Lemma 25.12.2. Let $\mathcal{C}$ be a site with fibre products. Let $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be a subset. Assume

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

2. 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 25.11.3 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 25.3.1.

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 25.12.1.

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

25.12.2.1
\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 25.3.1. 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 (25.12.2.1), 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 (25.12.2.1) 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:

1. 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,

2. 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

3. 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 25.12.1).

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 $0 \leq a, b \leq n + 1$. $\square$

 For example, if $i$ is nondegenerate, then $m = m_0$ and $\psi = \text{id}_{[m]}$.

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