Lemma 13.16.9. 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. $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined,

3. $RF : D(\mathcal{A}) \to D(\mathcal{B})$ 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. $\square$

Comment #4282 by David Benjamin Lim on

There seems to be a typo in this lemma. We surely mean $RF : D^+(\mathcal{A}) \to D^+(\mathcal{B})$ (and similarly for the one on unbounded derived categories) no?

Comment #4446 by on

For some reason your comment made me laugh! Thanks and fixed here.

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