Lemma 13.38.1. Let $\mathcal{D}$ be a triangulated category with direct sums which is compactly generated. Let $H : \mathcal{D} \to \textit{Ab}$ be a contravariant cohomological functor which transforms direct sums into products. Then $H$ is representable.

## 13.38 Brown representability

A reference for the material in this section is [Neeman-Grothendieck].

**Proof.**
Let $E_ i$, $i \in I$ be a set of compact objects such that $\bigoplus _{i \in I} E_ i$ generates $\mathcal{D}$. We may and do assume that the set of objects $\{ E_ i\} $ is preserved under shifts. Consider pairs $(i, a)$ where $i \in I$ and $a \in H(E_ i)$ and set

Since $H(X_1) = \prod _{(i, a)} H(E_ i)$ we see that $(a)_{(i, a)}$ defines an element $a_1 \in H(X_1)$. Set $H_1 = \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(- , X_1)$. By Yoneda's lemma (Categories, Lemma 4.3.5) the element $a_1$ defines a natural transformation $H_1 \to H$.

We are going to inductively construct $X_ n$ and transformations $a_ n : H_ n \to H$ where $H_ n = \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, X_ n)$. Namely, we apply the procedure above to the functor $\mathop{\mathrm{Ker}}(H_ n \to H)$ to get an object

and a transformation $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, K_{n + 1}) \to \mathop{\mathrm{Ker}}(H_ n \to H)$. By Yoneda's lemma the composition $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, K_{n + 1}) \to H_ n$ gives a morphism $K_{n + 1} \to X_ n$. We choose a distinguished triangle

in $\mathcal{D}$. The element $a_ n \in H(X_ n)$ maps to zero in $H(K_{n + 1})$ by construction. Since $H$ is cohomological we can lift it to an element $a_{n + 1} \in H(X_{n + 1})$.

We claim that $X = \text{hocolim} X_ n$ represents $H$. Applying $H$ to the defining distinguished triangle

we obtain an exact sequence

Thus there exists an element $a \in H(X)$ mapping to $(a_ n)$ in $\prod H(X_ n)$. Hence a natural transformation $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(- , X) \to H$ such that

commutes. For each $i$ the map $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E_ i, X) \to H(E_ i)$ is surjective, by construction of $X_1$. On the other hand, by construction of $X_ n \to X_{n + 1}$ the kernel of $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E_ i, X_ n) \to H(E_ i)$ is killed by the map $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E_ i, X_ n) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E_ i, X_{n + 1})$. Since

by Lemma 13.33.9 we see that $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E_ i, X) \to H(E_ i)$ is injective.

To finish the proof, consider the subcategory

As $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, X) \to H$ is a transformation between cohomological functors, the subcategory $\mathcal{D}'$ is a strictly full, saturated, triangulated subcategory of $\mathcal{D}$ (details omitted; see proof of Lemma 13.6.3). Moreover, as both $H$ and $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, X)$ transform direct sums into products, we see that direct sums of objects of $\mathcal{D}'$ are in $\mathcal{D}'$. Thus derived colimits of objects of $\mathcal{D}'$ are in $\mathcal{D}'$. Since $\{ E_ i\} $ is preserved under shifts, we see that $E_ i$ is an object of $\mathcal{D}'$ for all $i$. It follows from Lemma 13.37.3 that $\mathcal{D}' = \mathcal{D}$ and the proof is complete. $\square$

Proposition 13.38.2. Let $\mathcal{D}$ be a triangulated category with direct sums which is compactly generated. Let $F : \mathcal{D} \to \mathcal{D}'$ 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 13.38.1 we find an object $X$ of $\mathcal{D}$ such that $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, X) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(F(W), Y)$. The existence of the adjoint follows from Categories, Lemma 4.24.2. Exactness follows from Lemma 13.7.1. $\square$

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