Lemma 24.7.3. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $K$ be a hypercovering of $X$. Let $k \geq 0$ be an integer. Let $u : Z \to K_ k$ be a covering in $\text{SR}(\mathcal{C}, X)$. Then there exists a morphism of hypercoverings $f: L \to K$ such that $L_ k \to K_ k$ factors through $u$.

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

$\prod \nolimits _{\alpha : [k] \to [n]} Z \longrightarrow \prod \nolimits _{\alpha : [k] \to [n]} Y$

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

$L = K \times _{\mathop{\mathrm{Hom}}\nolimits (C[k], Y)} \mathop{\mathrm{Hom}}\nolimits (C[k], Z).$

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

$\xymatrix{ L_ k \ar[r] \ar[d] & \prod _{\alpha : [k] \to [k]} Z \ar[r]^-{\text{pr}_{\text{id}_{[k]}}} \ar[d] & Z \ar[d] \\ K_ k \ar[r] & \prod _{\alpha : [k] \to [k]} Y \ar[r]^-{\text{pr}_{\text{id}_{[k]}}} & Y }$

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 24.7.1. Condition (1) is satisfied by assumption. For (2), the morphism

$\mathop{\mathrm{Hom}}\nolimits (C[k], Z)_0 \to \mathop{\mathrm{Hom}}\nolimits (C[k], Y)_0$

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

$\xymatrix{ \prod \nolimits _{\alpha : [k] \to [1]} Z \ar[r] \ar[d] & Z \times Z \ar[d] \\ \prod \nolimits _{\alpha : [k] \to [1]} Y \ar[r] & Y \times Y }$

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

$\prod \nolimits _{\alpha : [k] \to [1]} Z \longrightarrow \prod \nolimits _{\alpha : [k] \to [1]\text{ not onto}} Z \times \prod \nolimits _{\alpha : [k] \to [1]\text{ onto}} Y$

which is a product of coverings and hence a covering by Lemma 24.3.2.

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

24.7.3.1
$$\label{hypercovering-equation-map} \mathop{\mathrm{Hom}}\nolimits (C[k], T)_{n + 1} \to (\text{cosk}_ n \text{sk}_ n \mathop{\mathrm{Hom}}\nolimits (C[k], T))_{n + 1}$$

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

$\prod \nolimits _{s \in S} Z \longrightarrow \prod \nolimits _{s \in S'} Z \times \prod \nolimits _{s \not\in \tau (S')} Y$

which is a product of coverings and hence a covering by Lemma 24.3.2. 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

$f_{i, \delta ^ n_{j - 1} \circ \beta } = f_{j, \delta _ i^ n \circ \beta }$

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

$S' = \{ 0, \ldots , n + 1\} \times \text{Map}([k], [n]) / \sim$

where the equivalence relation is generated by the equivalences

$(i, \delta ^ n_{j - 1} \circ \beta ) \sim (j, \delta _ i^ n \circ \beta )$

for $0 \leq i < j \leq n + 1$ and $\beta : [k] \to [n - 1]$. A computation (omitted) shows that the morphism (24.7.3.1) 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

$\delta ^{n + 1}_ i \circ \alpha = \delta ^{n + 1}_ j \circ \alpha '$

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. $\square$

Comment #1236 by Amit Hogadi on

We are given a covering $u:Z\to K_k=Y$ in this lemma. There may be no morphism from $Y\to Z$. One probably needs to exchange the places of $Y$ and $Z$ in the proof in several places (e.g. lines 4, 5, 9, 14 of the proof).

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