# Equivalences between categories $\mathcal{K}^b(\text{Injectives})$ and $\mathcal{D}^b(\mathcal A)$ if $\mathcal{A}$ has enough injectives

I have the following question:

Let $\mathcal{A}$ be a abelian category and $\mathcal{I}$ be the full subcategory of injective objexts of $\mathcal{A}$. Assume that $\mathcal{A}$ has enough injectives. Let $\mathcal{K}^b(\mathcal{I})$ be the homotopy category of (bounded)-complexes of $\mathcal{I}$, and $\mathcal{D}^b(\mathcal A)$ be the bounded derived category of $\mathcal{A}$. Is it true that $\mathcal{K}^b(\mathcal{I})$ and $\mathcal{D}^b(\mathcal A)$ are equivalent? Thanks very much!

• See theorem 1 in nLab:derived category. – Pece Feb 1 '15 at 14:34
• Thank you! Is there any counter-example? (Pierre Schapira's book requires more condition). – Khako Feb 1 '15 at 14:58
• It's true provided you replace "bounded complex" with "bounded below cochain complex" and assume something like AB5*. It's also true if you interpret "bounded" cohomologically rather than in terms of cochain complexes. Otherwise I imagine you will need a hypothesis on cohomological dimension. – Zhen Lin Feb 1 '15 at 17:17