices $N(I_ n)$ in each degree $n$, see Simplicial, Lemma 14.18.2. The coproduct of two objects $\{ U_ i\} _{i \in I}$ and $\{ U_ j\} _{j \in J}$ of $\text{SR}(\mathcal{C})$ is given by $\{ U_ l\} _{l \in I \amalg J}$ with obvious notation. Hence a splitting of $K$ must be given by $N(K_ n) = \{ U_ i\} _{i \in N(I_ n)}$. The equivalence of (1) and (2) now follows by unwinding the definitions. The equivalence of (2) and (3) follows from the fact that any surjection $\varphi : [m] \to [n]$ is a composition of morphisms $\sigma ^ k_ j$ with $k = n, n + 1, \ldots , m - 1$.
$\square$

Lemma 24.12.2. Let $\mathcal{C}$ be a site with fibre products. Let $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be a subset. Assume

any object $U$ of $\mathcal{C}$ has a covering $\{ U_ j \to U\} _{j \in J}$ with $U_ j \in \mathcal{B}$, and

if $\{ U_ j \to U\} _{j \in J}$ is a covering with $U_ j \in \mathcal{B}$ and $\{ U' \to U\} $ is a morphism with $U' \in \mathcal{B}$, then $\{ U_ j \to U\} _{j \in J} \amalg \{ U' \to U\} $ is a covering.

Then for any $X$ in $\mathcal{C}$ there is a hypercovering $K$ of $X$ such that $K_ n = \{ U_{n, i}\} _{i \in I_ n}$ with $U_{n, i} \in \mathcal{B}$ for all $i \in I_ n$.

**Proof.**
A warmup for this proof is the proof of Lemma 24.11.3 and we encourage the reader to read that proof first.

First we replace $\mathcal{C}$ by the site $\mathcal{C}/X$. After doing so we may assume that $X$ is the final object of $\mathcal{C}$ and that $\mathcal{C}$ has all finite limits (Categories, Lemma 4.18.4).

Let $n \geq 0$. Let us say that an *$n$-truncated $\mathcal{B}$-hypercovering of $X$* is given by an $n$-truncated simplicial object $K$ of $\text{SR}(\mathcal{C})$ such that for $i \in I_ a$, $0 \leq a \leq n$ we have $U_{a, i} \in \mathcal{B}$ and such that $K_0$ is a covering of $X$ and $K_{a + 1} \to (\text{cosk}_ a \text{sk}_ a K)_{a + 1}$ for $a = 0, \ldots , n - 1$ is a covering as in Definition 24.3.1.

Since $X$ has a covering $\{ U_{0, i} \to X\} _{i \in I_0}$ with $U_ i \in \mathcal{B}$ by assumption, we get a $0$-truncated $\mathcal{B}$-hypercovering of $X$. Observe that any $0$-truncated $\mathcal{B}$-hypercovering of $X$ is split, see Lemma 24.12.1.

The lemma follows if we can prove for $n \geq 0$ that given a split $n$-truncated $\mathcal{B}$-hypercovering $K$ of $X$ we can extend it to a split $(n + 1)$-truncated $\mathcal{B}$-hypercovering of $X$.

Construction of the extension. Consider the $(n + 1)$-truncated simplicial object $K' = \text{sk}_{n + 1}(\text{cosk}_ n K)$ of $\text{SR}(\mathcal{C})$. Write

\[ K'_{n + 1} = \{ U'_{n + 1, i}\} _{i \in I'_{n + 1}} \]

Since $K = \text{sk}_ n K'$ we have $K_ a = K'_ a$ for $0 \leq a \leq n$. For every $i' \in I'_{n + 1}$ we choose a covering

24.12.2.1
\begin{equation} \label{hypercovering-equation-choose-covering-B} \{ g_{n + 1, j} : U_{n + 1, j} \to U'_{n + 1, i'}\} _{j \in J_{i'}} \end{equation}

with $U_{n + 1, j} \in \mathcal{B}$ for $j \in J_{i'}$. This is possible by our assumption on $\mathcal{B}$ in the lemma. For $0 \leq m \leq n$ denote $N_ m \subset I_ m$ the subset of nondegenerate indices. We set

\[ I_{n + 1} = \coprod \nolimits _{\varphi : [n + 1] \to [m]\text{ surjective, }0\leq m \leq n} N_ m \amalg \coprod \nolimits _{i' \in I'_{n + 1}} J_{i'} \]

