The Stacks project

Lemma 13.20.1. Let $\mathcal{A}$ be an abelian category. Let $I \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be an injective object. Let $I^\bullet $ be a bounded below complex of injectives in $\mathcal{A}$.

  1. $I^\bullet $ computes $RF$ relative to $\text{Qis}^{+}(\mathcal{A})$ for any exact functor $F : K^{+}(\mathcal{A}) \to \mathcal{D}$ into any triangulated category $\mathcal{D}$.

  2. $I$ is right acyclic for any additive functor $F : \mathcal{A} \to \mathcal{B}$ into any abelian category $\mathcal{B}$.

Proof. Part (2) is a direct consequences of part (1) and Definition 13.15.3. To prove (1) let $\alpha : I^\bullet \to K^\bullet $ be a quasi-isomorphism into a complex. By Lemma 13.18.6 we see that $\alpha $ has a left inverse. Hence the category $I^\bullet /\text{Qis}^{+}(\mathcal{A})$ is essentially constant with value $\text{id} : I^\bullet \to I^\bullet $. Thus also the ind-object

\[ I^\bullet /\text{Qis}^{+}(\mathcal{A}) \longrightarrow \mathcal{D}, \quad (I^\bullet \to K^\bullet ) \longmapsto F(K^\bullet ) \]

is essentially constant with value $F(I^\bullet )$. This proves (1), see Definitions 13.14.2 and 13.14.10. $\square$

Comments (2)

Comment #2336 by Keenan Kidwell on

I think that, for the existence of a left inverse to in , one should cite 13.18.6 instead of 13.18.7.

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Your email address will not be published.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 05TH. Beware of the difference between the letter 'O' and the digit '0'.