[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. $\square$

This is a rather annoying question but which notion of representability do we use in the formulation of Lemma 0A8F? The usual one is defined for functors to the category of sets and not to Ab...

Comment #6755 by on

We do indeed mean that $H$ as a functor into the category of sets is representable, in other words that $forget \circ H$ is representable where $forget : \textit{Ab} \to \textit{Sets}$ is the forgetful functor. This comes up frequently. Since no other notion of representable has been defined, I think no confusion can result. But let me know if you think otherwise.

Suppose $\mathcal{A}$ is an additive category and $H, H' : \mathcal{A} \to \textit{Ab}$ are two additive functors. As a related exercise you can show that if $t : forget \circ H \to forget \circ H'$ is a transformation of functors, then $t$ lifts to $t : H \to H'$.

Thanks: the exercise explains that there is no difference between "representable" functors from an additive category to Ab defined as isomorphic to functors associated to objects and the usual definition when we use the forgetful functor from Ab to Sets.

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