Lemma 20.41.2. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $\mathcal{B}$ be a basis for the topology on $Y$.

1. Assume $K$ is in $D(\mathcal{O}_ X)$ such that for $V \in \mathcal{B}$ we have $H^ i(f^{-1}(V), K) = 0$ for $i < 0$. Then $Rf_*K$ has vanishing cohomology sheaves in negative degrees, $H^ i(f^{-1}(V), K) = 0$ for $i < 0$ for all opens $V \subset Y$, and the rule $V \mapsto H^0(f^{-1}V, K)$ is a sheaf on $Y$.

2. Assume $K, L$ are in $D(\mathcal{O}_ X)$ such that for $V \in \mathcal{B}$ we have $\mathop{\mathrm{Ext}}\nolimits ^ i(K|_{f^{-1}V}, L|_{f^{-1}V}) = 0$ for $i < 0$. Then $\mathop{\mathrm{Ext}}\nolimits ^ i(K|_{f^{-1}V}, L|_{f^{-1}V}) = 0$ for $i < 0$ for all opens $V \subset Y$ and the rule $V \mapsto \mathop{\mathrm{Hom}}\nolimits (K|_{f^{-1}V}, L|_{f^{-1}V})$ is a sheaf on $Y$.

Proof. Lemma 20.32.6 tells us $H^ i(Rf_*K)$ is the sheaf associated to the presheaf $V \mapsto H^ i(f^{-1}(V), K) = H^ i(V, Rf_*K)$. The assumptions in (1) imply that $Rf_*K$ has vanishing cohomology sheaves in degrees $< 0$. We conclude that for any open $V \subset Y$ the cohomology group $H^ i(V, Rf_*K)$ is zero for $i < 0$ and is equal to $H^0(V, H^0(Rf_*K))$ for $i = 0$. This proves (1).

To prove (2) apply (1) to the complex $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, L)$ using Lemma 20.38.1 to do the translation. $\square$

There are also:

• 2 comment(s) on Section 20.41: Glueing complexes

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).