Lemma 13.26.12. Let $\mathcal{A}$ be an abelian category with enough injectives. Let $\mathcal{I}^ f \subset \text{Fil}^ f(\mathcal{A})$ denote the strictly full additive subcategory whose objects are the filtered injective objects. The canonical functor

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

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

$\xymatrix{ K^{+}(\mathcal{I}^ f) \ar[d]_{\text{gr}^ p} \ar[r] & DF^{+}(\mathcal{A}) \ar[d]_{\text{gr}^ p} \\ K^{+}(\mathcal{I}) \ar[r] & D^{+}(\mathcal{A}) } \quad \xymatrix{ K^{+}(\mathcal{I}^ f) \ar[d]^{\text{forget }F} \ar[r] & DF^{+}(\mathcal{A}) \ar[d]^{\text{forget }F} \\ K^{+}(\mathcal{I}) \ar[r] & D^{+}(\mathcal{A}) }$

are commutative, where $\mathcal{I} \subset \mathcal{A}$ is the strictly full additive subcategory whose objects are the injective objects.

Proof. The functor $K^{+}(\mathcal{I}^ f) \to DF^{+}(\mathcal{A})$ is essentially surjective by Lemma 13.26.9. It is fully faithful by Lemma 13.26.11. It is an exact functor by our definitions regarding distinguished triangles. The commutativity of the squares is immediate. $\square$

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).