Lemma 25.11.3. Let $X$ be a topological space. Let $\mathcal{B}$ be a basis for the topology of $X$. There exists a hypercovering $(I, \{ U_ i\} )$ of $X$ such that each $U_ i$ is an element of $\mathcal{B}$.

Proof. Let $n \geq 0$. Let us say that an $n$-truncated hypercovering of $X$ is given by an $n$-truncated simplicial set $I$ and for each $i \in I_ a$, $0 \leq a \leq n$ an open $U_ i$ of $X$ such that the conditions defining a hypercovering hold whenever they make sense. In other words we require the inclusion relations and covering conditions only when all simplices that occur in them are $a$-simplices with $a \leq n$. The lemma follows if we can prove that given a $n$-truncated hypercovering $(I, \{ U_ i\} )$ with all $U_ i \in \mathcal{B}$ we can extend it to an $(n + 1)$-truncated hypercovering without adding any $a$-simplices for $a \leq n$. This we do as follows. First we consider the $(n + 1)$-truncated simplicial set $I'$ defined by $I' = \text{sk}_{n + 1}(\text{cosk}_ n I)$. Recall that

$I'_{n + 1} = \left\{ \begin{matrix} (i_0, \ldots , i_{n + 1}) \in (I_ n)^{n + 2} \text{ such that} \\ d^ n_{b - 1}(i_ a) = d^ n_ a(i_ b) \text{ for all }0\leq a < b\leq n + 1 \end{matrix} \right\}$

If $i' \in I'_{n + 1}$ is degenerate, say $i' = s^ n_ a(i)$ then we set $U_{i'} = U_ i$ (this is forced on us anyway by the second condition). We also set $J_{i'} = \{ i'\}$ in this case. If $i' \in I'_{n + 1}$ is nondegenerate, say $i' = (i_0, \ldots , i_{n + 1})$, then we choose a set $J_{i'}$ and an open covering

25.11.3.1
\begin{equation} \label{hypercovering-equation-choose-covering} U_{i_0} \cap \ldots \cap U_{i_{n + 1}} = \bigcup \nolimits _{i \in J_{i'}} U_ i, \end{equation}

with $U_ i \in \mathcal{B}$ for $i \in J_{i'}$. Set

$I_{n + 1} = \coprod \nolimits _{i' \in I'_{n + 1}} J_{i'}$

There is a canonical map $\pi : I_{n + 1} \to I'_{n + 1}$ which is a bijection over the set of degenerate simplices in $I'_{n + 1}$ by construction. For $i \in I_{n + 1}$ we define $d^{n + 1}_ a(i) = d^{n + 1}_ a(\pi (i))$. For $i \in I_ n$ we define $s^ n_ a(i) \in I_{n + 1}$ as the unique simplex lying over the degenerate simplex $s^ n_ a(i) \in I'_{n + 1}$. We omit the verification that this defines an $(n + 1)$-truncated hypercovering of $X$. $\square$

Comment #3776 by on

I believe that the construction presented in this tag (Tag 01H6) can be rewardingly factored as the composition of two constructions:

1. First construct a "semi-simplicial hypercovering" in a straightforward manner (no case distinctions needed).
2. Then apply the functor from semi-simplicial sets to simplicial sets (or rather, from semi-simplicial objects in $\mathrm{SR}(X_\text{Zar},X)$ to simplicial objects in $\mathrm{SR}(X_\text{Zar},X)$).

This way it's obvious that the use of the law of excluded middle, in the test whether $i'$ is degenerate or not, can be avoided. (I know that avoiding the law of excluded middle is not important for the Stacks Project, but it's important to my current research.) The price is that the composition of these two constructions doesn't quite coincide with the construction presented in the tag, since self-intersections such as $U_i \cap U_i$ yield a couple of additional simplices in my proposed construction.\ref{}

Comment #3906 by on

In the Stacks project we have (so far) intentionally avoided using semi-simplicial things... I really would like to avoid discussing them... So I'm going to leave this proof alone for now.

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).