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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)