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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

  1. every object of $\mathcal{A}$ is right acyclic for $F$,
  2. $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is everywhere defined,
  3. $RF : D(\mathcal{A}) \to D(\mathcal{A})$ is everywhere defined,
  4. every complex computes $RF$, in other words, the canonical map $F(K^\bullet) \to RF(K^\bullet)$ is an isomorphism for all complexes, and
  5. $R^iF = 0$ 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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