[Theorem 3.1, Neeman-Grothendieck].

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.

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

$X_1 = \bigoplus \nolimits _{(i, a)} E_ i$

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

$K_{n + 1} = \bigoplus \nolimits _{(i, k),\ k \in \mathop{\mathrm{Ker}}(H_ n(E_ i) \to H(E_ i))} E_ i$

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

$K_{n + 1} \to X_ n \to X_{n + 1} \to K_{n + 1}[1]$

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

$\bigoplus X_ n \to \bigoplus X_ n \to X \to \bigoplus X_ n[1]$

we obtain an exact sequence

$\prod H(X_ n) \leftarrow \prod H(X_ n) \leftarrow H(X)$

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

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, X_1) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, X_2) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, X_3) \to \ldots \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, X) \to H$

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

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E_ i, X) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E_ i, X_ n)$

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

$\mathcal{D}' = \{ Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \mid \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(Y[n], X) \to H(Y[n]) \text{ is an isomorphism for all }n\}$

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. $\square$

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).