The Stacks project

Lemma 17.22.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules. If $\mathcal{F}$ is of finite type then the canonical map

\[ \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})_ x \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\mathcal{F}_ x, \mathcal{G}_ x) \]

is injective. If $\mathcal{F}$ is finitely presented, this canonical morphism is an isomorphism.

Proof. The map sends the equivalence class of $(U, \varphi )$ in $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})_ x$, where $x \in U \subset X$ is open and $\varphi \in \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{F}|_ U, \mathcal{G}|_ U)$, to the the induced map on stalks at $x$, namely $\varphi _ x : \mathcal{F}_ x \to \mathcal{G}_ x$.

Suppose $\mathcal{F}$ is of finite type. Pick a representative $(U, \varphi )$ of an element $\sigma $ in the kernel of the map, i.e., $\varphi _ x = 0$. Shrinking $U$ if necessary, choose sections $s^1, \ldots , s^ n \in \mathcal{F}(U)$ generating $\mathcal{F}|_ U$. Since $\varphi _ x(s^ i_ x) = 0$ and we are dealing with a finite number of sections, we can find an open neighborhood $V \subset U$ of $x$ such that $\varphi _ V(s^ i|_ V)=0$ for all $i = 1, \ldots , n$. Since $s^ i|_ V$, $i = 1, \ldots , n$ generate $\mathcal{F}|_ V$ this means that $\varphi |_ V = 0$. Since $(U, \varphi )$ is equivalent to $(V, \varphi |_ V)$ we conclude $\sigma = 0$ and injectivity of the map follows.

Next, assume $\mathcal{F}$ is finitely presented. By localizing on $X$ we may assume that $\mathcal{F}$ has a presentation

\[ \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{O}_ X \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{O}_ X \to \mathcal{F} \to 0. \]

By Lemma 17.22.2 this gives an exact sequence $ 0 \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \to \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{G} \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{G}. $ Taking stalks we get an exact sequence $ 0 \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})_ x \to \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{G}_ x \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{G}_ x $ and the result follows since $\mathcal{F}_ x$ sits in an exact sequence $ \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{O}_{X, x} \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{O}_{X, x} \to \mathcal{F}_ x \to 0 $ which induces the exact sequence $ 0 \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\mathcal{F}_ x, \mathcal{G}_ x) \to \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{G}_ x \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{G}_ x $ which is the same as the one above. $\square$

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