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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #1236 by Amit Hogadi on
Comment #1249 by Johan on