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

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

$H^ p(V, f_*\mathcal{I}) = 0$ 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

Comment #2406 by Johan on