## 22.32 Variant of derived Hom

Let $\mathcal{A}$ be an abelian category. Consider the differential graded category $\text{Comp}^{dg}(\mathcal{A})$ of complexes of $\mathcal{A}$, see Example 22.26.6. Let $K^\bullet $ be a complex of $\mathcal{A}$. Set

\[ (E, \text{d}) = \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{A})}(K^\bullet , K^\bullet ) \]

and consider the functor of differential graded categories

\[ \text{Comp}^{dg}(\mathcal{A}) \longrightarrow \text{Mod}^{dg}_{(E, \text{d})}, \quad X^\bullet \longmapsto \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{A})}(K^\bullet , X^\bullet ) \]

of Lemma 22.26.10.

Lemma 22.32.1. In the situation above. If the right derived functor $R\mathop{\mathrm{Hom}}\nolimits (K^\bullet , -)$ of $\mathop{\mathrm{Hom}}\nolimits (K^\bullet , -) : K(\mathcal{A}) \to D(\textit{Ab})$ is everywhere defined on $D(\mathcal{A})$, then we obtain a canonical exact functor

\[ R\mathop{\mathrm{Hom}}\nolimits (K^\bullet , -) : D(\mathcal{A}) \longrightarrow D(E, \text{d}) \]

of triangulated categories which reduces to the usual one on taking associated complexes of abelian groups.

**Proof.**
Note that we have an associated functor $K(\mathcal{A}) \to K(\text{Mod}_{(E, \text{d})})$ by Lemma 22.26.10. We claim this functor is an exact functor of triangulated categories. Namely, let $f : A^\bullet \to B^\bullet $ be a map of complexes of $\mathcal{A}$. Then a computation shows that

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{A})}(K^\bullet , C(f)^\bullet ) = C\left( \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{A})}(K^\bullet , A^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}^{dg}(\mathcal{A})}(K^\bullet , B^\bullet ) \right) \]

where the right hand side is the cone in $\text{Mod}_{(E, \text{d})}$ defined earlier in this chapter. This shows that our functor is compatible with cones, hence with distinguished triangles. Let $X^\bullet $ be an object of $K(\mathcal{A})$. Consider the category of quasi-isomorphisms $s : X^\bullet \to Y^\bullet $. We are given that the functor $(s : X^\bullet \to Y^\bullet ) \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(K^\bullet , Y^\bullet )$ is essentially constant when viewed in $D(\textit{Ab})$. But since the forgetful functor $D(E, \text{d}) \to D(\textit{Ab})$ is compatible with taking cohomology, the same thing is true in $D(E, \text{d})$. This proves the lemma.
$\square$

**Warning:** Although the lemma holds as stated and may be useful as stated, the differential algebra $E$ isn't the “correct” one unless $H^ n(E) = \mathop{\mathrm{Ext}}\nolimits ^ n_{D(\mathcal{A})}(K^\bullet , K^\bullet )$ for all $n \in \mathbf{Z}$.

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