## Tag `015F`

Chapter 13: Derived Categories > Section 13.17: Higher derived functors

Lemma 13.17.9. Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor of abelian categories. Then

- every object of $\mathcal{A}$ is right acyclic for $F$,
- $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is everywhere defined,
- $RF : D(\mathcal{A}) \to D(\mathcal{A})$ is everywhere defined,
- every complex computes $RF$, in other words, the canonical map $F(K^\bullet) \to RF(K^\bullet)$ is an isomorphism for all complexes, and
- $R^iF = 0$ for $i \not = 0$.

Proof.This is true because $F$ transforms acyclic complexes into acyclic complexes and quasi-isomorphisms into quasi-isomorphisms. Details omitted. $\square$

The code snippet corresponding to this tag is a part of the file `derived.tex` and is located in lines 5734–5746 (see updates for more information).

```
\begin{lemma}
\label{lemma-right-derived-exact-functor}
Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor of
abelian categories. Then
\begin{enumerate}
\item every object of $\mathcal{A}$ is right acyclic for $F$,
\item $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is everywhere defined,
\item $RF : D(\mathcal{A}) \to D(\mathcal{A})$ is everywhere defined,
\item every complex computes $RF$, in other words, the canonical
map $F(K^\bullet) \to RF(K^\bullet)$ is an isomorphism for all complexes, and
\item $R^iF = 0$ for $i \not = 0$.
\end{enumerate}
\end{lemma}
\begin{proof}
This is true because $F$ transforms acyclic complexes into acyclic complexes
and quasi-isomorphisms into quasi-isomorphisms. Details omitted.
\end{proof}
```

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