## 24.11 Hypercoverings of spaces

The theory above is mildly interesting even in the case of topological spaces. In this case we can work out what a hypercovering is and see what the result actually says.

Let $X$ be a topological space. Consider the site $X_{Zar}$ of Sites, Example 7.6.4. Recall that an object of $X_{Zar}$ is simply an open of $X$ and that morphisms of $X_{Zar}$ correspond simply to inclusions. So what is a hypercovering of $X$ for the site $X_{Zar}$?

Let us first unwind Definition 24.2.1. An object of $\text{SR}(X_{Zar}, X)$ is simply given by a set $I$ and for each $i \in I$ an open $U_ i \subset X$. Let us denote this by $\{ U_ i\} _{i \in I}$ since there can be no confusion about the morphism $U_ i \to X$. A morphism $\{ U_ i\} _{i \in I} \to \{ V_ j\} _{j \in J}$ between two such objects is given by a map of sets $\alpha : I \to J$ such that $U_ i \subset V_{\alpha (i)}$ for all $i \in I$. When is such a morphism a covering? This is the case if and only if for every $j \in J$ we have $V_ j = \bigcup _{i\in I, \ \alpha (i) = j} U_ i$ (and is a covering in the site $X_{Zar}$).

Using the above we get the following description of a hypercovering in the site $X_{Zar}$. A hypercovering of $X$ in $X_{Zar}$ is given by the following data

a simplicial set $I$ (see Simplicial, Section 14.11), and

for each $n \geq 0$ and every $i \in I_ n$ an open set $U_ i \subset X$.

We will denote such a collection of data by the notation $(I, \{ U_ i\} )$. In order for this to be a hypercovering of $X$ we require the following properties

for $i \in I_ n$ and $0 \leq a \leq n$ we have $U_ i \subset U_{d^ n_ a(i)}$,

for $i \in I_ n$ and $0 \leq a \leq n$ we have $U_ i = U_{s^ n_ a(i)}$,

we have

24.11.0.1
\begin{equation} \label{hypercovering-equation-covering-X} X = \bigcup \nolimits _{i \in I_0} U_ i, \end{equation}

for every $i_0, i_1 \in I_0$, we have

24.11.0.2
\begin{equation} \label{hypercovering-equation-covering-two} U_{i_0} \cap U_{i_1} = \bigcup \nolimits _{i \in I_1, \ d^1_0(i) = i_0, \ d^1_1(i) = i_1} U_ i, \end{equation}

for every $n \geq 1$ and every $(i_0, \ldots , i_{n + 1}) \in (I_ n)^{n + 2}$ such that $d^ n_{b - 1}(i_ a) = d^ n_ a(i_ b)$ for all $0\leq a < b\leq n + 1$ we have

24.11.0.3
\begin{equation} \label{hypercovering-equation-covering-general} U_{i_0} \cap \ldots \cap U_{i_{n + 1}} = \bigcup \nolimits _{i \in I_{n + 1}, \ d^{n + 1}_ a(i) = i_ a, \ a = 0, \ldots , n + 1} U_ i, \end{equation}

each of the open coverings (24.11.0.1), (24.11.0.2), and (24.11.0.3) is an element of $\text{Cov}(X_{Zar})$ (this is a set theoretic condition, bounding the size of the index sets of the coverings).

Conditions (24.11.0.1) and (24.11.0.2) should be familiar from the chapter on sheaves on spaces for example, and condition (24.11.0.3) is the natural generalization.

In topology we often want to find hypercoverings of $X$ which have the property that all the $U_ i$ come from a given basis for the topology of $X$ and that all the coverings (24.11.0.2) and (24.11.0.3) are from a given cofinal collection of coverings. Here are two example lemmas.

Lemma 24.11.3. 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

24.11.3.1
\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 $X$.
$\square$

Lemma 24.11.4. Let $X$ be a topological space. Let $\mathcal{B}$ be a basis for the topology of $X$. Assume that

$X$ is quasi-compact,

each $U \in \mathcal{B}$ is quasi-compact open, and

the intersection of any two quasi-compact opens in $X$ is quasi-compact.

Then there exists a hypercovering $(I, \{ U_ i\} )$ of $X$ with the following properties

each $U_ i$ is an element of the basis $\mathcal{B}$,

each of the $I_ n$ is a finite set, and in particular

each of the coverings (24.11.0.1), (24.11.0.2), and (24.11.0.3) is finite.

**Proof.**
This follows directly from the construction in the proof of Lemma 24.11.3 if we choose finite coverings by elements of $\mathcal{B}$ in (24.11.3.1). Details omitted.
$\square$

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