The Stacks project

Proof. Choose a set $I$ and for each $i \in I$ an $\mathcal{O}_ X$-module of finite presentation and a homomorphism of $\mathcal{O}_ X$-modules $\varphi _ i : \mathcal{F}_ i \to \mathcal{F}$ with the following property: For any $\psi : \mathcal{G} \to \mathcal{F}$ with $\mathcal{G}$ of finite presentation there is an $i \in I$ such that there exists an isomorphism $\alpha : \mathcal{F}_ i \to \mathcal{G}$ with $\varphi _ i = \psi \circ \alpha $. It is clear from Modules, Lemma 17.9.8 that such a set exists (see also its proof). We denote $\mathcal{I}$ the category with $\mathop{\mathrm{Ob}}\nolimits (\mathcal{I}) = I$ and given $i, i' \in I$ we set

We claim that $\mathcal{I}$ is a filtered category and that $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i$.

Let $i, i' \in I$. Then we can consider the morphism

which is the direct sum of $\varphi _ i$ and $\varphi _{i'}$. Since a direct sum of finitely presented $\mathcal{O}_ X$-modules is finitely presented we see that there exists some $i'' \in I$ such that $\varphi _{i''} : \mathcal{F}_{i''} \to \mathcal{F}$ is isomorphic to the displayed arrow towards $\mathcal{F}$ above. Since there are commutative diagrams

we see that there are morphisms $i \to i''$ and $i' \to i''$ in $\mathcal{I}$. Next, suppose that we have $i, i' \in I$ and morphisms $\alpha , \beta : i \to i'$ (corresponding to $\mathcal{O}_ X$-module maps $\alpha , \beta : \mathcal{F}_ i \to \mathcal{F}_{i'}$). In this case consider the coequalizer

Note that $\mathcal{G}$ is an $\mathcal{O}_ X$-module of finite presentation. Since by definition of morphisms in the category $\mathcal{I}$ we have $\varphi _{i'} \circ \alpha = \varphi _{i'} \circ \beta $ we see that we get an induced map $\psi : \mathcal{G} \to \mathcal{F}$. Hence again the pair $(\mathcal{G}, \psi )$ is isomorphic to the pair $(\mathcal{F}_{i''}, \varphi _{i''})$ for some $i''$. Hence we see that there exists a morphism $i' \to i''$ in $\mathcal{I}$ which equalizes $\alpha $ and $\beta $. Thus we have shown that the category $\mathcal{I}$ is filtered.

We still have to show that the colimit of the diagram is $\mathcal{F}$. By definition of the colimit, and by our definition of the category $\mathcal{I}$ there is a canonical map

Pick $x \in X$. Let us show that $\varphi _ x$ is an isomorphism. Recall that

see Sheaves, Section 6.29. First we show that the map $\varphi _ x$ is injective. Suppose that $s \in \mathcal{F}_{i, x}$ is an element such that $s$ maps to zero in $\mathcal{F}_ x$. Then there exists a quasi-compact open $U$ such that $s$ comes from $s \in \mathcal{F}_ i(U)$ and such that $\varphi _ i(s) = 0$ in $\mathcal{F}(U)$. By Lemma 28.22.2 we can find a finite type quasi-coherent subsheaf $\mathcal{K} \subset \mathop{\mathrm{Ker}}(\varphi _ i)$ which restricts to the quasi-coherent $\mathcal{O}_ U$-submodule of $\mathcal{F}_ i$ generated by $s$: $\mathcal{K}|_ U = \mathcal{O}_ U\cdot s \subset \mathcal{F}_ i|_ U$. Clearly, $\mathcal{F}_ i/\mathcal{K}$ is of finite presentation and the map $\varphi _ i$ factors through the quotient map $\mathcal{F}_ i \to \mathcal{F}_ i/\mathcal{K}$. Hence we can find an $i' \in I$ and a morphism $\alpha : \mathcal{F}_ i \to \mathcal{F}_{i'}$ in $\mathcal{I}$ which can be identified with the quotient map $\mathcal{F}_ i \to \mathcal{F}_ i/\mathcal{K}$. Then it follows that the section $s$ maps to zero in $\mathcal{F}_{i'}(U)$ and in particular in $(\mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i)_ x = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_{i, x}$. The injectivity follows. Finally, we show that the map $\varphi _ x$ is surjective. Pick $s \in \mathcal{F}_ x$. Choose a quasi-compact open neighbourhood $U \subset X$ of $x$ such that $s$ corresponds to a section $s \in \mathcal{F}(U)$. Consider the map $s : \mathcal{O}_ U \to \mathcal{F}$ (multiplication by $s$). By Lemma 28.22.4 there exists an $\mathcal{O}_ X$-module $\mathcal{G}$ of finite presentation and an $\mathcal{O}_ X$-module map $\mathcal{G} \to \mathcal{F}$ such that $\mathcal{G}|_ U \to \mathcal{F}|_ U$ is identified with $s : \mathcal{O}_ U \to \mathcal{F}$. Again by definition of $\mathcal{I}$ there exists an $i \in I$ such that $\mathcal{G} \to \mathcal{F}$ is isomorphic to $\varphi _ i : \mathcal{F}_ i \to \mathcal{F}$. Clearly there exists a section $s' \in \mathcal{F}_ i(U)$ mapping to $s \in \mathcal{F}(U)$. This proves surjectivity and the proof of the lemma is complete. $\square$