The Stacks project

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 $n \in \mathbf{Z}$.


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09LJ. Beware of the difference between the letter 'O' and the digit '0'.