Lemma 13.16.6. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor between abelian categories and assume $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is everywhere defined.

1. The functors $R^ iF$, $i \geq 0$ come equipped with a canonical structure of a $\delta$-functor from $\mathcal{A} \to \mathcal{B}$, see Homology, Definition 12.12.1.

2. If every object of $\mathcal{A}$ is a subobject of a right acyclic object for $F$, then $\{ R^ iF, \delta \} _{i \geq 0}$ is a universal $\delta$-functor, see Homology, Definition 12.12.3.

Proof. The functor $\mathcal{A} \to \text{Comp}^{+}(\mathcal{A})$, $A \mapsto A[0]$ is exact. The functor $\text{Comp}^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ is a $\delta$-functor, see Lemma 13.12.1. The functor $RF : D^{+}(\mathcal{A}) \to D^{+}(\mathcal{B})$ is exact. Finally, the functor $H^0 : D^{+}(\mathcal{B}) \to \mathcal{B}$ is a homological functor, see Definition 13.11.3. Hence we get the structure of a $\delta$-functor from Lemma 13.4.22 and Lemma 13.4.21. Part (2) follows from Homology, Lemma 12.12.4 and the description of acyclics in Lemma 13.16.4. $\square$

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.

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 05TE. Beware of the difference between the letter 'O' and the digit '0'.