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. $\square$

Comment #2335 by Keenan Kidwell on

In the final sentence of the statement of the lemma, shouldn't $\Gamma(U,-)$ for $U\subseteq X$ open be $\Gamma(V,-)$ for $V\subseteq Y$ open?

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