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.

## 19.15 Brown representability and Grothendieck abelian categories

In this section we quickly prove a representability theorem for derived categories of Grothendieck abelian categories. The reader should first read the case of compactly generated triangulated categories in Derived Categories, Section 13.38. After that, instead of reading this section, it makes sense to consult the literature for more general results of this nature, for example see [Franke], [Neeman], [Krause], or take a look at Derived Categories, Section 13.39.

**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. $\square$

Proposition 19.15.2. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $\mathcal{D}$ be a triangulated category. Let $F : D(\mathcal{A}) \to \mathcal{D}$ be an exact functor of triangulated categories which transforms direct sums into direct sums. Then $F$ has an exact right adjoint.

**Proof.**
For an object $Y$ of $\mathcal{D}$ consider the contravariant functor

This is a cohomological functor as $F$ is exact and transforms direct sums into products as $F$ transforms direct sums into direct sums. Thus by Lemma 19.15.1 we find an object $X$ of $D(\mathcal{A})$ such that $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(W, X) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(F(W), Y)$. The existence of the adjoint follows from Categories, Lemma 4.24.2. Exactness follows from Derived Categories, Lemma 13.7.1. $\square$

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