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The Stacks project

Lemma 20.11.10. Let f : X \to Y be a morphism of ringed spaces. Let \mathcal{I} be an injective \mathcal{O}_ X-module. Then

  1. \check{H}^ p(\mathcal{V}, f_*\mathcal{I}) = 0 for all p > 0 and any open covering \mathcal{V} : V = \bigcup _{j \in J} V_ j of Y.

  2. H^ p(V, f_*\mathcal{I}) = 0 for all p > 0 and every open V \subset Y.

In other words, f_*\mathcal{I} is right acyclic for \Gamma (V, -) (see Derived Categories, Definition 13.15.3) for any V \subset Y open.

Proof. Set \mathcal{U} : f^{-1}(V) = \bigcup _{j \in J} f^{-1}(V_ j). It is an open covering of X and

\check{\mathcal{C}}^\bullet (\mathcal{V}, f_*\mathcal{I}) = \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}).

This is true because

f_*\mathcal{I}(V_{j_0 \ldots j_ p}) = \mathcal{I}(f^{-1}(V_{j_0 \ldots j_ p})) = \mathcal{I}(f^{-1}(V_{j_0}) \cap \ldots \cap f^{-1}(V_{j_ p})) = \mathcal{I}(U_{j_0 \ldots j_ p}).

Thus the first statement of the lemma follows from Lemma 20.11.1. The second statement follows from the first and Lemma 20.11.8. \square


Comments (2)

Comment #2335 by Keenan Kidwell on

In the final sentence of the statement of the lemma, shouldn't for open be for open?


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