Lemma 25.12.4. Let $f : \mathcal{C} \to \mathcal{D}$ be a morphism of sites given by the functor $u : \mathcal{D} \to \mathcal{C}$. Assume $\mathcal{D}$ and $\mathcal{C}$ have equalizers and fibre products and $u$ commutes with them. If a simplicial object $K$ of $\text{SR}(\mathcal{D})$ is a hypercovering, then $u(K)$ is a hypercovering.

Proof. If we write $K_ n = \{ U_{n, i}\} _{i \in I_ n}$ as in the introduction to this section, then $u(K)$ is the object of $\text{SR}(\mathcal{C})$ given by $u(K_ n) = \{ u(U_ i)\} _{i \in I_ n}$. By Sites, Lemma 7.13.5 we have $f^{-1}h_ U^\# = h_{u(U)}^\#$ for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. This means that $f^{-1}F(K_ n)^\# = F(u(K_ n))^\#$ for all $n$. Let us check the conditions (1), (2), (3) for $u(K)$ to be a hypercovering from Definition 25.6.1. Since $f^{-1}$ is an exact functor, we find that

$F(u(K_0))^\# = f^{-1}F(K_0)^\# \to f^{-1}* = *$

is surjective as a pullback of a surjective map and we get (1). Similarly,

$F(u(K_1))^\# = f^{-1}F(K_1)^\# \to f^{-1} (F(K_0) \times F(K_0))^\# = F(u(K_0))^\# \times F(u(K_0))^\#$

is surjective as a pullback and we get (2). For condition (3), in order to conclude by the same method it suffices if

$F((\text{cosk}_ n \text{sk}_ n u(K))_{n + 1})^\# = f^{-1}F((\text{cosk}_ n \text{sk}_ n K)_{n + 1})^\#$

The above shows that $f^{-1}F(-) = F(u(-))$. Thus it suffices to show that $u$ commutes with the limits used in defining $(\text{cosk}_ n \text{sk}_ n K)_{n + 1}$ for $n \geq 1$. By Simplicial, Remark 14.19.11 these limits are finite connected limits and $u$ commutes with these by assumption. $\square$

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