For $j \in I_{n + 1}$ we set

\[ U_{n + 1, j} = \left\{ \begin{matrix} U_{m, i}
& \text{if}
& j = (\varphi , i)
& \text{where}
& \varphi : [n + 1] \to [m], i \in N_ m
\\ U_{n + 1, j}
& \text{if}
& j \in J_{i'}
& \text{where}
& i' \in I'_{n + 1}
\end{matrix} \right. \]

with obvious notation. We set $K_{n + 1} = \{ U_{n + 1, j}\} _{j \in I_{n + 1}}$. By construction $U_{n + 1, j}$ is an element of $\mathcal{B}$ for all $j \in I_{n + 1}$. Let us define compatible maps

\[ I_{n + 1} \to I'_{n + 1} \quad \text{and}\quad K_{n + 1} \to K'_{n + 1} \]

Namely, the first map is given by $(\varphi , i) \mapsto \alpha '(\varphi )(i)$ and $(j \in J_{i'}) \mapsto i'$. For the second map we use the morphisms

\[ f'_{\varphi , i} : U_{m, i} \to U'_{n + 1, \alpha '(\varphi )(i)} \quad \text{and}\quad g_{n + 1, j} : U_{n + 1, j} \to U'_{n + 1, i'} \]

We claim the morphism

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

is a covering as in Definition 24.3.1. Namely, if $i' \in I'_{n + 1}$, then either $i'$ is nondegenerate and the inverse image of $i'$ in $I_{n + 1}$ is equal to $J_{i'}$ and we get a covering of $U'_{n + 1, i'}$ by our choice (24.12.2.1), or $i'$ is degenerate and the inverse image of $i'$ in $I_{n + 1}$ is $J_{i'} \amalg \{ (\varphi , i)\} $ for a unique pair $(\varphi , i)$ and we get a covering by our choice (24.12.2.1) and assumption (2) of the lemma.

To finish the proof we have to define the morphisms $K(\varphi ) : K_{n + 1} \to K_ m$ corresponding to morphisms $\varphi : [m] \to [n + 1]$, $0 \leq m \leq n$ and the morphisms $K(\varphi ) : K_ m \to K_{n + 1}$ corresponding to morphisms $\varphi : [n + 1] \to [m]$, $0 \leq m \leq n$ satisfying suitable composition relations. For the first kind we use the composition

\[ K_{n + 1} \to K'_{n + 1} \xrightarrow {K'(\varphi )} K'_ m = K_ m \]

to define $K(\varphi ) : K_{n + 1} \to K_ m$. For the second kind, suppose given $\varphi : [n + 1] \to [m]$, $0 \leq m \leq n$. We define the corresponding morphism $K(\varphi ) : K_ m \to K_{n + 1}$ as follows:

for $i \in I_ m$ there is a unique surjective map $\psi : [m] \to [m_0]$ and a unique $i_0 \in I_{m_0}$ nondegenerate such that $\alpha (\psi )(i_0) = i$^{1},

we set $\varphi _0 = \psi _0 \circ \varphi : [n + 1] \to [m_0]$ and we map $i \in I_ m$ to $(\varphi _0, i_0) \in I_{n + 1}$, in other words, $\alpha (\varphi )(i) = (\varphi _0, i_0)$, and

the morphism $f_{\varphi , i} : U_{m, i} \to U_{n + 1, \alpha (\varphi )(i)} = U_{m_0, i_0}$ is the inverse of the isomorphism $f_{\psi , i_0} : U_{m_0, i_0} \to U_{m, i}$ (see Lemma 24.12.1).

We omit the straightforward but cumbersome verification that this defines a split $(n + 1)$-truncated $\mathcal{B}$-hypercovering of $X$ extending the given $n$-truncated one. In fact, everything is clear from the above, except for the verification that the morphisms $K(\varphi )$ compose correctly for all $\varphi : [a] \to [b]$ with $0 \leq a, b \leq n + 1$.
$\square$

Lemma 24.12.3. Let $\mathcal{C}$ be a site with equalizers and fibre products. Let $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be a subset. Assume that any object of $\mathcal{C}$ has a covering whose members are elements of $\mathcal{B}$. Then there is a hypercovering $K$ such that $K_ n = \{ U_ i\} _{i \in I_ n}$ with $U_ i \in \mathcal{B}$ for all $i \in I_ n$.

