Lemma 7.38.2. Let $\mathcal{C}$ be a site and let $\{ p_ i\} _{i\in I}$ be a conservative family of points. Then

1. Given any map of sheaves $\varphi : \mathcal{F} \to \mathcal{G}$ we have $\forall i, \varphi _{p_ i}$ injective implies $\varphi$ injective.

2. Given any map of sheaves $\varphi : \mathcal{F} \to \mathcal{G}$ we have $\forall i, \varphi _{p_ i}$ surjective implies $\varphi$ surjective.

3. Given any pair of maps of sheaves $\varphi _1, \varphi _2 : \mathcal{F} \to \mathcal{G}$ we have $\forall i, \varphi _{1, p_ i} = \varphi _{2, p_ i}$ implies $\varphi _1 = \varphi _2$.

4. Given a finite diagram $\mathcal{G} : \mathcal{J} \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$, a sheaf $\mathcal{F}$ and morphisms $q_ j : \mathcal{F} \to \mathcal{G}_ j$ then $(\mathcal{F}, q_ j)$ is a limit of the diagram if and only if for each $i$ the stalk $(\mathcal{F}_{p_ i}, (q_ j)_{p_ i})$ is one.

5. Given a finite diagram $\mathcal{F} : \mathcal{J} \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$, a sheaf $\mathcal{G}$ and morphisms $e_ j : \mathcal{F}_ j \to \mathcal{G}$ then $(\mathcal{G}, e_ j)$ is a colimit of the diagram if and only if for each $i$ the stalk $(\mathcal{G}_{p_ i}, (e_ j)_{p_ i})$ is one.

Proof. We will use over and over again that all the stalk functors commute with any finite limits and colimits and hence with products, fibred products, etc. We will also use that injective maps are the monomorphisms and the surjective maps are the epimorphisms. A map of sheaves $\varphi : \mathcal{F} \to \mathcal{G}$ is injective if and only if $\mathcal{F} \to \mathcal{F} \times _\mathcal {G}\mathcal{F}$ is an isomorphism. Hence (1). Similarly, $\varphi : \mathcal{F} \to \mathcal{G}$ is surjective if and only if $\mathcal{G} \amalg _\mathcal {F} \mathcal{G} \to \mathcal{G}$ is an isomorphism. Hence (2). The maps $a, b : \mathcal{F} \to \mathcal{G}$ are equal if and only if $\mathcal{F} \times _{a, \mathcal{G}, b}\mathcal{F} \to \mathcal{F} \times \mathcal{F}$ is an isomorphism. Hence (3). The assertions (4) and (5) follow immediately from the definitions and the remarks at the start of this proof. $\square$

Comment #8356 by ZL on

A minor problem. In the proof for $(3)$, let $\mathcal{F}=\mathcal{G}$ and $a=b=\mathrm{id}$, we have $\mathcal{F}\times _{a,\mathcal{G},b}\mathcal{F}=\mathcal{F}$ which is not isomorphic to $\mathcal{F}\times \mathcal{F}$. Maybe we can use the fact that $a=b$ iff $\mathrm{ker}(a,b)\rightarrow \mathcal{F}$ is an isomorphism?

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