Lemma 25.3.9. 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
\[ \amalg _{i \in I} h_{U_ i} \to h_ U \]
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 25.2.3 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 25.3.1, and the result follows from the first fact mentioned in this proof.
$\square$
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