**Proof.**
This proof is almost the same as the proof of Lemma 24.12.2. We will only explain the differences.

Let $n \geq 1$. Let us say that an *$n$-truncated $\mathcal{B}$-hypercovering* is given by an $n$-truncated simplicial object $K$ of $\text{SR}(\mathcal{C})$ such that for $i \in I_ a$, $0 \leq a \leq n$ we have $U_{a, i} \in \mathcal{B}$ and such that

$F(K_0)^\# \to *$ is surjective,

$F(K_1)^\# \to F(K_0)^\# \times F(K_0)^\# $ is surjective,

$F(K_{a + 1})^\# \to F((\text{cosk}_ a \text{sk}_ a K)_{a + 1})^\# $ for $a = 1, \ldots , n - 1$ is surjective.

We first explicitly construct a split $1$-truncated $\mathcal{B}$-hypercovering.

Take $I_0 = \mathcal{B}$ and $K_0 = \{ U\} _{U \in \mathcal{B}}$. Then (1) holds by our assumption on $\mathcal{B}$. Set

\[ \Omega = \{ (U, V, W, a, b) \mid U, V, W \in \mathcal{B}, a : U \to V, b : U \to W\} \]

Then we set $I_1 = I_0 \amalg \Omega $. For $i \in I_1$ we set $U_{1, i} = U_{0, i}$ if $i \in I_0$ and $U_{1, i} = U$ if $i = (U, V, W, a, b) \in \Omega $. The map $K(\sigma ^0_0) : K_0 \to K_1$ corresponds to the inclusion $\alpha (\sigma ^0_0) : I_0 \to I_1$ and the identity $f_{\sigma ^0_0, i} : U_{0, i} \to U_{1, i}$ on objects. The maps $K(\delta ^1_0), K(\delta ^1_1) : K_1 \to K_0$ correspond to the two maps $I_1 \to I_0$ which are the identity on $I_0 \subset I_1$ and map $(U, V, W, a, b) \in \Omega \subset I_1$ to $V$, resp. $W$. The corresponding morphisms $f_{\delta ^1_0, i}, f_{\delta ^1_1, i} : U_{1, i} \to U_{0, i}$ are the identity if $i \in I_0$ and $a, b$ in case $i = (U, V, W, a, b) \in \Omega $. The reason that (2) holds is that any section of $F(K_0)^\# \times F(K_0)^\# $ over an object $U$ of $\mathcal{C}$ comes, after replacing $U$ by the members of a covering, from a map $U \to F(K_0) \times F(K_0)$. This in turn means we have $V, W \in \mathcal{B}$ and two morphisms $U \to V$ and $U \to W$. Further replacing $U$ by the members of a covering we may assume $U \in \mathcal{B}$ as desired.

The lemma follows if we can prove that given a split $n$-truncated $\mathcal{B}$-hypercovering $K$ for $n \geq 1$ we can extend it to a split $(n + 1)$-truncated $\mathcal{B}$-hypercovering. Here the argument proceeds exactly as in the proof of Lemma 24.12.2. We omit the precise details, except for the following comments. First, we do not need assumption (2) in the proof of the current lemma as we do not need the morphism $K_{n + 1} \to (\text{cosk}_ n K)_{n + 1}$ to be covering; we only need it to induce a surjection on associated sheaves of sets which follows from Sites, Lemma 7.12.4. Second, the assumption that $\mathcal{C}$ has fibre products and equalizers guarantees that $\text{SR}(\mathcal{C})$ has fibre products and equalizers and $F$ commutes with these (Lemma 24.2.3). This suffices assure us the coskeleton functors used exist (see Simplicial, Remark 14.19.11 and Categories, Lemma 4.18.2).
$\square$

Lemma 24.12.4. Let $f : \mathcal{C} \to \mathcal{D}$ be a morphism of sites given by the functor $u : \mathcal{D} \to \mathcal{C}$. Assume $\mathcal{D}$ and $\mathcal{C}$ have equalizers and fibre products and $u$ commutes with them. If a simplicial object $K$ of $\text{SR}(\mathcal{D})$ is a hypercovering, then $u(K)$ is a hypercovering.

