Lemma 30.10.6. Let $X$ be a locally Noetherian scheme. Let $\mathcal{F}$, $\mathcal{G}$ be coherent $\mathcal{O}_ X$-modules. Let $U \subset X$ be open and let $\varphi : \mathcal{F}|_ U \to \mathcal{G}|_ U$ be an $\mathcal{O}_ U$-module map. Then there exists a coherent submodule $\mathcal{F}' \subset \mathcal{F}$ agreeing with $\mathcal{F}$ over $U$ such that $\varphi$ extends to $\varphi ' : \mathcal{F}' \to \mathcal{G}$.

Proof. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the coherent sheaf of ideals cutting out the reduced induced scheme structure on $X \setminus U$. If $X$ is Noetherian, then Lemma 30.10.5 tells us that we can take $\mathcal{F}' = \mathcal{I}^ n\mathcal{F}$ for some $n$. The general case will follow from this using Zorn's lemma.

Consider the set of triples $(U', \mathcal{F}', \varphi ')$ where $U \subset U' \subset X$ is open, $\mathcal{F}' \subset \mathcal{F}|_{U'}$ is a coherent subsheaf agreeing with $\mathcal{F}$ over $U$, and $\varphi ' : \mathcal{F}' \to \mathcal{G}|_{U'}$ restricts to $\varphi$ over $U$. We say $(U'', \mathcal{F}'', \varphi '') \geq (U', \mathcal{F}', \varphi ')$ if and only if $U'' \supset U'$, $\mathcal{F}''|_{U'} = \mathcal{F}'$, and $\varphi ''|_{U'} = \varphi '$. It is clear that if we have a totally ordered collection of triples $(U_ i, \mathcal{F}_ i, \varphi _ i)$, then we can glue the $\mathcal{F}_ i$ to a subsheaf $\mathcal{F}'$ of $\mathcal{F}$ over $U' = \bigcup U_ i$ and extend $\varphi$ to a map $\varphi ' : \mathcal{F}' \to \mathcal{G}|_{U'}$. Hence any totally ordered subset of triples has an upper bound. Finally, suppose that $(U', \mathcal{F}', \varphi ')$ is any triple but $U' \not= X$. Then we can choose an affine open $W \subset X$ which is not contained in $U'$. By the result of the first paragraph we can extend the subsheaf $\mathcal{F}'|_{W \cap U'}$ and the restriction $\varphi '|_{W \cap U'}$ to some subsheaf $\mathcal{F}'' \subset \mathcal{F}|_ W$ and map $\varphi '' : \mathcal{F}'' \to \mathcal{G}|_ W$. Of course the agreement between $(\mathcal{F}', \varphi ')$ and $(\mathcal{F}'', \varphi '')$ over $W \cap U'$ exactly means that we can extend this to a triple $(U' \cup W, \mathcal{F}''', \varphi ''')$. Hence any maximal triple $(U', \mathcal{F}', \varphi ')$ (which exist by Zorn's lemma) must have $U' = X$ and the proof is complete. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).