Lemma 25.8.1. 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 $U \subset V$ be simplicial sets, with $U_ n, V_ n$ finite nonempty for all $n$. Assume that $U$ has finitely many nondegenerate simplices. Suppose $n \geq 0$ and $x \in V_ n$, $x \not\in U_ n$ are such that

1. $V_ i = U_ i$ for $i < n$,

2. $V_ n = U_ n \cup \{ x\}$,

3. any $z \in V_ j$, $z \not\in U_ j$ for $j > n$ is degenerate.

Then the morphism

$\mathop{\mathrm{Hom}}\nolimits (V, K)_0 \longrightarrow \mathop{\mathrm{Hom}}\nolimits (U, K)_0$

of $\text{SR}(\mathcal{C}, X)$ is a covering.

Proof. If $n = 0$, then it follows easily that $V = U \amalg \Delta [0]$ (see below). In this case $\mathop{\mathrm{Hom}}\nolimits (V, K)_0 = \mathop{\mathrm{Hom}}\nolimits (U, K)_0 \times K_0$. The result, in this case, then follows from Lemma 25.3.2.

Let $a : \Delta [n] \to V$ be the morphism associated to $x$ as in Simplicial, Lemma 14.11.3. Let us write $\partial \Delta [n] = i_{(n-1)!} \text{sk}_{n - 1} \Delta [n]$ for the $(n - 1)$-skeleton of $\Delta [n]$. Let $b : \partial \Delta [n] \to U$ be the restriction of $a$ to the $(n - 1)$ skeleton of $\Delta [n]$. By Simplicial, Lemma 14.21.7 we have $V = U \amalg _{\partial \Delta [n]} \Delta [n]$. By Simplicial, Lemma 14.17.5 we get that

$\xymatrix{ \mathop{\mathrm{Hom}}\nolimits (V, K)_0 \ar[r] \ar[d] & \mathop{\mathrm{Hom}}\nolimits (U, K)_0 \ar[d] \\ \mathop{\mathrm{Hom}}\nolimits (\Delta [n], K)_0 \ar[r] & \mathop{\mathrm{Hom}}\nolimits (\partial \Delta [n], K)_0 }$

is a fibre product square. Thus it suffices to show that the bottom horizontal arrow is a covering. By Simplicial, Lemma 14.21.11 this arrow is identified with

$K_ n \to (\text{cosk}_{n - 1} \text{sk}_{n - 1} K)_ n$

and hence is a covering by definition of a hypercovering. $\square$

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).