Lemma 6.30.17. Let $(f, f^\sharp ) : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. Let $\mathcal{G}$ be a sheaf of $\mathcal{O}_ Y$-modules. Let $\mathcal{B}_ Y$ be a basis for the topology on $Y$. Let $\mathcal{B}_ X$ be a basis for the topology on $X$. Suppose given for every $V \in \mathcal{B}_ Y$, and $U \in \mathcal{B}_ X$ such that $f(U) \subset V$ a $\mathcal{O}_ Y(V)$-module map

$\varphi _ V^ U : \mathcal{G}(V) \longrightarrow \mathcal{F}(U)$

compatible with restriction mappings. Here the $\mathcal{O}_ Y(V)$-module structure on $\mathcal{F}(U)$ comes from the $\mathcal{O}_ X(U)$-module structure via the map $f^\sharp _ V : \mathcal{O}_ Y(V) \to \mathcal{O}_ X(f^{-1}V) \to \mathcal{O}_ X(U)$. Then there is a unique $f$-map of sheaves of modules (see Definition 6.21.7 and the discussion of $f$-maps in Section 6.26) $\varphi : \mathcal{G} \to \mathcal{F}$ recovering $\varphi _ V^ U$ as the composition

$\mathcal{G}(V) \xrightarrow {\varphi _ V} \mathcal{F}(f^{-1}(V)) \xrightarrow {\text{restrc.}} \mathcal{F}(U)$

for every pair $(U, V)$ as above.

Proof. Similar to the above and omitted. $\square$

Comment #4912 by Tim Holzschuh on

Very small and unnecessary detail: The last morphism in the lemma is called "restrc." whereas the "corresponding" morphism in the lemma before is called "restr.".

There are also:

• 6 comment(s) on Section 6.30: Bases and sheaves

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).