The Stacks project

Lemma 17.11.6. Let $X$ be a ringed space. Let $I$ be a preordered set and let $(\mathcal{F}_ i, \varphi _{ii'})$ be a system over $I$ consisting of sheaves of $\mathcal{O}_ X$-modules (see Categories, Section 4.21). Assume

  1. $I$ is directed,

  2. $\mathcal{G}$ is an $\mathcal{O}_ X$-module of finite presentation, and

  3. $X$ has a cofinal system of open coverings $\mathcal{U} : X = \bigcup _{j\in J} U_ j$ with $J$ finite and $U_ j \cap U_{j'}$ quasi-compact for all $j, j' \in J$.

Then we have

\[ \mathop{\mathrm{colim}}\nolimits _ i \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, \mathcal{F}_ i) = \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i). \]

Proof. An element of the left hand side is given by the equivalence classe of a pair $(i, \alpha _ i)$ where $i \in I$ and $\alpha _ i : \mathcal{G} \to \mathcal{F}_ i$ is a morphism of $\mathcal{O}_ X$-modules, see Categories, Section 4.19. Postcomposing with the coprojection $p_ i : \mathcal{F}_ i \to \mathop{\mathrm{colim}}\nolimits _{i' \in I} \mathcal{F}_{i'}$ we get $\alpha = p_ i \circ \alpha _ i$ in the right hand side. We obtain a map

\[ \mathop{\mathrm{colim}}\nolimits _ i \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, \mathcal{F}_ i) \to \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i) \]

Let us show this map is injective. Let $\alpha _ i$ be as above such that $\alpha = p_ i \circ \alpha _ i$ is zero. By the assumption that $\mathcal{G}$ is of finite presentation, for every $x \in X$ we can choose an open neighbourhood $U_ x \subset X$ of $x$ and a finite set $s_{x, 1}, \ldots , s_{x, n_ x} \in \mathcal{G}(U_ x)$ generating $\mathcal{G}|_{U_ x}$. These sections map to zero in the stalk $(\mathop{\mathrm{colim}}\nolimits _{i'} \mathcal{F}_ i)_ x = \mathop{\mathrm{colim}}\nolimits _{i'} \mathcal{F}_{i', x}$. Hence for each $x$ we can pick $i(x) \geq i$ such that after replacing $U_ x$ by a smaller open we have that $s_{x, 1}, \ldots , s_{x, n_ x}$ map to zero in $\mathcal{F}_{i(x)}(U_ x)$. Then $X = \bigcup U_ x$. By condition (3) we can refine this open covering by a finite open covering $X = \bigcup _{j \in J} U_ j$. For $j \in J$ pick $x_ j \in X$ with $U_ j \subset U_{x_ j}$. Set $i' = \max (i(x_ j); j \in J)$. Then $\mathcal{G}|_{U_ j}$ is generated by the sections $s_{x_ j, k}$ which are mapped to zero in $\mathcal{F}_{i(x)}$ and hence in $\mathcal{F}_{i'}$. Hence the composition $\mathcal{G} \to \mathcal{F}_ i \to \mathcal{F}_{i'}$ is zero as desired.

Proof of surjectivity. Let $\alpha $ be an element of the right hand side. For every point $x \in X$ we may choose an open neighbourhood $U \subset X$ and finitely many sections $s_ j \in \mathcal{G}(U)$ which generate $\mathcal{G}$ over $U$ and finitely many relations $\sum f_{kj} s_ j = 0$, $k = 1, \ldots , n$ with $f_{kj} \in \mathcal{O}_ X(U)$ which generate the kernel of $\bigoplus _{j = 1, \ldots , m} \mathcal{O}_ U \to \mathcal{G}$. After possibly shrinking $U$ to a smaller open neighbourhood of $x$ we may assume there exists an index $i \in I$ such that the sections $\alpha (s_ j)$ all come from sections $s_ j' \in \mathcal{F}_ i(U)$. After possibly shrinking $U$ to a smaller open neighbourhood of $x$ and increasing $i$ we may assume the relations $\sum f_{kj} s'_ j = 0$ hold in $\mathcal{F}_ i(U)$. Hence we see that $\alpha |_ U$ lifts to a morphism $\mathcal{G}|_ U \to \mathcal{F}_ i|_ U$ for some index $i \in I$.

By condition (3) and the preceding arguments, we may choose a finite open covering $X = \bigcup _{j = 1, \ldots , m} U_ j$ such that (a) $\mathcal{G}|_{U_ j}$ is generated by finitely many sections $s_{jk} \in \mathcal{G}(U_ j)$, (b) the restriction $\alpha |_{U_ j}$ comes from a morphism $\alpha _ j : \mathcal{G} \to \mathcal{F}_{i_ j}$ for some $i_ j \in I$, and (c) the intersections $U_ j \cap U_{j'}$ are all quasi-compact. For every pair $(j, j') \in \{ 1, \ldots , m\} ^2$ and any $k$ we can find we can find an index $i \geq \max (i_ j, i_{j'})$ such that

\[ \varphi _{i_ ji}(\alpha _ j(s_{jk}|_{U_ j \cap U_{j'}})) = \varphi _{i_{j'}i}(\alpha _{j'}(s_{jk}|_{U_ j \cap U_{j'}})) \]

see Sheaves, Lemma 6.29.1 (2). Since there are finitely many of these pairs $(j, j')$ and finitely many $s_{jk}$ we see that we can find a single $i$ which works for all of them. For this index $i$ all of the maps $\varphi _{i_ ji} \circ \alpha _ j$ agree on the overlaps $U_ j \cap U_{j'}$ as the sections $s_{jk}$ generate $\mathcal{G}$ over this overlap. Hence we get a morphism $\mathcal{G} \to \mathcal{F}_ i$ as desired. $\square$

Comments (2)

Comment #4323 by Minh-Tien Tran on

I think the proof only show that the map is surjective. It is not very clear to me why this map is also an injection.

Comment #4479 by on

THanks for pointing this out. This was a bit of a brain teaser. We never use the lemma in full strenghth (namely we use it only when the can be chosen quasi-compact as well), but since we claimed it was true we should prove the whole thing as well... The fix is here.

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