Lemma 25.7.1. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $K, L, M$ be simplicial objects of $\text{SR}(\mathcal{C}, X)$. Let $a : K \to L$, $b : M \to L$ be morphisms. Assume

1. $K$ is a hypercovering of $X$,

2. the morphism $M_0 \to L_0$ is a covering, and

3. for all $n \geq 0$ in the diagram

$\xymatrix{ M_{n + 1} \ar[dd] \ar[rr] \ar[rd]^\gamma & & (\text{cosk}_ n \text{sk}_ n M)_{n + 1} \ar[dd] \\ & L_{n + 1} \times _{(\text{cosk}_ n \text{sk}_ n L)_{n + 1}} (\text{cosk}_ n \text{sk}_ n M)_{n + 1} \ar[ld] \ar[ru] & \\ L_{n + 1} \ar[rr] & & (\text{cosk}_ n \text{sk}_ n L)_{n + 1} }$

the arrow $\gamma$ is a covering.

Then the fibre product $K \times _ L M$ is a hypercovering of $X$.

Proof. The morphism $(K \times _ L M)_0 = K_0 \times _{L_0} M_0 \to K_0$ is a base change of a covering by (2), hence a covering, see Lemma 25.3.2. And $K_0 \to \{ X \to X\}$ is a covering by (1). Thus $(K \times _ L M)_0 \to \{ X \to X\}$ is a covering by Lemma 25.3.2. Hence $K \times _ L M$ satisfies the first condition of Definition 25.3.3.

We still have to check that

$K_{n + 1} \times _{L_{n + 1}} M_{n + 1} = (K \times _ L M)_{n + 1} \longrightarrow (\text{cosk}_ n \text{sk}_ n (K \times _ L M))_{n + 1}$

is a covering for all $n \geq 0$. We abbreviate as follows: $A = (\text{cosk}_ n \text{sk}_ n K)_{n + 1}$, $B = (\text{cosk}_ n \text{sk}_ n L)_{n + 1}$, and $C = (\text{cosk}_ n \text{sk}_ n M)_{n + 1}$. The functor $\text{cosk}_ n \text{sk}_ n$ commutes with fibre products, see Simplicial, Lemma 14.19.13. Thus the right hand side above is equal to $A \times _ B C$. Consider the following commutative diagram

$\xymatrix{ K_{n + 1} \times _{L_{n + 1}} M_{n + 1} \ar[r] \ar[d] & M_{n + 1} \ar[d] \ar[rd]_\gamma \ar[rrd] & & \\ K_{n + 1} \ar[r] \ar[rd] & L_{n + 1} \ar[rrd] & L_{n + 1} \times _ B C \ar[l] \ar[r] & C \ar[d] \\ & A \ar[rr] & & B }$

This diagram shows that

$K_{n + 1} \times _{L_{n + 1}} M_{n + 1} = (K_{n + 1} \times _ B C) \times _{(L_{n + 1} \times _ B C), \gamma } M_{n + 1}$

Now, $K_{n + 1} \times _ B C \to A \times _ B C$ is a base change of the covering $K_{n + 1} \to A$ via the morphism $A \times _ B C \to A$, hence is a covering. By assumption (3) the morphism $\gamma$ is a covering. Hence the morphism

$(K_{n + 1} \times _ B C) \times _{(L_{n + 1} \times _ B C), \gamma } M_{n + 1} \longrightarrow K_{n + 1} \times _ B C$

is a covering as a base change of a covering. The lemma follows as a composition of coverings is a covering. $\square$

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