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The Stacks project

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{restr.}} \mathcal{F}(U)

for every pair (U, V) as above.

Proof. Similar to the above and omitted. \square


Comments (2)

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.".

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

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