The Stacks project

Proof. This proof is very similar to the proof of Lemma 13.38.1. We may replace $\mathcal{E}$ by $\bigcup _{i \in \mathbf{Z}} \mathcal{E}[i]$ and assume that $\mathcal{E}$ is preserved by shifts. Consider pairs $(E, a)$ where $E \in \mathcal{E}$ and $a \in H(E)$ and set

Since $H(X_1) = \prod _{(E, a)} H(E)$ we see that $(a)_{(E, a)}$ 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 $a_{n + 1} \in H(X_{n + 1})$.

Set $X = \text{hocolim} X_ n$. 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. We claim that $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, X) \to H(-)$ is an isomorphism.

Let $E \in \mathcal{E}$. Let us show that

is injective. Namely, let $\alpha : E \to \bigoplus X_ n$. Then by assumption (2) we obtain a factorization $\alpha = (\bigoplus \beta _ n) \circ \gamma$. Since $E_ n \to X_ n \to X_{n + 1}$ is zero by construction, we see that the composition $\bigoplus E_ n \to \bigoplus X_ n \to \bigoplus X_ n$ is equal to $\bigoplus \beta _ n$. Hence also the composition $E \to \bigoplus X_ n \to \bigoplus X_ n$ is equal to $\alpha$. This proves the stated injectivity and hence also

is injective. It follows that we have an exact sequence

for all $E \in \mathcal{E}$.

Let $E \in \mathcal{E}$ and let $f : E \to X$ be a morphism. By the previous paragraph, we may choose $\alpha : E \to \bigoplus X_ n$ lifting $f$. Then by assumption (2) we obtain a factorization $\alpha = (\bigoplus \beta _ n) \circ \gamma$. For each $n$ there is a morphism $\delta _ n : E_ n \to X_1$ such that $\delta _ n$ and $\beta _ n$ map to the same element of $H(E_ n)$. Then the compositions

are equal by construction of $X_ n \to X_{n + 1}$. It follows that

are the same too. Observing that $\bigoplus X_1 \to X$ factors as $\bigoplus X_1 \to X_1 \to X$, we conclude that

is surjective. Since by construction the map $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, X_1) \to H(E)$ is surjective and by construction the kernel of this map is annihilated by $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, X_1) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, X)$ we conclude that $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, X) \to H(E)$ is a bijection for all $E \in \mathcal{E}$.

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 $\mathcal{E}$ is preserved by shifts, we conclude that $\mathcal{E} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$ by the result of the previous paragraph. To finish the proof we have to show that $\mathcal{D}' = \mathcal{D}$.

Let $Y$ be an object of $\mathcal{D}$ and set $H(-) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, Y)$. Then $H$ is a cohomological functor which transforms direct sums into products. By the construction in the first part of the proof we obtain a morphism $\mathop{\mathrm{colim}}\nolimits X_ n = X \to Y$ such that $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, X) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, Y)$ is bijective for all $E \in \mathcal{E}$. Then assumption (1) tells us that $X \to Y$ is an isomorphism! On the other hand, by construction $X_1, X_2, \ldots$ are in $\mathcal{D}'$ and so is $X$. Thus $Y \in \mathcal{D}'$ and the proof is complete. $\square$

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 13.39.1 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. $\square$