The Stacks project

20.3 Derived functors

We briefly explain how to get right derived functors using resolution functors. For the unbounded derived functors, please see Section 20.28.

Let $(X, \mathcal{O}_ X)$ be a ringed space. The category $\textit{Mod}(\mathcal{O}_ X)$ is abelian, see Modules, Lemma 17.3.1. In this chapter we will write

\[ K(\mathcal{O}_ X) = K(\textit{Mod}(\mathcal{O}_ X)) \quad \text{and} \quad D(\mathcal{O}_ X) = D(\textit{Mod}(\mathcal{O}_ X)). \]

and similarly for the bounded versions for the triangulated categories introduced in Derived Categories, Definition 13.8.1 and Definition 13.11.3. By Derived Categories, Remark 13.24.3 there exists a resolution functor

\[ j = j_ X : K^{+}(\textit{Mod}(\mathcal{O}_ X)) \longrightarrow K^{+}(\mathcal{I}) \]

where $\mathcal{I}$ is the strictly full additive subcategory of $\textit{Mod}(\mathcal{O}_ X)$ consisting of injective sheaves. For any left exact functor $F : \textit{Mod}(\mathcal{O}_ X) \to \mathcal{B}$ into any abelian category $\mathcal{B}$ we will denote $RF$ the right derived functor described in Derived Categories, Section 13.20 and constructed using the resolution functor $j_ X$ just described:
\begin{equation} \label{cohomology-equation-RF} RF = F \circ j_ X' : D^{+}(X) \longrightarrow D^{+}(\mathcal{B}) \end{equation}

see Derived Categories, Lemma 13.25.1 for notation. Note that we may think of $RF$ as defined on $\textit{Mod}(\mathcal{O}_ X)$, $\text{Comp}^{+}(\textit{Mod}(\mathcal{O}_ X))$, $K^{+}(X)$, or $D^{+}(X)$ depending on the situation. According to Derived Categories, Definition 13.16.2 we obtain the $i$th right derived functor
\begin{equation} \label{cohomology-equation-RFi} R^ iF = H^ i \circ RF : \textit{Mod}(\mathcal{O}_ X) \longrightarrow \mathcal{B} \end{equation}

so that $R^0F = F$ and $\{ R^ iF, \delta \} _{i \geq 0}$ is universal $\delta $-functor, see Derived Categories, Lemma 13.20.4.

Here are two special cases of this construction. Given a ring $R$ we write $K(R) = K(\text{Mod}_ R)$ and $D(R) = D(\text{Mod}_ R)$ and similarly for bounded versions. For any open $U \subset X$ we have a left exact functor $ \Gamma (U, -) : \textit{Mod}(\mathcal{O}_ X) \longrightarrow \text{Mod}_{\mathcal{O}_ X(U)} $ which gives rise to
\begin{equation} \label{cohomology-equation-total-derived-cohomology} R\Gamma (U, -) : D^{+}(X) \longrightarrow D^{+}(\mathcal{O}_ X(U)) \end{equation}

by the discussion above. We set $H^ i(U, -) = R^ i\Gamma (U, -)$. If $U = X$ we recover ( If $f : X \to Y$ is a morphism of ringed spaces, then we have the left exact functor $ f_* : \textit{Mod}(\mathcal{O}_ X) \longrightarrow \textit{Mod}(\mathcal{O}_ Y) $ which gives rise to the derived pushforward
\begin{equation} \label{cohomology-equation-total-derived-direct-image} Rf_* : D^{+}(X) \longrightarrow D^{+}(Y) \end{equation}

The $i$th cohomology sheaf of $Rf_*\mathcal{F}^\bullet $ is denoted $R^ if_*\mathcal{F}^\bullet $ and called the $i$th higher direct image in accordance with ( The two displayed functors above are exact functors of derived categories.

Abuse of notation: When the functor $Rf_*$, or any other derived functor, is applied to a sheaf $\mathcal{F}$ on $X$ or a complex of sheaves it is understood that $\mathcal{F}$ has been replaced by a suitable resolution of $\mathcal{F}$. To facilitate this kind of operation we will say, given an object $\mathcal{F}^\bullet \in D(\mathcal{O}_ X)$, that a bounded below complex $\mathcal{I}^\bullet $ of injectives of $\textit{Mod}(\mathcal{O}_ X)$ represents $\mathcal{F}^\bullet $ in the derived category if there exists a quasi-isomorphism $\mathcal{F}^\bullet \to \mathcal{I}^\bullet $. In the same vein the phrase “let $\alpha : \mathcal{F}^\bullet \to \mathcal{G}^\bullet $ be a morphism of $D(\mathcal{O}_ X)$” does not mean that $\alpha $ is represented by a morphism of complexes. If we have an actual morphism of complexes we will say so.

Comments (1)

Comment #1810 by Keenan Kidwell on

In the second sentence of the text block following the displayed equation , "...are exact functor..." should be "...are exact functors..."

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0716. Beware of the difference between the letter 'O' and the digit '0'.