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

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

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

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DAV. Beware of the difference between the letter 'O' and the digit '0'.