Lemma 25.7.3. 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 25.7.1. 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 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
25.7.3.1
\begin{equation} \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} \end{equation}
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 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
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 (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
\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
Comments (2)
Comment #1236 by Amit Hogadi on
Comment #1249 by Johan on