Weak version of [Theorem A, Krause]

Lemma 13.39.1. Let $\mathcal{D}$ be a triangulated category with direct sums. Suppose given a set $\mathcal{E}$ of objects of $\mathcal{D}$ such that

1. if $X$ is a nonzero object of $\mathcal{D}$, then there exists an $E \in \mathcal{E}$ and a nonzero map $E \to X$, and

2. given objects $X_ n$, $n \in \mathbf{N}$ of $\mathcal{D}$, $E \in \mathcal{E}$, and $\alpha : E \to \bigoplus X_ n$, there exist $E_ n \in \mathcal{E}$ and $\beta _ n : E_ n \to X_ n$ and a morphism $\gamma : E \to \bigoplus E_ n$ such that $\alpha = (\bigoplus \beta _ n) \circ \gamma$.

Let $H : \mathcal{D} \to \textit{Ab}$ be a contravariant cohomological functor which transforms direct sums into products. Then $H$ is representable.

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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GYG. Beware of the difference between the letter 'O' and the digit '0'.