Lemma 13.31.2. Let $\mathcal{A}$ be an abelian category. Let $I^\bullet$ be a complex. The following are equivalent

1. $I^\bullet$ is K-injective,

2. for every quasi-isomorphism $M^\bullet \to N^\bullet$ the map

$\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(N^\bullet , I^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet , I^\bullet )$

is bijective, and

3. for every complex $N^\bullet$ the map

$\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(N^\bullet , I^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(N^\bullet , I^\bullet )$

is an isomorphism.

Proof. Assume (1). Then (2) holds because the functor $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}( - , I^\bullet )$ is cohomological and the cone on a quasi-isomorphism is acyclic.

Assume (2). A morphism $N^\bullet \to I^\bullet$ in $D(\mathcal{A})$ is of the form $fs^{-1} : N^\bullet \to I^\bullet$ where $s : M^\bullet \to N^\bullet$ is a quasi-isomorphism and $f : M^\bullet \to I^\bullet$ is a map. By (2) this corresponds to a unique morphism $N^\bullet \to I^\bullet$ in $K(\mathcal{A})$, i.e., (3) holds.

Assume (3). If $M^\bullet$ is acyclic then $M^\bullet$ is isomorphic to the zero complex in $D(\mathcal{A})$ hence $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(M^\bullet , I^\bullet ) = 0$, whence $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet , I^\bullet ) = 0$ by (3), i.e., (1) holds. $\square$

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