The Stacks project

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

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

This diagram shows that

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

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

Proof. You can either verify this directly, or use Lemma 25.7.1 above and check that $L \to \{ X \to X\} $ has property (3). $\square$

Proof. Denote $Y = K_ k$. Let $C[k]$ be the cosimplicial set defined in Simplicial, Example 14.5.6. We will use the description of $\mathop{\mathrm{Hom}}\nolimits (C[k], Y)$ and $\mathop{\mathrm{Hom}}\nolimits (C[k], Z)$ given in Simplicial, Lemma 14.15.2. There is a canonical morphism $K \to \mathop{\mathrm{Hom}}\nolimits (C[k], Y)$ corresponding to $\text{id} : K_ k = Y \to Y$. Consider the morphism $\mathop{\mathrm{Hom}}\nolimits (C[k], Z) \to \mathop{\mathrm{Hom}}\nolimits (C[k], Y)$ which on degree $n$ terms is the morphism

using the given morphism $Z \to Y$ on each factor. Set

The morphism $L_ k \to K_ k$ sits in to a commutative diagram

Since the composition of the two bottom arrows is the identity we conclude that we have the desired factorization.

We still have to show that $L$ is a hypercovering of $X$. To see this we will use Lemma 25.7.1. Condition (1) is satisfied by assumption. For (2), the morphism

is a covering because it is isomorphic to $Z \to Y$ as there is only one morphism $[k] \to [0]$.

Let us consider condition (3) for $n = 0$. Then, since $(\text{cosk}_0 T)_1 = T \times T$ (Simplicial, Example 14.19.1) and since $\mathop{\mathrm{Hom}}\nolimits (C[k], Z)_1 = \prod _{\alpha : [k] \to [1]} Z$ we obtain the diagram

with horizontal arrows corresponding to the projection onto the factors corresponding to the two nonsurjective $\alpha $. Thus the arrow $\gamma $ is the morphism

which is a product of coverings and hence a covering by Lemma 25.3.2.

Let us consider condition (3) for $n > 0$. We claim there is an injective map $\tau : S' \to S$ of finite sets, such that for any object $T$ of $\text{SR}(\mathcal{C}, X)$ the morphism

is isomorphic to the projection $\prod _{s \in S} T \to \prod _{s' \in S'} T$ functorially in $T$. If this is true, then we see, arguing as in the previous paragraph, that the arrow $\gamma $ is the morphism

which is a product of coverings and hence a covering by Lemma 25.3.2. By construction, we have $\mathop{\mathrm{Hom}}\nolimits (C[k], T)_{n + 1} = \prod _{\alpha : [k] \to [n + 1]} T$ (see Simplicial, Lemma 14.15.2). Correspondingly we take $S = \text{Map}([k], [n + 1])$. On the other hand, Simplicial, Lemma 14.19.5, provides a description of points of $(\text{cosk}_ n \text{sk}_ n \mathop{\mathrm{Hom}}\nolimits (C[k], T))_{n + 1}$ as sequences $(f_0, \ldots , f_{n + 1})$ of points of $\mathop{\mathrm{Hom}}\nolimits (C[k], T)_ n$ satisfying $d^ n_{j - 1} f_ i = d^ n_ i f_ j$ for $0 \leq i < j \leq n + 1$. We can write $f_ i = (f_{i, \alpha })$ with $f_{i, \alpha }$ a point of $T$ and $\alpha \in \text{Map}([k], [n])$. The conditions translate into

for any $0 \leq i < j \leq n + 1$ and $\beta : [k] \to [n - 1]$. Thus we see that

where the equivalence relation is generated by the equivalences

for $0 \leq i < j \leq n + 1$ and $\beta : [k] \to [n - 1]$. A computation (omitted) shows that the morphism (25.7.3.1) corresponds to the map $S' \to S$ which sends $(i, \alpha )$ to $\delta ^{n + 1}_ i \circ \alpha \in S$. (It may be a comfort to the reader to see that this map is well defined by part (1) of Simplicial, Lemma 14.2.3.) To finish the proof it suffices to show that if $\alpha , \alpha ' : [k] \to [n]$ and $0 \leq i < j \leq n + 1$ are such that

then we have $\alpha = \delta ^ n_{j - 1} \circ \beta $ and $\alpha ' = \delta _ i^ n \circ \beta $ for some $\beta : [k] \to [n - 1]$. This is easy to see and omitted. $\square$

Proof. Write $K_ n = \{ U_ i \to X\} _{i \in I}$. Thus the map $u$ is a morphism of presheaves of sets $u : \mathcal{F} \to \amalg h_{u_ i}$. The assumption on $u$ means that for every $i \in I$ there exists a covering $\{ U_{ij} \to U_ i\} _{j \in I_ i}$ of the site $\mathcal{C}$ and a morphism of presheaves $t_{ij} : h_{U_{ij}} \to \mathcal{F}$ such that $u \circ t_{ij}$ is the map $h_{U_{ij}} \to h_{U_ i}$ coming from the morphism $U_{ij} \to U_ i$. Set $J = \amalg _{i \in I} I_ i$, and let $\alpha : J \to I$ be the obvious map. For $j \in J$ denote $V_ j = U_{\alpha (j)j}$. Set $Z = \{ V_ j \to X\} _{j \in J}$. Finally, consider the morphism $u' : Z \to K_ n$ given by $\alpha : J \to I$ and the morphisms $V_ j = U_{\alpha (j)j} \to U_{\alpha (j)}$ above. Clearly, this is a covering in the category $\text{SR}(\mathcal{C}, X)$, and by construction $F(u') : F(Z) \to F(K_ n)$ factors through $u$. Thus the result follows from Lemma 25.7.3 above. $\square$