**Proof.**
If we write $K_ n = \{ U_{n, i}\} _{i \in I_ n}$ as in the introduction to this section, then $u(K)$ is the object of $\text{SR}(\mathcal{C})$ given by $u(K_ n) = \{ u(U_ i)\} _{i \in I_ n}$. By Sites, Lemma 7.13.5 we have $f^{-1}h_ U^\# = h_{u(U)}^\# $ for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. This means that $f^{-1}F(K_ n)^\# = F(u(K_ n))^\# $ for all $n$. Let us check the conditions (1), (2), (3) for $u(K)$ to be a hypercovering from Definition 24.6.1. Since $f^{-1}$ is an exact functor, we find that

\[ F(u(K_0))^\# = f^{-1}F(K_0)^\# \to f^{-1}* = * \]

is surjective as a pullback of a surjective map and we get (1). Similarly,

\[ F(u(K_1))^\# = f^{-1}F(K_1)^\# \to f^{-1} (F(K_0) \times F(K_0))^\# = F(u(K_0))^\# \times F(u(K_0))^\# \]

is surjective as a pullback and we get (2). For condition (3), in order to conclude by the same method it suffices if

\[ F((\text{cosk}_ n \text{sk}_ n u(K))_{n + 1})^\# = f^{-1}F((\text{cosk}_ n \text{sk}_ n K)_{n + 1})^\# \]

The above shows that $f^{-1}F(-) = F(u(-))$. Thus it suffices to show that $u$ commutes with the limits used in defining $(\text{cosk}_ n \text{sk}_ n K)_{n + 1}$ for $n \geq 1$. By Simplicial, Remark 14.19.11 these limits are finite connected limits and $u$ commutes with these by assumption.
$\square$

Lemma 24.12.5. Let $\mathcal{C}$, $\mathcal{D}$ be sites. Let $u : \mathcal{D} \to \mathcal{C}$ be a continuous functor. Assume $\mathcal{D}$ and $\mathcal{C}$ have fibre products and $u$ commutes with them. Let $Y \in \mathcal{D}$ and $K \in \text{SR}(\mathcal{D}, Y)$ a hypercovering of $Y$. Then $u(K)$ is a hypercovering of $u(Y)$.

**Proof.**
This is easier than the proof of Lemma 24.12.4 because the notion of being a hypercovering of an object is stronger, see Definitions 24.3.3 and 24.3.1. Namely, $u$ sends coverings to coverings by the definition of a morphism of sites. It suffices to check $u$ commutes with the limits used in defining $(\text{cosk}_ n \text{sk}_ n K)_{n + 1}$ for $n \geq 1$. This is clear because the induced functor $\mathcal{D}/Y \to \mathcal{C}/X$ commutes with all finite limits (and source and target have all finite limits by Categories, Lemma 4.18.4).
$\square$

Lemma 24.12.6. Let $\mathcal{C}$ be a site. Let $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be a subset. Assume

$\mathcal{C}$ has fibre products,

for all $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ there exists a finite covering $\{ U_ i \to X\} _{i \in I}$ with $U_ i \in \mathcal{B}$,

if $\{ U_ i \to X\} _{i \in I}$ is a finite covering with $U_ i \in \mathcal{B}$ and $U \to X$ is a morphism with $U \in \mathcal{B}$, then $\{ U_ i \to X\} _{i \in I} \amalg \{ U \to X\} $ is a covering.

Then for every $X$ there exists a hypercovering $K$ of $X$ such that each $K_ n = \{ U_{n, i} \to X\} _{i \in I_ n}$ with $I_ n$ finite and $U_{n, i} \in \mathcal{B}$.

**Proof.**
This lemma is the analogue of Lemma 24.11.4 for sites. To prove the lemma we follow exactly the proof of Lemma 24.12.2 paying attention to the following two points

We choose our initial covering $\{ U_{0, i} \to X\} _{i \in I_0}$ with $U_{0, i} \in \mathcal{B}$ such that the index set $I_0$ is finite, and

in choosing the coverings (24.12.2.1) we choose $J_{i'}$ finite.

The reader sees easily that with these modifications we end up with finite index sets $I_ n$ for all $n$.
$\square$

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