Proposition 13.23.1. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives. Denote $\mathcal{I} \subset \mathcal{A}$ the strictly full additive subcategory whose objects are the injective objects of $\mathcal{A}$. The functor

$K^{+}(\mathcal{I}) \longrightarrow D^{+}(\mathcal{A})$

is exact, fully faithful and essentially surjective, i.e., an equivalence of triangulated categories.

Proof. It is clear that the functor is exact. It is essentially surjective by Lemma 13.18.3. Fully faithfulness is a consequence of Lemma 13.18.8. $\square$

## Comments (1)

Comment #9469 by on

I guess one should mention the fact that $K^+(\mathcal{I})$ is a full triangulated subcategory of $K^+(\mathcal{A})$. This follows from Lemma 13.4.16, since the cone of a morphism of complexes of injective objects is a complex of injective objects by Homology, Lemma 12.27.

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