Lemma 25.7.4. 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 $n \geq 0$ be an integer. Let $u : \mathcal{F} \to F(K_ n)$ be a morphism of presheaves which becomes surjective on sheafification. Then there exists a morphism of hypercoverings $f: L \to K$ such that $F(f_ n) : F(L_ n) \to F(K_ n)$ factors through $u$.

Proof. Write $K_ n = \{ U_ i \to X\} _{i \in I}$. Thus the map $u$ is a morphism of presheaves of sets $u : \mathcal{F} \to \amalg h_{u_ i}$. The assumption on $u$ means that for every $i \in I$ there exists a covering $\{ U_{ij} \to U_ i\} _{j \in I_ i}$ of the site $\mathcal{C}$ and a morphism of presheaves $t_{ij} : h_{U_{ij}} \to \mathcal{F}$ such that $u \circ t_{ij}$ is the map $h_{U_{ij}} \to h_{U_ i}$ coming from the morphism $U_{ij} \to U_ i$. Set $J = \amalg _{i \in I} I_ i$, and let $\alpha : J \to I$ be the obvious map. For $j \in J$ denote $V_ j = U_{\alpha (j)j}$. Set $Z = \{ V_ j \to X\} _{j \in J}$. Finally, consider the morphism $u' : Z \to K_ n$ given by $\alpha : J \to I$ and the morphisms $V_ j = U_{\alpha (j)j} \to U_{\alpha (j)}$ above. Clearly, this is a covering in the category $\text{SR}(\mathcal{C}, X)$, and by construction $F(u') : F(Z) \to F(K_ n)$ factors through $u$. Thus the result follows from Lemma 25.7.3 above. $\square$

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).