Lemma 19.15.1. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $H : D(\mathcal{A}) \to \textit{Ab}$ be a contravariant cohomological functor which transforms direct sums into products. Then $H$ is representable.
Proof. Let $R, F, G, RG$ be as in Lemma 19.14.4 and consider the functor $H \circ F : D(\text{Mod}_ R) \to \textit{Ab}$. Observe that since $F$ is a left adjoint it sends direct sums to direct sums and hence $H \circ F$ transforms direct sums into products. On the other hand, the derived category $D(\text{Mod}_ R)$ is generated by a single compact object, namely $R$. By Derived Categories, Lemma 13.38.1 we see that $H \circ F$ is representable, say by $L \in D(\text{Mod}_ R)$. Choose a distinguished triangle
in $D(\text{Mod}_ R)$. Then $F(M) = 0$ because $F \circ RG = \text{id}$. Hence $H(F(M)) = 0$ hence $\mathop{\mathrm{Hom}}\nolimits (M, L) = 0$. It follows that $L \to RG(F(L))$ is the inclusion of a direct summand, see Derived Categories, Lemma 13.4.11. For $A$ in $D(\mathcal{A})$ we obtain
where the arrow has a left inverse functorial in $A$. In other words, we find that $H$ is the direct summand of a representable functor. Since $D(\mathcal{A})$ is Karoubian (Derived Categories, Lemma 13.4.14) we conclude. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)