**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

\[ X_1 = \bigoplus \nolimits _{(E, a)} E \]

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

\[ K_{n + 1} = \bigoplus \nolimits _{(E, k),\ k \in \mathop{\mathrm{Ker}}(H_ n(E) \to H(E))} E \]

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 $a_{n + 1} \in H(X_{n + 1})$.

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

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

\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, \bigoplus X_ n) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, \bigoplus X_ n) \]

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

\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, \bigoplus X_ n[1]) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, \bigoplus X_ n[1]) \]

is injective. It follows that we have an exact sequence

\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, \bigoplus X_ n) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, \bigoplus X_ n) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, X) \to 0 \]

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

\[ E_ n \to X_ n \to X_{n + 1} \quad \text{and}\quad E_ n \to X_1 \to X_{n + 1} \]

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

\[ \bigoplus E_ n \to \bigoplus X_ n \to X \quad \text{and}\quad \bigoplus E_ n \to \bigoplus X_1 \to X \]

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

\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, X_1) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(E, X) \]

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

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

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 $\mathcal{E}$ 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.
$\square$

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