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

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

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

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 $X$. $\square$

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## Comments (1)

Comment #3776 by Ingo Blechschmidt on