Lemma 13.37.3. Let $\mathcal{D}$ be a triangulated category with direct sums. Let $E_ i$, $i \in I$ be a family of compact objects of $\mathcal{D}$ such that $\bigoplus E_ i$ generates $\mathcal{D}$. Then every object $X$ of $\mathcal{D}$ can be written as

$X = \text{hocolim} X_ n$

where $X_1$ is a direct sum of shifts of the $E_ i$ and each transition morphism fits into a distinguished triangle $Y_ n \to X_ n \to X_{n + 1} \to Y_ n[1]$ where $Y_ n$ is a direct sum of shifts of the $E_ i$.

Proof. Set $X_1 = \bigoplus _{(i, m, \varphi )} E_ i[m]$ where the direct sum is over all triples $(i, m, \varphi )$ such that $i \in I$, $m \in \mathbf{Z}$ and $\varphi : E_ i[m] \to X$. Then $X_1$ comes equipped with a canonical morphism $X_1 \to X$. Given $X_ n \to X$ we set $Y_ n = \bigoplus _{(i, m, \varphi )} E_ i[m]$ where the direct sum is over all triples $(i, m, \varphi )$ such that $i \in I$, $m \in \mathbf{Z}$, and $\varphi : E_ i[m] \to X_ n$ is a morphism such that $E_ i[m] \to X_ n \to X$ is zero. Choose a distinguished triangle $Y_ n \to X_ n \to X_{n + 1} \to Y_ n[1]$ and let $X_{n + 1} \to X$ be any morphism such that $X_ n \to X_{n + 1} \to X$ is the given one; such a morphism exists by our choice of $Y_ n$. We obtain a morphism $\text{hocolim} X_ n \to X$ by the construction of our maps $X_ n \to X$. Choose a distinguished triangle

$C \to \text{hocolim} X_ n \to X \to C[1]$

Let $E_ i[m] \to C$ be a morphism. Since $E_ i$ is compact, the composition $E_ i[m] \to C \to \text{hocolim} X_ n$ factors through $X_ n$ for some $n$, say by $E_ i[m] \to X_ n$. Then the construction of $Y_ n$ shows that the composition $E_ i[m] \to X_ n \to X_{n + 1}$ is zero. In other words, the composition $E_ i[m] \to C \to \text{hocolim} X_ n$ is zero. This means that our morphism $E_ i[m] \to C$ comes from a morphism $E_ i[m] \to X[-1]$. The construction of $X_1$ then shows that such morphism lifts to $\text{hocolim} X_ n$ and we conclude that our morphism $E_ i[m] \to C$ is zero. The assumption that $\bigoplus E_ i$ generates $\mathcal{D}$ implies that $C$ is zero and the proof is done. $\square$

Comment #4349 by awllower on

I think in the proof the composition $E_ i[m] \to X \to \text{hocolim} X_ n$ is supposed to be $E_ i[m] \to C \to \text{hocolim} X_ n$?

Comment #8351 by TheSentence on

I think we should consider the case that $Y_{n}=0$ for some $n$. Whenever $Y_{n}=0$ for some $n$, we can prove $X_{n} \rightarrow X$ is isomorphism, which means we can stop the step for $n$.

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