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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