Lemma 25.12.3. 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 25.12.2. 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

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 25.12.2. 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 25.2.3). This suffices assure us the coskeleton functors used exist (see Simplicial, Remark 14.19.11 and Categories, Lemma 4.18.2). $\square$

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