Lemma 19.12.2. Let $\mathcal{A}$ be a Grothendieck abelian category. There exists a cardinal $\kappa$ such that given any acyclic complex $M^\bullet$ we have

1. if $M^\bullet$ is nonzero, there is a nonzero subcomplex $N^\bullet$ which is bounded above, acyclic, and $|N^ n| \leq \kappa$,

2. there exists a surjection of complexes

$\bigoplus \nolimits _{i \in I} M_ i^\bullet \longrightarrow M^\bullet$

where $M_ i^\bullet$ is bounded above, acyclic, and $|M_ i^ n| \leq \kappa$.

Proof. Choose a generator $U$ of $\mathcal{A}$. Denote $c$ the function of Lemma 19.12.1. Set $\kappa = \sup \{ c^ n(|U|), n = 1, 2, 3, \ldots \}$. Let $n \in \mathbf{Z}$ and let $\psi : U \to M^ n$ be a morphism. In order to prove (1) and (2) it suffices to prove there exists a subcomplex $N^\bullet \subset M^\bullet$ which is bounded above, acyclic, and $|N^ m| \leq \kappa$, such that $\psi$ factors through $N^ n$. To do this set $N^ n = \mathop{\mathrm{Im}}(\psi )$, $N^{n + 1} = \mathop{\mathrm{Im}}(U \to M^ n \to M^{n + 1})$, and $N^ m = 0$ for $m \geq n + 2$. Suppose we have constructed $N^ m \subset M^ m$ for all $m \geq k$ such that

1. $\text{d}(N^ m) \subset N^{m + 1}$, $m \geq k$,

2. $\mathop{\mathrm{Im}}(N^{m - 1} \to N^ m) = \mathop{\mathrm{Ker}}(N^ m \to N^{m + 1})$ for all $m \geq k + 1$, and

3. $|N^ m| \leq c^{\max \{ n - m, 0\} }(|U|)$.

for some $k \leq n$. Because $M^\bullet$ is acyclic, we see that the subobject $\text{d}^{-1}(\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})) \subset M^{k - 1}$ surjects onto $\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})$. Thus we can choose $N^{k - 1} \subset M^{k - 1}$ surjecting onto $\mathop{\mathrm{Ker}}(N^ k \to N^{k + 1})$ with $|N^{k - 1}| \leq c^{n - k + 1}(|U|)$ by Lemma 19.12.1. The proof is finished by induction on $k$. $\square$

Comment #8590 by nkym on

Unnecessary period at the end of (3).

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