Lemma 7.10.16. Let $\mathcal{C}$ be a site. Let $\mathcal{F} \to \mathcal{G}$ be a map of presheaves of sets on $\mathcal{C}$. Denote $\mathcal{B}$ the set of $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ such that $\mathcal{F}(U) \to \mathcal{G}(U)$ is bijective. If every object of $\mathcal{C}$ has a covering by elements of $\mathcal{B}$, then $\mathcal{F}^\# \to \mathcal{G}^\# $ is an isomorphism.

**Proof.**
Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Let us prove that $\mathcal{F}^\# (U) \to \mathcal{G}^\# (U)$ is surjective. To do this we will use Lemma 7.10.15 without further mention. For any $s \in \mathcal{G}^\# (U)$ there exists a covering $\{ U_ i \to U\} $ and sections $s_ i \in \mathcal{G}(U_ i)$ such that

$s|_{U_ i}$ is the image of $s_ i$ via $\mathcal{G} \to \mathcal{G}^\# $, and

for every $i, j$ there exists a covering $\{ U_{ijk} \to U_ i \times _ U U_ j\} $ of $\mathcal{D}$ such that the pullbacks of $s_ i$ and $s_ j$ to each $U_{ijk}$ agree.

By assumption, for each $i$ we may choose a covering $\{ U_{i, a} \to U_ i\} $ with $U_{i, a} \in \mathcal{B}$. Then $\{ U_{i, a} \to U\} $ is a covering. Denoting $s_{i, a}$ the image of $s_ i$ in $\mathcal{G}(U_{i, a})$ we see that the pullbacks of $s_{i, a}$ and $s_{j, b}$ to the members of the covering

agree. Hence we may assume that $U_ i \in \mathcal{B}$. Repeating the argument, we may also assume $U_{ijk} \in \mathcal{B}$ for all $i, j, k$ (details omitted). Then since $\mathcal{F}(U_ i) \to \mathcal{G}(U_ i)$ is bijective by our definition of $\mathcal{B}$ in the statement of the lemma, we get unique $t_ i \in \mathcal{F}(U_ i)$ mapping to $s_ i$. The pullbacks of $t_ i$ and $t_ j$ to each $U_{ijk}$ agree in $\mathcal{F}(U_{ijk})$ because $U_{ijk} \in \mathcal{B}$ and because we have the agreement for $t_ i$ and $t_ j$. Then the lemma tells us there exists a unique $t \in \mathcal{F}^\# (U)$ such that $t|_{U_ i}$ is the image of $t_ i$. It follows that $t$ maps to $s$. We omit the proof that $\mathcal{F}^\# (U) \to \mathcal{G}^\# (U)$ is injective. $\square$

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